Section 5.2: Lecture 10/26Hide/Show in Tooltips

Subsection 5.2.1: Volume Part I

Ex 1: Find the volume of the solid generated when he region between $y = x^{2} - 1$ , $y = x + 1$ , $x = - 1$ , and $x = 2$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = - 1, b = 2 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} x (top - bottom) d x$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (0) & = {(0)}^{2} - 1 = - 1 \\ g (0) & = (0) + 1 = 1 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} g (x) is the top and f (x) is the bottom \end{matrix}} \begin{array}{r} b/c 1 (the output of g (x)) is higher than - 1 (the output of f (x)) \end{array}$
	$Step 8: Volume = 2 π \int_{a}^{b} x (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{- 1}^{2} x (x + 1 - (x^{2} - 1)) d x \\ = 2 π \int_{- 1}^{2} x (x + 1) - x (x^{2} - 1) d x \\ = 2 π \int_{- 1}^{2} \underset{⏟}{\begin{array}{c} x^{2} + x \end{array}} \begin{array}{r} distributed \end{array} - (\underset{⏟}{\begin{array}{c} x^{3} - x \end{array}} \begin{array}{r} distributed \end{array}) d x \\ = 2 π (\underset{⏟}{\begin{array}{c} \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int x^{2} d x & = (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} \frac{1}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int x d x & = \int x^{1} d x \\ = (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{4} x^{4} \end{array}} \begin{array}{r} \begin{aligned} \int x^{3} d x & = (\frac{1}{3 + 1} x^{3 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} \frac{1}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int x d x & = \int x^{1} d x \\ = (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array})_{- 1}^{2}) \\ = 2 π (\frac{1}{3} {(2)}^{3} + \frac{1}{2} {(2)}^{2} - (\frac{1}{4} {(2)}^{4} - \frac{1}{2} {(2)}^{2}) - (\frac{1}{3} {(- 1)}^{3} + \frac{1}{2} {(- 1)}^{2} - (\frac{1}{4} {(- 1)}^{4} - \frac{1}{2} {(- 1)}^{2}))) \end{aligned}$

Ex 2: Find the volume of the solid generated when he region between $y = x^{2}$ , $y = x$ , $x = 0$ , and $x = 1$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 0, b = 1 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (0.5) & = {(0.5)}^{2} = 0.3 \\ g (0.5) & = (0.5) = 0.5 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} g (x) is the outer and f (x) is the inner \end{matrix}} \begin{array}{r} b/c 0.5 (the output of g (x)) is farther from 0 than 0.3 (the output of f (x)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{0}^{1} {(x)}^{2} - {(x^{2})}^{2} d x \\ = π \int_{0}^{1} x^{2} - (x^{2}) (x^{2}) d x \\ π \int_{0}^{1} x^{2} - \underset{⏟}{\begin{array}{c} x^{4} \end{array}} \begin{matrix} \begin{aligned} x^{2} x^{2} & = x^{2 + 2} \end{aligned} \end{matrix} d x \\ = π (\underset{⏟}{\begin{array}{c} \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int x^{2} d x & = (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{5} x^{5} \end{array}} \begin{array}{r} \begin{aligned} \int x^{4} d x & = (\frac{1}{4 + 1} x^{4 + 1}) \end{aligned} \end{array})_{0}^{1}) \\ = π (\frac{1}{3} {(1)}^{3} - (\frac{1}{5} {(1)}^{5}) - (\frac{1}{3} {(0)}^{3} - (\frac{1}{5} {(0)}^{5}))) \end{aligned}$

Ex 3: Find the volume of the solid generated when he region between $y = 2 x + 1$ , $y = x^{2} + 1$ , $x = 0$ , and $x = 2$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 0, b = 2 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} x (top - bottom) d x$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (1) & = 2 (1) + 1 = 3 \\ g (1) & = {(1)}^{2} + 1 = 2 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the top and g (x) is the bottom \end{matrix}} \begin{array}{r} b/c 3 (the output of f (x)) is higher than 2 (the output of g (x)) \end{array}$
	$Step 8: Volume = 2 π \int_{a}^{b} x (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{0}^{2} x (2 x + 1 - (x^{2} + 1)) d x \\ = 2 π \int_{0}^{2} x (2 x + 1) - x (x^{2} + 1) d x \\ = 2 π \int_{0}^{2} \underset{⏟}{\begin{array}{c} 2 x^{2} + x \end{array}} \begin{array}{r} distributed \end{array} - (\underset{⏟}{\begin{array}{c} x^{3} + x \end{array}} \begin{array}{r} distributed \end{array}) d x \\ = 2 π (\underset{⏟}{\begin{array}{c} \frac{2}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int 2 x^{2} d x & = 2 \int x^{2} d x \\ = 2 (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} \frac{1}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int x d x & = \int x^{1} d x \\ = (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{4} x^{4} \end{array}} \begin{array}{r} \begin{aligned} \int x^{3} d x & = (\frac{1}{3 + 1} x^{3 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} \frac{1}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int x d x & = \int x^{1} d x \\ = (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array})_{0}^{2}) \\ = 2 π (\frac{2}{3} {(2)}^{3} + \frac{1}{2} {(2)}^{2} - (\frac{1}{4} {(2)}^{4} + \frac{1}{2} {(2)}^{2}) - (\frac{2}{3} {(0)}^{3} + \frac{1}{2} {(0)}^{2} - (\frac{1}{4} {(0)}^{4} + \frac{1}{2} {(0)}^{2}))) \end{aligned}$

Ex 4: Find the volume of the solid generated when he region between $y = 4 x^{2}$ , $y = x^{3}$ , $x = 0$ , and $x = 4$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 0, b = 4 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (1) & = 4 {(1)}^{2} = 4 \\ g (1) & = {(1)}^{3} = 1 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the outer and g (x) is the inner \end{matrix}} \begin{array}{r} b/c 4 (the output of f (x)) is farther from 0 than 1 (the output of g (x)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{0}^{4} {(4 x^{2})}^{2} - {(x^{3})}^{2} d x \\ = π \int_{0}^{2} 4 x^{2} (4 x^{2}) - x^{3} x^{3} d x \\ π \int_{0}^{2} 16 \underset{⏟}{\begin{array}{c} x^{4} \end{array}} \begin{matrix} \begin{aligned} x^{2} x^{2} & = x^{2 + 2} \end{aligned} \end{matrix} - \underset{⏟}{\begin{array}{c} x^{6} \end{array}} \begin{matrix} \begin{aligned} x^{3} x^{3} & = x^{3 + 3} \end{aligned} \end{matrix} d x \\ = π (\underset{⏟}{\begin{array}{c} \frac{16}{5} x^{5} \end{array}} \begin{array}{r} \begin{aligned} \int 16 x^{4} d x & = 16 \int x^{4} d x \\ = 16 (\frac{1}{4 + 1} x^{4 + 1}) \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{7} x^{7} \end{array}} \begin{array}{r} \begin{aligned} \int x^{6} d x & = (\frac{1}{6 + 1} x^{6 + 1}) \end{aligned} \end{array})_{0}^{4}) \\ = π (\frac{16}{5} {(4)}^{5} - (\frac{1}{7} {(4)}^{7}) - (\frac{16}{5} {(0)}^{5} - (\frac{1}{7} {(0)}^{7}))) \end{aligned}$

Ex 5: Find the volume of the solid generated when he region between $y = x^{2} - 1$ , $y = - x + 1$ , $x = - 2$ , and $x = 1$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = - 2, b = 1 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} x (top - bottom) d x$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (- 1) & = {(- 1)}^{2} - 1 = 0 \\ g (- 1) & = - (- 1) + 1 = 2 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} g (x) is the top and f (x) is the bottom \end{matrix}} \begin{array}{r} b/c 2 (the output of g (x)) is higher than 0 (the output of f (x)) \end{array}$
	$Step 8: Volume = 2 π \int_{a}^{b} x (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{- 2}^{1} x (- x + 1 - (x^{2} - 1)) d x \\ = 2 π \int_{- 2}^{1} x (- x + 1) - x (x^{2} - 1) d x \\ = 2 π \int_{- 2}^{1} \underset{⏟}{\begin{array}{c} - x^{2} + x \end{array}} \begin{array}{r} distributed \end{array} - (\underset{⏟}{\begin{array}{c} x^{3} - x \end{array}} \begin{array}{r} distributed \end{array}) d x \\ = 2 π (\underset{⏟}{\begin{array}{c} - \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int - 1 x^{2} d x & = - 1 \int x^{2} d x \\ = - 1 (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} \frac{1}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int x d x & = \int x^{1} d x \\ = (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{4} x^{4} \end{array}} \begin{array}{r} \begin{aligned} \int x^{3} d x & = (\frac{1}{3 + 1} x^{3 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} \frac{1}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int x d x & = \int x^{1} d x \\ = (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array})_{- 2}^{1}) \\ = 2 π (- \frac{1}{3} {(1)}^{3} + \frac{1}{2} {(1)}^{2} - (\frac{1}{4} {(1)}^{4} - \frac{1}{2} {(1)}^{2}) - (- \frac{1}{3} {(- 2)}^{3} + \frac{1}{2} {(- 2)}^{2} - (\frac{1}{4} {(- 2)}^{4} - \frac{1}{2} {(- 2)}^{2}))) \end{aligned}$

Ex 6: Find the volume of the solid generated when he region between $y = x^{2}$ , $y = 2 x$ , $x = 0$ , and $x = 2$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 0, b = 2 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (1) & = {(1)}^{2} = 1 \\ g (1) & = 2 (1) = 1 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} g (x) is the outer and f (x) is the inner \end{matrix}} \begin{array}{r} b/c 1 (the output of g (x)) is farther from 0 than 1 (the output of f (x)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{0}^{2} {(2 x)}^{2} - {(x^{2})}^{2} d x \\ = π \int_{0}^{2} 2 x (2 x) - x^{2} x^{2} d x \\ π \int_{0}^{2} 4 \underset{⏟}{\begin{array}{c} x^{2} \end{array}} \begin{matrix} \begin{aligned} x x & = x^{1} x^{1} \\ x^{1} x^{1} & = x^{1 + 1} \end{aligned} \end{matrix} - \underset{⏟}{\begin{array}{c} x^{4} \end{array}} \begin{matrix} \begin{aligned} x^{2} x^{2} & = x^{2 + 2} \end{aligned} \end{matrix} d x \\ = π (\underset{⏟}{\begin{array}{c} \frac{4}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int 4 x^{2} d x & = 4 \int x^{2} d x \\ = 4 (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{5} x^{5} \end{array}} \begin{array}{r} \begin{aligned} \int x^{4} d x & = (\frac{1}{4 + 1} x^{4 + 1}) \end{aligned} \end{array})_{0}^{2}) \\ = π (\frac{4}{3} {(2)}^{3} - (\frac{1}{5} {(2)}^{5}) - (\frac{4}{3} {(0)}^{3} - (\frac{1}{5} {(0)}^{5}))) \end{aligned}$

Ex 7: Find the volume of the solid generated when he region between $y = x^{2}$ , $y = 0$ , $x = 1$ , and $x = 5$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 1, b = 5 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (2) & = {(2)}^{2} = 4 \\ g (2) & = 0 = 0 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the outer and g (x) is the inner \end{matrix}} \begin{array}{r} b/c 4 (the output of f (x)) is farther from 0 than 0 (the output of g (x)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{1}^{5} {(x^{2})}^{2} - {(0)}^{2} d x \\ = π \int_{1}^{5} \underset{⏟}{\begin{array}{c} x^{4} \end{array}} \begin{array}{r} \begin{aligned} {(x^{2})}^{2} & = x^{2 (2)} \end{aligned} \end{array} d x \\ = π (\underset{⏟}{\begin{array}{c} \frac{1}{5} x^{5} \end{array}} \begin{array}{r} \begin{aligned} \int x^{4} d x & = (\frac{1}{4 + 1} x^{4 + 1}) \end{aligned} \end{array}_{1}^{5}) \\ = π (\frac{1}{5} {(5)}^{5} - (\frac{1}{5} {(1)}^{5})) \end{aligned}$

Ex 8: Find the volume of the solid generated when he region between $y = e^{2 x}$ , $y = 0$ , $x = - \ln (3)$ , and $x = \ln (3)$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = - \ln (3), b = \ln (3) \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (0) & = e^{2 (0)} = 1 \\ g (0) & = 0 = 0 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the outer and g (x) is the inner \end{matrix}} \begin{array}{r} b/c 1 (the output of f (x)) is farther from 0 than 0 (the output of g (x)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{- \ln (3)}^{\ln (3)} {(e^{2 x})}^{2} - {(0)}^{2} d x \\ = π \int_{- \ln (3)}^{\ln (3)} \underset{⏟}{\begin{array}{c} e^{4 x} \end{array}} \begin{array}{r} \begin{aligned} {(e^{2 x})}^{2} & = e^{2 x (2)} \end{aligned} \end{array} d x \\ = π (\underset{⏟}{\begin{array}{c} \frac{e^{4 x}}{4} \end{array}} \begin{array}{r} \begin{matrix} \int e^{4 x} d x = \frac{e^{4 x}}{4} \end{matrix} \end{array}_{- \ln (3)}^{\ln (3)}) \\ = π (\frac{e^{4 (\ln (3))}}{4} - (\frac{e^{4 (- \ln (3))}}{4})) \end{aligned}$

Ex 9: Find the volume of the solid generated when he region between $y = x^{2 / 3}$ and $y = 1$ and $x = 0$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} \underset{⏟}{\begin{array}{c} x^{2 / 3} \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} 1 \end{array}} \begin{array}{r} g (x) \end{array} \\ x & = 1 \\ a & = 0, b = 1 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (0.5) & = {(0.5)}^{2 / 3} = 0.6 \\ g (0.5) & = 1 = 1 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} g (x) is the outer and f (x) is the inner \end{matrix}} \begin{array}{r} b/c 1 (the output of g (x)) is farther from 0 than 0.6 (the output of f (x)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{0}^{1} {(1)}^{2} - {(x^{2 / 3})}^{2} d x \\ = π \int_{0}^{1} 1 - \underset{⏟}{\begin{array}{c} x^{\frac{4}{3}} \end{array}} \begin{array}{r} \begin{aligned} {(x^{2 / 3})}^{2} & = x^{2 / 3 (2)} \end{aligned} \end{array} d x \\ = π (\underset{⏟}{\begin{array}{c} 1 x \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c x + C \\ \int 1 d x & = 1 x + C \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{3}{7} x^{\frac{7}{3}} \end{array}} \begin{array}{r} \begin{aligned} \int x^{\frac{4}{3}} d x & = (\frac{1}{\frac{4}{3} + 1} x^{\frac{4}{3} + 1}) \end{aligned} \end{array})_{0}^{1}) \\ = π (1 (1) - (\frac{3}{7} {(1)}^{\frac{7}{3}}) - (1 (0) - (\frac{3}{7} {(0)}^{\frac{7}{3}}))) \end{aligned}$

Ex 10: Find the volume of the solid generated when he region between $y = \frac{1}{x}$ , $x = 0$ , $y = 1$ , and $y = 3$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{aligned} y & = \frac{1}{x} \\ x & = \frac{1}{y} \\ f (y) & = \frac{1}{y} \\ x & = 0 \\ g (y) & = 0 \end{aligned}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 1, b = 3 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (2) & = \frac{1}{(2)} = 0.3 \\ g (2) & = 0 = 0 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (y) is the outer and g (y) is the inner \end{matrix}} \begin{array}{r} b/c 0.3 (the output of f (y)) is farther from 0 than 0 (the output of g (y)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{1}^{3} {(\frac{1}{y})}^{2} - {(0)}^{2} d y \\ = π \int_{1}^{3} \frac{1^{2}}{y^{2}} d y \\ = π \int_{1}^{3} \frac{1}{y^{2}} d y \\ = π \int_{1}^{3} \underset{⏟}{\begin{array}{c} y^{-} 2 \end{array}} \begin{array}{r} \begin{aligned} \frac{1}{y^{2}} & = 1 y^{- 2} \end{aligned} \end{array} d y \\ = π (\underset{⏟}{\begin{array}{c} - 1 y^{- 1} \end{array}} \begin{array}{r} \begin{aligned} \int y^{- 2} d x & = (\frac{1}{- 2 + 1} y^{- 2 + 1}) \end{aligned} \end{array}_{1}^{3}) \\ = π (- 1 {(3)}^{- 1} - (- 1 {(1)}^{- 1})) \end{aligned}$

Ex 11: Find the volume of the solid generated when he region between $y = - | x - 4 | + 4$ and $y = 0$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{aligned} y & = - \| x - 4 \| + 4 \\ y & = {\begin{cases} - (4 - x) + 4 & if x \leq 4 \\ - (x - 4) + 4 if x \geq 4 \end{cases} \\ = {\begin{cases} - 4 + x + 4 & if x \leq 4 \\ - x + 4 + 4 & if x \geq 4 \end{cases} \\ = {\begin{cases} x & if x \leq 4 \\ - x + 8 & if x \geq 4 \end{cases} \\ f (x) & = - \| x - 4 \| + 4 \\ y & = 0 \\ g (x) & = 0 \end{aligned}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} \underset{⏟}{\begin{array}{c} - \| x - 4 \| + 4 \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} 0 \end{array}} \begin{array}{r} g (x) \end{array} \\ if x \leq 4, \underset{⏟}{\begin{array}{c} x \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} 0 \end{array}} \begin{array}{r} g (x) \end{array} \\ if x \geq 4, \underset{⏟}{\begin{array}{c} - x + 8 \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} 0 \end{array}} \begin{array}{r} g (x) \end{array} \\ - x & = - 8 \\ x - 8 & = 0 \\ a & = 0, b = 8 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (1) & = - \| (1) - 4 \| + 4 = 1 \\ g (1) & = 0 = 0 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the outer and g (x) is the inner \end{matrix}} \begin{array}{r} b/c 1 (the output of f (x)) is farther from 0 than 0 (the output of g (x)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{\begin{array}{r} 0 \end{array}}^{4} x^{2} - 0^{2} d y + π \int_{4}^{8} (- x + 8)^{2} - 0^{2} d y \\ = π \int_{\begin{array}{r} 0 \end{array}}^{4} x^{2} d y + π \int_{4}^{8} \underset{⏟}{\begin{array}{c} x^{2} - 16 x + 64 \end{array}} \begin{array}{r} distributed \end{array} d y \\ = π (\underset{⏟}{\begin{array}{c} \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int x^{2} d x & = (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array}_{0}^{4}) \\ + π (\underset{⏟}{\begin{array}{c} \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int x^{2} d x & = (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} 8 x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int 16 x d x & = 16 \int x^{1} d x \\ = 16 (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 64 x \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c x + C \\ \int 64 d x & = 64 x + C \end{aligned} \end{array}_{4}^{8}) \\ = π (\frac{1}{3} {(4)}^{3}) - π (\frac{1}{3} {(0)}^{3}) \\ + π (\frac{1}{3} {(8)}^{3} - 8 {(8)}^{2} + 64 (8)) \\ - π (\frac{1}{3} {(4)}^{3} - 8 {(4)}^{2} + 64 (4)) \end{aligned}$

Ex 12: Find the volume of the solid generated when he region between $y = \frac{5}{x}$ and $y = - x + 6$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} \underset{⏟}{\begin{array}{c} \frac{5}{x} \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} - x + 6 \end{array}} \begin{array}{r} g (x) \end{array} \\ - x (x) + 6 (x) & = \frac{5}{x} x \\ - x^{2} + 6 x & = \underset{⏟}{\begin{array}{c} 5 \end{array}} \begin{array}{r} \begin{matrix} \frac{x}{x} = 1 \end{matrix} \end{array} \\ x^{2} - 6 x + 5 & = 0 \\ (x - 1) (x - 5) & = 0 \\ x - 1 & = 0, x - 5 = 0 \\ x & = 1, 5 \\ a & = 1, b = 5 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} x (top - bottom) d x$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (2) & = \frac{5}{(2)} = 2.5 \\ g (2) & = - (2) + 6 = 4 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} g (x) is the top and f (x) is the bottom \end{matrix}} \begin{array}{r} b/c 4 (the output of g (x)) is higher than 2.5 (the output of f (x)) \end{array}$
	$Step 8: Volume = 2 π \int_{a}^{b} x (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{1}^{5} x (- x + 6 - (\frac{5}{x})) d x \\ = 2 π \int_{1}^{5} x (- x + 6) - \frac{5}{x} x d x \\ = 2 π \int_{1}^{5} \underset{⏟}{\begin{array}{c} x^{2} - 6 x \end{array}} \begin{array}{r} distributed \end{array} - \underset{⏟}{\begin{array}{c} 5 \end{array}} \begin{array}{r} \begin{matrix} \frac{x}{x} = 1 \end{matrix} \end{array} d x \\ = 2 π (\underset{⏟}{\begin{array}{c} - \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int - 1 x^{2} d x & = - 1 \int x^{2} d x \\ = - 1 (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 3 x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int 6 x d x & = 6 \int x^{1} d x \\ = 6 (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} 5 x \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c x + C \\ \int 5 d x & = 5 x + C \end{aligned} \end{array})_{1}^{5}) \\ = 2 π (- \frac{1}{3} {(5)}^{3} + 3 {(5)}^{2} - (5 (5)) - (- \frac{1}{3} {(1)}^{3} + 3 {(1)}^{2} - (5 (1)))) \end{aligned}$

Ex 13: Find the volume of the solid generated when he region between $y = e^{x}$ , $y = x + 1$ , $x = 0$ , and $x = 4$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 0, b = 4 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} x (top - bottom) d x$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (1) & = e^{(1)} = 2.7 \\ g (1) & = (1) + 1 = 2 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the top and g (x) is the bottom \end{matrix}} \begin{array}{r} b/c 2.7 (the output of f (x)) is higher than 2 (the output of g (x)) \end{array}$
	$Step 8: Volume = 2 π \int_{a}^{b} x (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{0}^{4} x (e^{x} - (x + 1)) d x \\ = 2 π \int_{1}^{5} x e^{x} - x (x^{2} + x) d x \\ = 2 π \int_{1}^{5} x e^{x} - (\underset{⏟}{\begin{array}{c} x + 1 \end{array}} \begin{array}{r} distributed \end{array}) d x \\ = 2 π (\underset{⏟}{\begin{array}{c} x e^{x} - e^{x} \end{array}} \begin{array}{r} integration by parts \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int x^{2} d x & = (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} \frac{1}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int x d x & = \int x^{1} d x \\ = (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array})_{0}^{4}) \\ = 2 π ((4) e^{(4)} - e^{(4)} - (\frac{1}{3} {(4)}^{3} + \frac{1}{2} {(4)}^{2}) - ((0) e^{(0)} - e^{(0)} - (\frac{1}{3} {(0)}^{3} + \frac{1}{2} {(0)}^{2}))) \end{aligned}$

Ex 14: Find the volume of the solid generated when he region between $y = \frac{1}{x}$ , $x = 0$ , $y = 1$ , and $y = 3$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{aligned} y & = \frac{1}{x} \\ x & = \frac{1}{y} \\ f (y) & = \frac{1}{y} \\ x & = 0 \\ g (y) & = 0 \end{aligned}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 1, b = 3 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} y (top - bottom) d y$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (2) & = \frac{1}{(2)} = 0.5 \\ g (2) & = 0 = 0 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (y) is the top and g (y) is the bottom \end{matrix}} \begin{array}{r} b/c 0.5 (the output of f (y)) is higher than 0 (the output of g (y)) \end{array}$
	$Step 8: Volume = 2 π \int_{a}^{b} x (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{1}^{3} y (\frac{1}{y} - (0)) d y \\ = 2 π \int_{1}^{3} \underset{⏟}{\begin{array}{c} 1 \end{array}} \begin{array}{r} \begin{matrix} \frac{y}{y} = 1 \end{matrix} \end{array} d x \\ = 2 π (\underset{⏟}{\begin{array}{c} 1 y \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c y + C \\ \int 1 d x & = 1 y + C \end{aligned} \end{array}_{1}^{3}) \\ = 2 π (1 (3) - (1 (1))) \end{aligned}$

Ex 15: Find the volume of the solid generated when he region between $x = 4 y^{3} - 2 y^{4}$ and $x = 0$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} \underset{⏟}{\begin{array}{c} 4 y^{3} - 2 y^{4} \end{array}} \begin{matrix} f (y) \end{matrix} & = \underset{⏟}{\begin{array}{c} 0 \end{array}} \begin{array}{r} g (y) \end{array} \\ 2 y^{3} (y - 2) & = 0 \\ 2 y^{3} & = 0, y - 2 = 0 \\ y^{3} & = 0, y = 2 \\ y & = 0, 2 \\ a & = 0, b = 2 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} y (top - bottom) d y$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (1) & = 4 {(1)}^{3} - 2 {(1)}^{4} = 2 \\ g (1) & = 0 = 0 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (y) is the top and g (y) is the bottom \end{matrix}} \begin{array}{r} b/c 2 (the output of f (y)) is higher than 0 (the output of g (y)) \end{array}$
	$Step 8: Volume = 2 π \int_{a}^{b} x (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{0}^{2} y (4 y^{3} - 2 y^{4} - (0)) d y \\ = 2 π \int_{1}^{3} y (4 y^{3}) - y (2 y^{4}) d y \\ = 2 π \int_{1}^{3} 4 \underset{⏟}{\begin{array}{c} y^{4} \end{array}} \begin{array}{r} \begin{aligned} y y^{3} & = y^{1} y^{3} \\ y^{1} y^{3} & = y^{1 + 3} \end{aligned} \end{array} - 2 \underset{⏟}{\begin{array}{c} y^{5} \end{array}} \begin{array}{r} \begin{aligned} y y^{4} & = y^{1} y^{4} \\ y^{1} y^{4} & = y^{1 + 4} \end{aligned} \end{array} d y \\ = 2 π (\underset{⏟}{\begin{array}{c} \frac{4}{5} y^{5} \end{array}} \begin{array}{r} \begin{aligned} \int 4 y^{4} d x & = 4 \int y^{4} d x \\ = 4 (\frac{1}{4 + 1} y^{4 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} \frac{1}{3} y^{6} \end{array}} \begin{array}{r} \begin{aligned} \int 2 y^{5} d x & = 2 \int y^{5} d x \\ = 2 (\frac{1}{5 + 1} y^{5 + 1}) \end{aligned} \end{array}_{0}^{2}) \\ = 2 π (\frac{4}{5} {(2)}^{5} - \frac{1}{3} {(2)}^{6} - (\frac{4}{5} {(0)}^{5} - \frac{1}{3} {(0)}^{6})) \end{aligned}$

Ex 16: Find the volume of the solid generated when he region between $x = y^{2}$ and $2 y - x + 3 = 0$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{aligned} x & = y^{2} \\ f (y) & = y^{2} \\ 2 y - x + 3 & = 0 \\ x & = 2 y + 3 \\ g (y) & = 2 y + 3 \end{aligned}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} \underset{⏟}{\begin{array}{c} y^{2} \end{array}} \begin{matrix} f (y) \end{matrix} & = \underset{⏟}{\begin{array}{c} 2 y + 3 \end{array}} \begin{array}{r} g (y) \end{array} \\ y^{2} - 2 y - 3 & = 0 \\ (y + 1) (y - 3) & = 0 \\ y + 1 & = 0, y - 3 = 0 \\ y & = - 1, 3 \\ a & = - 1, b = 3 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (0) & = {(0)}^{2} = 0 \\ g (0) & = 2 (0) + 3 = 3 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from 0 \\ ∙ The inner function is the function with the output closer to 0 \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} g (y) is the outer and f (y) is the inner \end{matrix}} \begin{array}{r} b/c 3 (the output of g (y)) is farther from 0 than 0 (the output of f (y)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x$	$\begin{aligned} Volume & = π \int_{- 1}^{3} {(2 y + 3)}^{2} - {(y^{2})}^{2} d y \\ = π \int_{- 1}^{3} \underset{⏟}{\begin{array}{c} 4 y^{2} + 12 y + 9 \end{array}} \begin{array}{r} distributed \end{array} - \underset{⏟}{\begin{array}{c} y^{4} \end{array}} \begin{array}{r} \begin{aligned} {(y^{2})}^{2} & = y^{2 (2)} \end{aligned} \end{array} d y \\ = π (\underset{⏟}{\begin{array}{c} \frac{4}{3} y^{3} \end{array}} \begin{array}{r} \begin{aligned} \int 4 y^{2} d x & = 4 \int y^{2} d x \\ = 4 (\frac{1}{2 + 1} y^{2 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 6 y^{2} \end{array}} \begin{array}{r} \begin{aligned} \int 12 y d x & = 12 \int y^{1} d x \\ = 12 (\frac{1}{1 + 1} y^{1 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 9 y \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c y + C \\ \int 9 d x & = 9 y + C \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{5} y^{5} \end{array}} \begin{array}{r} \begin{aligned} \int y^{4} d x & = (\frac{1}{4 + 1} y^{4 + 1}) \end{aligned} \end{array})_{- 1}^{3}) \\ = π (\frac{4}{3} {(3)}^{3} + 6 {(3)}^{2} + 9 (3) - (\frac{1}{5} {(3)}^{5}) - (\frac{4}{3} {(- 1)}^{3} + 6 {(- 1)}^{2} + 9 (- 1) - (\frac{1}{5} {(- 1)}^{5}))) \end{aligned}$

Ex 17: Find the volume of the solid generated when he region between $y = - | x - 4 | + 4$ and $y = 0$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{aligned} y & = - \| x - 4 \| + 4 \\ y & = {\begin{cases} - (4 - x) + 4 & if x \leq 4 \\ - (x - 4) + 4 if x \geq 4 \end{cases} \\ = {\begin{cases} - 4 + x + 4 & if x \leq 4 \\ - x + 4 + 4 & if x \geq 4 \end{cases} \\ = {\begin{cases} x & if x \leq 4 \\ - x + 8 & if x \geq 4 \end{cases} \\ f (x) & = - \| x - 4 \| + 4 \\ y & = 0 \\ g (x) & = 0 \end{aligned}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} \underset{⏟}{\begin{array}{c} - \| x - 4 \| + 4 \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} 0 \end{array}} \begin{array}{r} g (x) \end{array} \\ if x \leq 4, \underset{⏟}{\begin{array}{c} x \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} 0 \end{array}} \begin{array}{r} g (x) \end{array} \\ if x \geq 4, \underset{⏟}{\begin{array}{c} - x + 8 \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} 0 \end{array}} \begin{array}{r} g (x) \end{array} \\ - x & = - 8 \\ x - 8 & = 0 \\ a & = 0, b = 8 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about x axis & Rotating about y axis \\ f (x) and g (x) & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d x & 2 π \int_{a}^{b} x (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} y (top - bottom) d y & π \int_{a}^{b} {(outer)}^{2} - {(inner)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} x (top - bottom) d x$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (1) & = - \| (1) - 4 \| + 4 = 1 \\ g (1) & = 0 = 0 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the top and g (x) is the bottom \end{matrix}} \begin{array}{r} b/c 1 (the output of f (x)) is higher than 0 (the output of g (x)) \end{array}$
	$Step 8: Volume = 2 π \int_{a}^{b} x (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{\begin{array}{r} 0 \end{array}}^{4} x (x - 0) d x + 2 π \int_{4}^{8} x (- x + 8 - 0) d x \\ = π \int_{\begin{array}{r} 0 \end{array}}^{4} x^{2} d x + π \int_{4}^{8} \underset{⏟}{\begin{array}{c} - x^{2} + 8 x \end{array}} \begin{array}{r} distributed \end{array} d x \\ = π (\underset{⏟}{\begin{array}{c} \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int x^{2} d x & = (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array}_{0}^{4}) \\ + π (\underset{⏟}{\begin{array}{c} - \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int - 1 x^{2} d x & = - 1 \int x^{2} d x \\ = - 1 (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 4 x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int 8 x d x & = 8 \int x^{1} d x \\ = 8 (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array}_{4}^{8}) \\ = π (\frac{1}{3} {(4)}^{3}) - π (- \frac{1}{3} {(0)}^{3} + 4 {(0)}^{2}) \\ + π (- \frac{1}{3} {(8)}^{3} + 4 {(8)}^{2}) \\ - π (- \frac{1}{3} {(4)}^{3} + 4 {(4)}^{2}) \end{aligned}$

Subsection 5.2.2: Volume Part II

Ex 1: Find the volume of the solid generated when he region between $y = - x^{2} + 9$ and $x + y = 9$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{aligned} y & = - x^{2} + 9 \\ f (x) & = - x^{2} + 9 \\ x + y & = 9 \\ y & = 9 - x \\ g (x) & = 9 - x \end{aligned}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} \underset{⏟}{\begin{array}{c} - x^{2} + 9 \end{array}} \begin{matrix} f (x) \end{matrix} & = \underset{⏟}{\begin{array}{c} 9 - x \end{array}} \begin{array}{r} g (x) \end{array} \\ x^{2} + x & = 0 \\ x (x + 1) & = 0 \\ x & = 0, x + 1 = 0 \\ a & = 0, b = 1 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about y = c & Rotating about x = c \\ f (x) and g (x) & π \int_{a}^{b} {(outer - c)}^{2} - {(inner - c)}^{2} d x & 2 π \int_{a}^{b} radius (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} radius (top - bottom) d y & π \int_{a}^{b} {(outer - c)}^{2} - {(inner - c)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} radius (top - bottom) d x$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (0.5) & = - {(0.5)}^{2} + 9 = 8.8 \\ g (0.5) & = 9 - (0.5) = 8.5 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the top and g (x) is the bottom \end{matrix}} \begin{array}{r} b/c 8.8 (the output of f (x)) is higher than 8.5 (the output of g (x)) \end{array}$
	$Step 8: \begin{aligned} Determine the radius \\ \begin{aligned} ∙ If c > b, radius = c - x \\ ∙ If c < a, radius = x - c \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} radius = 8 - x \end{matrix}} \begin{array}{r} b/c 8 > 1 \end{array}$
	$Step 9: Volume = 2 π \int_{a}^{b} radius (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{0}^{1} (8 - x) (- x^{2} + 9 - (9 - x)) d x \\ = 2 π \int_{0}^{1} (8 - x) (- x^{2} + 9) - (8 - x) (9 - x) d x \\ = 2 π \int_{0}^{1} \underset{⏟}{\begin{array}{c} x^{3} - 8 x^{2} - 9 x + 72 \end{array}} \begin{array}{r} distributed \end{array} - (\underset{⏟}{\begin{array}{c} x^{2} - 17 x + 72 \end{array}} \begin{array}{r} distributed \end{array}) d x \\ = 2 π (\underset{⏟}{\begin{array}{c} \frac{1}{4} x^{4} \end{array}} \begin{array}{r} \begin{aligned} \int x^{3} d x & = (\frac{1}{3 + 1} x^{3 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} \frac{8}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int 8 x^{2} d x & = 8 \int x^{2} d x \\ = 8 (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} \frac{9}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int 9 x d x & = 9 \int x^{1} d x \\ = 9 (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 72 x \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c x + C \\ \int 72 d x & = 72 x + C \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int x^{2} d x & = (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} \frac{17}{2} x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int 17 x d x & = 17 \int x^{1} d x \\ = 17 (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 72 x \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c x + C \\ \int 72 d x & = 72 x + C \end{aligned} \end{array})_{0}^{1}) \\ = 2 π (\frac{1}{4} {(1)}^{4} - \frac{8}{3} {(1)}^{3} - \frac{9}{2} {(1)}^{2} + 72 (1) - (\frac{1}{3} {(1)}^{3} - \frac{17}{2} {(1)}^{2} + 72 (1))) \\ - 2 π (\frac{1}{4} {(0)}^{4} - \frac{8}{3} {(0)}^{3} - \frac{9}{2} {(0)}^{2} + 72 (0) - (\frac{1}{3} {(0)}^{3} - \frac{17}{2} {(0)}^{2} + 72 (0))) \end{aligned}$

Ex 2: Note: The WebAssign problem is different, so you need to double the answer Find the volume of the solid generated when he region between $y = x^{2 / 3}$ , $y = x^{2}$ , $x = 0$ , and $x = 1$ is rotated around the $x$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 0, b = 1 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about y = c & Rotating about x = c \\ f (x) and g (x) & π \int_{a}^{b} {(outer - c)}^{2} - {(inner - c)}^{2} d x & 2 π \int_{a}^{b} radius (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} radius (top - bottom) d y & π \int_{a}^{b} {(outer - c)}^{2} - {(inner - c)}^{2} d y \end{array} \end{aligned}$	$π \int_{a}^{b} {(outer - c)}^{2} - {(inner - c)}^{2} d x$
	$Step 5: Part 2: Determine the outer and inner functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (0.5) & = {(0.5)}^{2 / 3} = 0.6 \\ g (0.5) & = {(0.5)}^{2} = 0.3 \end{aligned}$
	$Step 7: \begin{aligned} Determine the outer and inner functions \\ \begin{aligned} ∙ The outer function is the function with the output farther from c \\ ∙ The inner function is the function with the output closer to c \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the outer and g (x) is the inner \end{matrix}} \begin{array}{r} b/c 0.6 (the output of f (x)) is farther from - 4 than 0.3 (the output of g (x)) \end{array}$
	$Step 8: Volume = π \int_{a}^{b} {(outer - c)}^{2} - {(inner - c)}^{2} d x$	$\begin{aligned} Volume & = π \int_{0}^{1} {(x^{2 / 3} - (- 4))}^{2} - {(x^{2} - (- 4))}^{2} d x \\ = π \int_{0}^{1} \underset{⏟}{\begin{array}{c} x^{4 / 3} - 8 x^{2 / 3} + 16 \end{array}} \begin{array}{r} distributed \end{array} - (\underset{⏟}{\begin{array}{c} x^{4} - 8 x^{2} + 16 \end{array}} \begin{array}{r} distributed \end{array}) d x \\ = π (\underset{⏟}{\begin{array}{c} \frac{3}{7} x^{\frac{7}{3}} \end{array}} \begin{array}{r} \begin{aligned} \int x^{\frac{4}{3}} d x & = (\frac{1}{\frac{4}{3} + 1} x^{\frac{4}{3} + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} \frac{24}{5} x^{\frac{5}{3}} \end{array}} \begin{array}{r} \begin{aligned} \int 8 x^{\frac{2}{3}} d x & = 8 \int x^{\frac{2}{3}} d x \\ = 8 (\frac{1}{\frac{2}{3} + 1} x^{\frac{2}{3} + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 16 x \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c x + C \\ \int 16 d x & = 16 x + C \end{aligned} \end{array} - (\underset{⏟}{\begin{array}{c} \frac{1}{5} x^{5} \end{array}} \begin{array}{r} \begin{aligned} \int x^{4} d x & = (\frac{1}{4 + 1} x^{4 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} \frac{8}{3} x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int 8 x^{2} d x & = 8 \int x^{2} d x \\ = 8 (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 16 x \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c x + C \\ \int 16 d x & = 16 x + C \end{aligned} \end{array})_{0}^{1}) \\ = π (\frac{3}{7} {(1)}^{\frac{7}{3}} - \frac{24}{5} {(1)}^{\frac{5}{3}} + 16 (1) - (\frac{1}{5} {(1)}^{5} - \frac{8}{3} {(1)}^{3} + 16 (1))) \\ - 2 π (\frac{3}{7} {(0)}^{\frac{7}{3}} - \frac{24}{5} {(0)}^{\frac{5}{3}} + 16 (0) - (\frac{1}{5} {(0)}^{5} - \frac{8}{3} {(0)}^{3} + 16 (0))) \end{aligned}$

Ex 3: Find the volume of the solid generated when he region between $y = x^{2} + 2$ , $y = 0$ , $x = 0$ , and $x = 2$ is rotated around the $y$ axis


	$Step 1: Part 1: Determine f (x), g (x), a, and b$
	$Step 2: \begin{aligned} Determine f (x) and g (x) or f (y) and g (y) if there are not given \\ \begin{aligned} ∙ To find f (x) and g (x) solve for y in the two given equations \\ ∙ To find f (y) and g (y) solve for x in the two given equations \end{aligned} \end{aligned}$	$\begin{array}{r} NA \end{array}$
	$Step 3: \begin{matrix} Find a and b if they are not given by solving f (x) = g (x) \\ a is the smaller x value and b is the larger x value \end{matrix}$	$\begin{aligned} a & = 0, b = 2 \end{aligned}$
	$Step 4: \begin{aligned} Determine which formula to use \\ \begin{array}{ccc} Rotating about y = c & Rotating about x = c \\ f (x) and g (x) & π \int_{a}^{b} {(outer - c)}^{2} - {(inner - c)}^{2} d x & 2 π \int_{a}^{b} radius (top - bottom) d x \\ f (y) and g (y) & 2 π \int_{a}^{b} radius (top - bottom) d y & π \int_{a}^{b} {(outer - c)}^{2} - {(inner - c)}^{2} d y \end{array} \end{aligned}$	$2 π \int_{a}^{b} radius (top - bottom) d x$
	$Step 5: Part 2: Determine the top and bottom functions$
	$Step 6: Plug in an \underset{⏟}{\begin{matrix} x value between a and b \end{matrix}} \begin{array}{r} called a test point \end{array} into f (x) and g (x)$	$\begin{aligned} f (1) & = {(1)}^{2} + 2 = 3 \\ g (1) & = 0 = 0 \end{aligned}$
	$Step 7: \begin{aligned} Determine the top and bottom functions \\ \begin{aligned} ∙ The top function is the function with the higher output \\ ∙ The bottom function is the function with the lower output \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} f (x) is the top and g (x) is the bottom \end{matrix}} \begin{array}{r} b/c 3 (the output of f (x)) is higher than 0 (the output of g (x)) \end{array}$
	$Step 8: \begin{aligned} Determine the radius \\ \begin{aligned} ∙ If c > b, radius = c - x \\ ∙ If c < a, radius = x - c \end{aligned} \end{aligned}$	$\underset{⏟}{\begin{matrix} radius = 3 - x \end{matrix}} \begin{array}{r} b/c 3 > 2 \end{array}$
	$Step 9: Volume = 2 π \int_{a}^{b} radius (top - bottom) d x$	$\begin{aligned} Volume & = 2 π \int_{0}^{2} (3 - x) (x^{2} + 2 - (0)) d x \\ = 2 π \int_{0}^{2} (3 - x) (x^{2} + 2) - (3 - x) (0) d x \\ = 2 π \int_{0}^{2} \underset{⏟}{\begin{array}{c} - x^{3} + 3 x^{2} - 2 x + 6 \end{array}} \begin{array}{r} distributed \end{array} d x \\ = 2 π (\underset{⏟}{\begin{array}{c} - \frac{1}{4} x^{4} \end{array}} \begin{array}{r} \begin{aligned} \int - 1 x^{3} d x & = - 1 \int x^{3} d x \\ = - 1 (\frac{1}{3 + 1} x^{3 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 1 x^{3} \end{array}} \begin{array}{r} \begin{aligned} \int 3 x^{2} d x & = 3 \int x^{2} d x \\ = 3 (\frac{1}{2 + 1} x^{2 + 1}) \end{aligned} \end{array} - \underset{⏟}{\begin{array}{c} 1 x^{2} \end{array}} \begin{array}{r} \begin{aligned} \int 2 x d x & = 2 \int x^{1} d x \\ = 2 (\frac{1}{1 + 1} x^{1 + 1}) \end{aligned} \end{array} + \underset{⏟}{\begin{array}{c} 6 x \end{array}} \begin{array}{r} \begin{aligned} \int c d x & = c x + C \\ \int 6 d x & = 6 x + C \end{aligned} \end{array}_{0}^{2}) \\ = 2 π (- \frac{1}{4} {(2)}^{4} + 1 {(2)}^{3} - 1 {(2)}^{2} + 6 (2)) \\ - 2 π (- \frac{1}{4} {(0)}^{4} + 1 {(0)}^{3} - 1 {(0)}^{2} + 6 (0)) \end{aligned}$