Exercises: Disks and Washers

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Exercises for the disk and washer methods.

The region $0 \leq y \leq \sqrt {x}$ with $x \leq 1$ , shown below, is revolved around the $x$ -axis. Use the disk method to find the volume of the solid of revolution.

The radius $R(x)$ will be a difference of $y$ -values because slices are indexed by the variable $x$ . Each slice will extend from $y=0$ to $y = \sqrt {x}$ , and so $R(x)$ must be the larger of these $y$ -values minus the smaller of these $y$ -values.

$R(x) = \answer {\sqrt {x}}$ $V = \int _{\answer {0}}^{\answer {1}} \pi (R(x))^2 dx = \answer {\frac {\pi }{2}}$

The region $0 \leq y \leq \sqrt {x}$ with $x \leq 1$ , shown below, is revolved around the axis $x=1$ . Use the disk method to find the volume of the solid of revolution.

The radius $R(y)$ will be a difference of $x$ -values because slices are indexed by the variable $y$ . Each slice will extend from $x = y^2$ to $x = 1$ , and so $R(y)$ must be the larger of these $x$ -values minus the smaller of these $x$ -values

$R(y) = \answer {1 - y^2}$ $V = \int _{\answer {0}}^{\answer {1}} \pi (R(y))^2 dy = \answer {\frac {8\pi }{15}}$

The region $0 \leq y \leq \sqrt {x}$ with $x \leq 1$ , shown below, is revolved around the axis $x=0$ . Use the washer method to find the volume of the solid of revolution.

Each radius will be a difference of $x$ -values because slices are indexed by the variable $y$ . The distance from the axis $x=0$ to the line $x=1$ is $1$ , and the distance from the axis $x=0$ to $x = y^2$ is $y^2$ .

$R_{\mathrm {outer}} (y) = \answer {1} \text { and } r_{\mathrm {inner}}(y) = \answer {y^2}$ $V = \int _{\answer {0}}^{\answer {1}} \pi \left [ (R_{\mathrm {outer}}(y))^2 - (r_{\mathrm {inner}}(y))^2 \right ] dy = \answer {\frac {4\pi }{5}}$

The region $0 \leq y \leq \sqrt {x}$ with $x \leq 1$ , shown below, is revolved around the axis $y=1$ . Use the washer method to find the volume of the solid of revolution.

Each radius will be a difference of $y$ -values because slices are indexed by the variable $x$ . The distance from the axis $y=1$ to the line $y=0$ is $1$ , and the distance from the axis $y=1$ to $y = \sqrt {x}$ is $1 - \sqrt {x}$ .

$R_{\mathrm {outer}} (x) = \answer {1} \text { and } r_{\mathrm {inner}}(x) = \answer {1-\sqrt {x}}$ $V = \int _{\answer {0}}^{\answer {1}} \pi \left [ (R_{\mathrm {outer}}(x))^2 - (r_{\mathrm {inner}}(x))^2 \right ] dx = \answer {\frac {5\pi }{6}}$

Sample Quiz Questions

The region in the plane bounded on the left by the curve $x=-y^2$ , on the right by the curve $x=y^2+2y+2$ , above by the line $y = 0$ , and below by the line $y = -2$ is revolved around the axis $x = 2$ . Compute the volume of the resulting solid. (Hints won’t reveal until after you choose a response.)

The axis $x = 2$ is perpendicular to the direction of slices using the integration variable $y$ , which indicates the washer method. The region lies to the left of the axis. One way to see this is to evaluate $x=-y^2$ at $y = -2$ , giving $x = -4$ , which is to the left of the axis $x = 2$ .

The integral to compute equals $\begin {aligned} V &= \int _{-2}^{0}\pi \left ((2-(-y^2))^2 - (2-(y^2+2y+2))^2\right )~ dy\\ & = \pi \int _{-2}^{0} (-4y^3+4)~ dy\\ & = \pi \left . \left (-y^4+4y\right ) \right |_{-2}^{0} = 24\pi . \end {aligned}$

The region in the plane bounded below by the curve $y=-2x^2+5x+2$ , above by the curve $y=-2x^2+2x+2$ , on the right by the line $x = 0$ , and on the left by the line $x = -1$ is revolved around the axis $y = 2$ . Compute the volume of the resulting solid. (Hints won’t reveal until after you choose a response.)

The axis $y = 2$ is perpendicular to the direction of slices using the integration variable $x$ , which indicates the washer method. The region lies below the axis. One way to see this is to evaluate $y=-2x^2+5x+2$ at $x = -1$ , giving $y = -5$ , which is below the axis $y = 2$ .

The integral to compute equals $\begin {aligned} V &= \int _{-1}^{0}\pi \left ((2-(-2x^2+5x+2))^2 - (2-(-2x^2+2x+2))^2\right )~ dx\\ & = \pi \int _{-1}^{0} (-12x^3+21x^2)~ dx\\ & = \pi \left . \left (-3x^4+7x^3\right ) \right |_{-1}^{0} = 10\pi . \end {aligned}$

The region in the plane given by $\left |{- \frac {x}{2} + \frac {1}{2} \sqrt {9 - 6 x^{2}}}\right | \leq y \leq \frac {x}{2} + \frac {1}{2} \sqrt {9 - 6 x^{2}}$ and $0 \leq x \leq \frac {2}{3} \sqrt {3}$ is revolved around the $x$ -axis. Compute the volume of the resulting solid. (Hints won’t reveal until after you choose a response.)

If the variable $x$ is used for slicing, then slices are perpendicular to the axis of rotation, which indicates the washer method should be used.

The inequalities for $y$ give the outer and inner radii, and $\left (\frac {x}{2} + \frac {1}{2} \sqrt {9 - 6 x^{2}}\right )^2 - \left ( \left |{- \frac {x}{2} + \frac {1}{2} \sqrt {9 - 6 x^{2}}}\right |\right )^2 = x \sqrt {9 - 6 x^{2}}.$ (Note that the absolute values go away when the radius is squared.)

To compute the integral $\int _{0}^{\frac {2}{3} \sqrt {3}} \pi x \sqrt {9 - 6 x^{2}}\, dx$ we can use the substitution $u = 9 - 6 x^{2}$ which implies the equality $du = \left (- 12 x\right )dx$ for the differentials. This gives the equality $\begin {aligned} \int \pi x \sqrt {9 - 6 x^{2}}\, dx & = \int \left (- \frac {\pi }{12} \sqrt {u}\right )\, du \\ & = - \frac {\pi }{18} u^{\frac {3}{2}}. \end {aligned}$ Reversing the substitution gives $\begin {aligned} \int _{0}^{\frac {2}{3} \sqrt {3}} \pi x \sqrt {9 - 6 x^{2}}\, dx & = \left . \left [- \frac {\pi }{18} \left (9 - 6 x^{2}\right )^{\frac {3}{2}} \right ] \right |_{0}^{\frac {2}{3} \sqrt {3}}\\ & = \left (- \frac {\pi }{18} \right ) - \left (- \frac {3}{2} \pi \right ) = \frac {13}{9} \pi . \end {aligned}$

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