The Disk and Washer Methods

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We practice setting up calculations related to the disk and washer methods.

Online Texts

Examples

Suppose the region below the graph $y = \sqrt {\sin x}$ and above the $x$ -axis between $x=0$ and $x=\pi$ is revolved around the $x$ -axis. Compute the volume of the resulting solid.

Because the axis of rotation lies perfectly along the boundary of the region, the diskwasher method can be used.
The radius $R$ is the length of a extending from the axis to the graph $y = \sqrt {\sin x}$ .
Thus we know that the radius $R$ must equal
$R(x) = \sqrt {\sin x} - 0 = \sqrt {\sin x}$ $R(y) = \arcsin y^2 - 0 = \arcsin y^2$
We conclude that $V = \int _{\answer {0}}^{\answer {\pi }} \pi \left ( \answer {\sqrt {\sin x}} \right )^2 d\answer {x} = \answer {2\pi }.$

Suppose the region between the graphs $y = x/2$ and $y=x^2/4$ is revolved around the axis $x=0$ . Compute the volume of the resulting solid.

Because the axis of rotation does not lie along the boundary of the region, the diskwasher method can be used.
In this case, radius will equal the length of a extending from the axis to the graphs $y = x/2$ and $y = x^2/4$ .
$R_{\mathrm {outer}} (x) = x/2$ and $r_{\mathrm {inner}}(x) = x^2/4$ $R_{\mathrm {outer}} (y) = 2 \sqrt {y}$ and $r_{\mathrm {inner}}(y) = 2 y$
We conclude that $V = \int _{\answer {0}}^{\answer {1}} \pi \left [ \left ( \answer {2 \sqrt {y}} \right )^2 - \left ( \answer {2 y} \right )^2 \right ] d\answer {y} = \answer {\frac {2 \pi }{3}}.$