The Shell Method

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We practice setting up setting up volume calculations using the shell method.

(Video) Calculus: Single Variable

Online Texts

Examples

The region defined by the inequalities $-\sqrt {1-x^2} \leq y \leq \sqrt {1-x^2}$ and $x \geq 0$ (shown below) is revolved around the $y$ -ais. Compute the volume using the shell method.

When the slicing variable is $x$ , the radius of a shell is the distance from an $x$ -slice to the axis $x = 0$ . Thus $r(x) = \answer {x} - \answer {0}.$
The height of an $x$ -slice is equal to
$h(x) = \sqrt {1-x^2}$ $h(x) = - \sqrt {1-x^2}$ $h(x) = \sqrt {1-x^2} - \left ( - \sqrt {1-x^2} \right ) = 2 \sqrt {1-x^2}$
The volume is equal to the integral of $2 \pi r h$ , so $V = \int _{\answer {0}}^{\answer {1}} \answer {4 \pi x \sqrt {1-x^2}} dx = \answer {\frac {4\pi }{3}}.$ (Note: to compute the integral, we can make the substitution $u = 1-x^2$ .)

The region between the curves $x = \sqrt {y}$ and $x = y + \sqrt {y}$ from $y=0$ to $y=1$ is revolved around the axis $y=1$ . Compute the volume of the resulting solid.

When the slicing variable is $y$ , the radius of a shell is the distance from a $y$ -slice to the axis $y = 1$ . Thus $r(y) = \answer {1} - \answer {y}.$
The “height” of a $y$ -slice is equal to
$h(y) = \sqrt {y}$ $h(y) = \sqrt {y} - (y + \sqrt {y}) = -y$ $h(y) = (y + \sqrt {y}) - \sqrt {y} = y$
The volume is equal to the integral of $2 \pi r h$ , so $V = \int _{\answer {0}}^{\answer {1}} \answer {2 \pi y(1-y)} dy = \answer {\frac {\pi }{3}}.$