Exercises: Choose Your Method

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Exercises choosing a method for computing volume.

The region in the plane bounded by $y = e^{-x/2}$ and the $x$ -axis for $0 \leq x \leq \ln 2$ is rotated about the $x$ -axis. The volume of the resulting solid of revolution is $V = \answer { \frac {\pi }{2}}.$ (Hints won’t be revealed until after you choose a response.)

If $x$ is used as the slicing variable, then slices are vertical and consequently perpendicular to the axis of rotation.

Furthermore one side of the region lies along the axis, so the disk method is appropriate in this case.

The distance from the axis to the upper edge of the region is $e^{-x/2}$ , so $\begin {aligned} V & = \int _0^{\ln 2} \pi \left ( e^{-x/2} \right )^2 dx = \pi \int _0^{\ln 2} e^{-x} dx \\ & = \left . - \pi e^{-x} \right |_{x=0}^{\ln 2} = \pi (-e^{-\ln 2} + e^0) = \pi \left ( - \frac {1}{2} + 1 \right ) = \frac {\pi }{2}. \end {aligned}$

The region in the plane bounded on the right by the curve $x = 2 - y^2$ , on the left by the curve $x = y^2$ , and on the bottom by $y = 0$ is revolved around the $y$ -axis. Compute the volume of the resulting solid. $V = \answer {\frac {8 \pi }{3}}.$

Compute the volume of the solid of revolution obtained by rotating the region between $x=0$ , $y=0$ , and $x=\sqrt {2+3y^2 - 5y^4}$ around the $y$ -axis. $V = \answer {2 \pi }.$

The region between the graph of $y = 1-x^2$ and the $x$ -axis is rotated around the line $y=1$ . What is the volume of the resulting solid? $V= \answer { \frac {8 \pi }{5}}.$

Find the volume obtained by rotating the region between the graph $x = \frac {1}{2} \sin (y^2)$ and the $y$ -axis for $0 \leq y \leq \sqrt {\pi }$ about the $x$ -axis. $V = \answer {\pi }.$

Sample Exam Questions

Calculate the volume of the solid obtained by rotating the area between the graphs of $\displaystyle y = \frac {1}{\sqrt {x^2-1}}$ and the $x$ -axis for $1 < x < \sqrt {5}$ around the $y$ -axis.

$\pi$ $4 \pi$ $6 \pi$ $8 \pi$ $3 \pi$ $2 \pi$

Let $f(x)$ be a continuous function that satisfies $f(0) = 0$ and $f(x) > 0$ for $x > 0$ . For every $b > 0$ , when the region between the graph of $y = f(x)$ , the $x$ -axis, and the line $x=b$ is rotated around the $x$ -axis, the volume of the resulting solid is $18 \pi b^2$ . What is $f(x)$ ? (Hints will not be revealed until after you choose a response.)

$\displaystyle 9x$ $\displaystyle 3x^2$ $\displaystyle 6 \sqrt {x}$ $\displaystyle 27 x^{3/2}$ $\displaystyle 9 x^2$ $\displaystyle \sqrt {3x}$

By the disk method, we have that $\int _0^b \pi (f(x))^2 dx = 18 \pi b^2$ for each $b > 0$ . Solve this equation for $b$ .

Differentiate both sides with respect to $b$ ; use the Fundamental Theorem of Calculus to differentiate the left-hand side.

Find the volume of the solid generated by revolving the region bounded above by $y = \sec x$ and bounded below by $y=0$ for $0 \leq x \leq \pi /3$ about the $x$ -axis.

$\pi$ $2 \pi$ $\pi \sqrt {3}$ $3 \pi$ $4 \pi$ none of these