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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 (Hints won’t be revealed until after you choose a response.)
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.
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.

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?

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.

### 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.)

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

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