Question

The region $R$ is bounded by the lines $x=2$, $x=3$, $y=-1$, and the curve $y=\frac{1}{2+x^3}$. Suppose a solid is formed by revolving $R$ about an axis. Use the disk/washer method to set up the integral (or expression involving sums of integrals) giving the volume of the solid formed when $R$ is revolved about $x=-5$.

Answer 1

The curve intersects the vertical lines when

$x=2 \implies y = \dfrac1{2+2^3} = \dfrac1{10}$

and

$x=3 \implies y = \dfrac1{2+3^3} = \dfrac1{29}$

Split up the region into washers of thickness $\Delta y$.

For $-1\le y\le\frac1{29}$, the inner and outer radii of each washer will be constant, with outer radius $x=3$ and inner radius $x=2$. Each washer in this interval will contribute a total volume of

$\pi (3^2 - 2^2) \Delta y = 5\pi\,\Delta y$

For $\frac1{29} \le y \le \frac1{10}$, the washers will have a varying outer radius of length

$y=\dfrac1{2+x^3} \implies x = \sqrt[3]{\dfrac1y - 2}$

and inner radius $x=2$. Their volumes are

$\pi \left(\left(\sqrt[3]{\dfrac1y - 2}\right)^2 - 2^2\right) \, \Delta y = \pi \left(\left(\dfrac1y - 2\right)^{2/3} - 4\right) \, \Delta y$

Let $\Delta y\to0$. Then as the number of washers goes to infinity, the total volume of the solid converges to the definite integral

$\displaystyle 5\pi \int_{-1}^{1/29} dy + \pi \int_{1/29}^{1/10} \left(\left(\dfrac1y - 2\right)^{2/3} - 4\right) \, dy$

The first integral is trivial, but the second one requires hypergeometric functions to evaluate exactly. With a calculator, we find the approximate volume to be 0.117884.