Question

The region
`R` is bounded by the lines
`x=2`,
`x=3`,
`y=-1`, and the curve
`y=(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 = \frac1{2+2^3} = \frac1{10}`

and

`x=3 \implies y = \frac1{2+3^3} = \frac1{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=\frac1{2+x^3} \implies x = \sqrt[3]{\frac1y - 2}`

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

`\pi \left(\left(\sqrt[3]{\frac1y - 2}\right)^2 - 2^2\right) \, \Delta y = \pi \left(\left(\frac1y - 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(\frac1y - 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.