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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.

1 Answer

3 votes

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.

User Steve Emmerson
by
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