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Andru [333]
3 years ago
5

Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by y=x^8 and y = 256

in the first quadrant about the y-axis.
Mathematics
1 answer:
Mkey [24]3 years ago
8 0

Using the shell method, the volume integral would be

\displaystyle 2\pi \int_0^2 x(256-x^8)\,\mathrm dx

That is, each shell has a radius of <em>x</em> (the distance from a given <em>x</em> in the interval [0, 2] to the axis of revolution, <em>x</em> = 0) and a height equal to the difference between the boundary curves <em>y</em> = <em>x</em> ⁸ and <em>y</em> = 256. Each shell contributes an infinitesimal volume of 2<em>π</em> (radius) (height) (thickness), so the total volume of the overall solid would be obtained by integrating over [0, 2].

The volume itself would be

\displaystyle 2\pi \int_0^2 x(256-x^8)\,\mathrm dx = 2\pi \left(128x^2-\frac1{10}x^{10}\right)\bigg|_{x=0}^{x=2} = \boxed{\frac{4096\pi}5}

Using the disk method, the integral for volume would be

\displaystyle \pi \int_0^{256} \left(\sqrt[8]{y}\right)^2\,\mathrm dy = \pi \int_0^{256} \sqrt[4]{y}\,\mathrm dy

where each disk would have a radius of <em>x</em> = ⁸√<em>y</em> (which comes from solving <em>y</em> = <em>x</em> ⁸ for <em>x</em>) and an infinitesimal height, such that each disk contributes an infinitesimal volume of <em>π</em> (radius)² (height). You would end up with the same volume, 4096<em>π</em>/5.

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