It's not that the problem is extremely hard, it's just got me second guessing myself all the time when I happen to think about it. (Problem from a test last Friday)(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

Find the volume of the solid formed by revolving the area bounded by [tex]y=x^2, y=4[/tex] about the x-axis.

2. Relevant equations

Disc method would be easiest since it keeps everything neat and tidy in terms of [tex]dx[/tex]

[tex]V = \pi\[ \int_a^b [r(x)]^2\,dx.\][/tex]

3. The attempt at a solution

My attempt (and I'm pretty sure I'm right on this) was to include the washer-method.

To find the volume of the area bounded by [tex]y=x^2, y=4[/tex] revolved around the x-axis:

The bounds of integration are easy enough to determine, just setting one function equal to the other yields the interval [-2,2]. I decided to work with the interval [0,2] since it would yield half of the volume of the entire figure, then double it (makes more sense to me).

So my process to solve was to find the volume revolved around the x-axis of the area bounded by [tex]y=4, x=2[/tex] which, using the disc method would be:

[tex]V = \pi\int_0^2 [4]^2\,dx. = 32\pi[/tex]

Then subtracting out the volume revolved around the x-axis of the area bounded by [tex]y=x^2, x=2[/tex] which, using the disc method would be:

[tex]V = \pi\int_0^2 [x^2]^2\,dx. = \frac{32\pi}{5}[/tex]

[tex]V= 32\pi - \frac{32\pi}{5} = \frac{128\pi}{5}[/tex]

Thus the entire volume would be [tex]2[\frac{128\pi}{5}] = \frac{256\pi}{5}[/tex]

Or you could combine both integrals and double them to yield the entire volume of the designated problem.

[tex]V = 2*[\pi\int_0^2 [4]^2 - [x^2]^2\,dx.] = \frac{256\pi}{5}[/tex]

This is my attempt, yea or nay?

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# Easy Solid of Revolution Problem

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