The region enclosed by $x=\frac{y^2}{4}$, $x=0$, $y=-4$, and $y=4$. I know my limits are $0$, $4$. And I have the integral set up. But I'm having issues finding the anti-derivative of the functions.
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If you have the integral, please show it. That will help with finding the antiderivative. – Ross Millikan Feb 15 '14 at 03:57
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Draw a picture. It seems reasonable to do the calculation by slicing. If we take a slice parallel to the $x$-axis at height $y$, the radius of the circle of cross-section is $x$, that is, $\frac{y^2}{4}$. Thus the volume is $$\int_{-4}^4 \pi\left(\frac{y^2}{4}\right)^2\,dy.$$ The integration is not difficult. Note that $\left(\frac{y^2}{4}\right)^2=\frac{y^4}{16}$.
Remark: I would prefer to use symmetry and note that the volume is $$2\int_{0}^4 \pi\left(\frac{y^2}{4}\right)^2\,dy.$$
André Nicolas
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So you used the disk method to solve? That's exactly where I got to, and for some reason when I integrate it, I'm not getting the right answer. – Pia Feb 15 '14 at 04:01
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I get $\frac{128}{5}\pi$. Without seeing what you have done, it is not possible to know what went wrong. – André Nicolas Feb 15 '14 at 04:03
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That's the answer. So for the antiderivative I got (y^5)/(4y). Is this correct? And then I plugged in 4. I didn't plug in 0 because that would just equal 0. – Pia Feb 15 '14 at 04:07
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Sure. We get $\frac{64}{5}\pi$. Then double because we are going from $y=-4$ to $y=4$. Or else integrate from $-4$ to $4$, we get $\pi\left(\frac{4^5}{16}-\frac{(-4)^5}{16}\right)$. – André Nicolas Feb 15 '14 at 04:14
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1Just saw your antiderivative, very wrong. An antiderivative of $y^4$ is $\frac{y^5}{5}$. So for our function it is $\frac{\pi}{16}\cdot\frac{y^5}{5}$. – André Nicolas Feb 15 '14 at 04:23
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