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Find the volume of the region inside the surface $z = x^2 + y^2$ and between $z = 0$ and $z = 10$.

Really the only thing I need help with in this problem is setting up the limits of integration.

for x:

$0 = x^2 ---> x = 0$ (lower limit)

$10 = x^2 ---> x = sqrt(10)$ (upper limit)

for y:

$0 = x^2 + y^2 ---> y = sqrt(10-x^2)$ (upper limit)

so:

$0 < x < sqrt(10)$

$0 < y < sqrt(10-x^2)$

then it should be the double integral of $x^2 + y^2$ with those limits dydx.

However, when I calculate this, my answer ends up negative. Did I mess up the integral or the limits?

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  • $\begingroup$ You can solve it in polar coordinates. $\int_{0}^{2\pi}\int_{0}^{\sqrt{10}} r^2.r dr d\theta$ $\endgroup$ – Satish Ramanathan Oct 30 '14 at 19:22
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Another way to think about this is that you want the volume generated by rotating $z = x^2$, $x = \sqrt{z} = f(z)$ about the z-axis from $z = 0 \rightarrow 10$:

$$\int_0^{10} \pi\ f(z)^2 \ dz \ \ = \ \int_0^{10} \pi z \ dz = \cdots$$

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  • $\begingroup$ So the answer is just 50π? That's an nice way of doing the problem. I am still curious as to why my method didn't work however. If you don't mind, what would the actual limits of integration be the way I was going about the problem? $\endgroup$ – infinitylord Oct 30 '14 at 19:39
  • $\begingroup$ Would I need a third integral that goes from z = x^2+y^2 to z = 10 and then drop the function? $\endgroup$ – infinitylord Oct 30 '14 at 19:50
  • $\begingroup$ Sure. $z$ is integrated from $x^2 + y^2 \rightarrow 10$, $y$ from $-\sqrt{10-x^2} \rightarrow \sqrt{10-x^2}$ and $x$ from $-\sqrt{10} \rightarrow \sqrt{10}$, so volume $V$, $$V = \int_{-\sqrt{10}}^{\sqrt{10}} \int_{-\sqrt{10-x^2}}^{\sqrt{10-x^2}} \int_{x^2 + y^2}^{10} 1 \ dz \ dy \ dx$$ Evaluating this is a bit tedious. But maybe a good exercise to convince yourself it's the same. $\endgroup$ – Simon S Oct 30 '14 at 19:50

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