I've got these surfaces: $$ z = 0\\ z = 4 - y^2 $$ And a cylinder: $$ x^2+y^2=4 $$ I need to find the volume enclosed by these figures.

As far as I understand the limits of integration for $z$ are from $0$ to $4$.

For $x$ it's from $-2$ to $2$ (since the radius of the cylinder is $2$ and its center matches the coordinate center)

And since there's the relation $z = 4 - y^2$, solving for $y$ we get the limits from $-\sqrt{4-z}$ to $\sqrt{4-z}$

I've set up this integral:

$$\int_{0}^{4}\int_{-2}^{2}\int_{-\sqrt{4-z}}^{\sqrt{4-z}}\, dy \, dx \, dz$$

But I realized it's wrong because it doesn't take into account the cylinder equation. And now I'm stuck. How can I plug in the cylinder equation under the integral?

Would it be right if I did something like this (solved the cylinder equation for x and multiplied by two to get the whole area):

$$\int_{0}^{4}\int_{-2}^{2}\int_{-\sqrt{4-z}}^{\sqrt{4-z}}2\sqrt{4-y^2} \, dy \, dx \, dz$$

Thank you in advance.


There is a better way. Notice that $z$ is bounded between $0$ and $4-y^2$, thus $0 \leq z \leq 4-y^2$ are the limits of integration for $z$. Since the region must be inside the cylinder this is the region in the $xy$ plane you must integrate. Using polar coordinates you can describe it as $0 \leq r \leq 2$ and $0 \leq \theta \leq 2 \pi$. Therefore the volume is

$$\int_0^2 \int_0^{2 \pi} \int_0^{4 - r^2 \sin^2 (\theta)} \, dz \, d\theta \, dr.$$

  • $\begingroup$ Thank you, Mark! I never looked at this problem from this perspective $\endgroup$ – serg66 Sep 28 '14 at 12:06

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