Find the volume of the region lying below z = 8 - x2 - y2 and above
the xy-plane.

Question

Find the volume of the region lying below z = 8 - x2 - y2 and above
the xy-plane.

anonymous · Answer

it is calc 3! double integral i think!

kinggeorge · Answer

First, you have to find where $z=8-x^2-y^2$ intersects the plane $z=0$. From there you can find the bounds with which to integrate.

anonymous · Answer

i know that is the general scheme, but i cant set it up so if you can thatd be sweet bro

kinggeorge · Answer

Well, if $z=0$, then $x^2+y^2=8$. In other words, this is a circle of radius $\sqrt{8}$ centered at the origin.

kinggeorge · Answer

Thus, we might want to switch to polar coordinates. Have you worked with polar yet?

anonymous · Answer

yes i was just about to say that, yes i have worked with them how would that look?

kinggeorge · Answer

You would want to integrate r first, and then $\theta$. Also, you know that $x^2+y^2=r^2$ and $\theta$ is the full rotation around. So the integral should look something like$$\int\limits_0^{2\pi} \int\limits_0^{\sqrt{8}} r(8-r^2) \quad dr \;\;d\theta$$

anonymous · Answer

wow man great job! thanks alot!

kinggeorge · Answer

You're welcome.