Use cylindrical coords to find the volume of the solid:

[hank]

The solid that is inside the surface r^2 + z^2 = 20 abut not above the surface z = r^2.

Can someone verify my setup?


\(\displaystyle V=\int^{2\pi}_{0} \int^{\sqrt{20}}_{0} \int^{\sqrt{20-r^2} - r^2}_{-\sqrt{20-r^2} } r dzdrd\theta\)

 

[tkhunny]

That appears to be sufficient for the section that is directly under z = r^2, but it's no good outside that.

In other words, if you had the limits of the second integral [0,2] you would have the center piece correct. You need to rethink it for r > 2.

 

[galactus]

You want the region between \(\displaystyle r^{2}+z^{2}=20\) and \(\displaystyle z=r^{2}\)

\(\displaystyle r^{2}+z^{2}=20\) intersects \(\displaystyle z=r^{2}\) in a circle of radius 2.

So, we get:

\(\displaystyle V=\int_{0}^{2\pi}\int_{0}^{2}\int_{r^{2}}^{\sqrt{20-r^{2}}}rdzdrd{\theta}=\int_{0}^{2\pi}\int_{0}^{2}\left(r\sqrt{20-r^{2}}-r^{3}\right)drd{\theta}\)

 

[tkhunny]

I still think that is missing \(\displaystyle 2 < r < \sqrt{20}\). There is more to this problem than r < 2.