# Triple integral in which height is less than r

I've been assigned to evaluate through cylindrical coordinates

$$\iiint_E{z}\;{dV}$$

in which $E$ is in

$$0\le z\le x^2+y^2\le9$$

I approached the problem by trying to set the limits of integration as

$$\int_0^{2\pi}\int_\sqrt{z}^3\int_0^{r^2}z\,r\;dz\,dr\,d\theta$$

however, once you do the interior integral, you're left with

$$\int_0^{2\pi}\int_\sqrt{z}^3\frac{r^5}{2}\;dr\,d\theta$$

which does not let you get a numerical answer

If I plot the inequality in mathematica I get an open cylinder esque shape with a hole in the bottom which disappears as I plot more points, so the inequality doesn't exist at $0$?

If I had to take a wild stab at it I suppose $\sqrt{z}$ could be replaced with zero as $z$ does approach it which would give me $\frac{243\pi}{2}$. Any help would be appreciated

• Your lower limit for r is 0. The boundary is the disk of radius 3 (at z=0). And then you measure what is above the disk and below the paraboloid z = r^2. – Doug M Mar 30 '16 at 2:43

You are correct. The lower limit of the second integral should be 0 in place of $\sqrt z$. You have already accounted for the fact that z is always $<r^2$ in the inner integral, so the middle integral would go from 0 to 3.