How do I set up the triple integral in cylindrical coordinates?

Question

I am trying to solve a problem from my textbook. I am uncertain of how to set up the triple integral.

Problem: Find the volume of the solid bounded by the graphs of r = $2\sin{3\theta}$, z = 8 + $\sqrt{x^2+y^2}$ and z = 0, in the first octant.

My solution:

I set up the triple integral as:

\begin{align} \int_0^{\pi/2}{\int_0^{2\sin{3\theta}}{\int_0^{8+r}{r \ dz \ dr \ d\theta}}} \end{align}

But my computed answer ($4\pi + 16/27$) is not correct.

Answer 1

Note that your limit for $\theta$ in $$\begin{align} \int_0^{\pi/2}{\int_0^{2\sin{3\theta}}{\int_0^{8+r}{r \ dz \ dr \ d\theta}}} \end{align}$$ goes beyond the loop of $$ R= 2\sin{3\theta}$$

Change your limit of $\theta$ to get

$$\begin{align} \int_0^{\pi/3}{\int_0^{2\sin{3\theta}}{\int_0^{8+r}{r \ dz \ dr \ d\theta}}}\end{align}$$