# Compute the volume of a solid bounded by a cone and cylinder

I'm having problems with computing the volume of the solid bounded by the cone $z = 3\sqrt{x^2 + y^2}$, the plane $z = 0$, and the cylinder $x^2 + (y-1)^2 = 1$.

I tried substituting cylindrical coordinates and computed $r = 2sin\theta$, then I integrated $\int_0^{2\pi}\int_0^{2sin\theta}\int_0^{3r} \ \text{r dz dr d} \ \theta$, but I ended up getting 0 for some reason. The correct answer is $\frac{32}{3}$. Anyone knows what went wrong and how I should go about solving this question?

Thanks!

The parameterized (polar) curve $$r(\theta) = 2 \sin \theta$$ can be written as the parameterized (Cartesian) curve $$(x(\theta), y(\theta)) = (2 \sin \theta \cos \theta, 2 \sin^2 \theta) = (\sin 2 \theta, \cos 2 \theta + 1);$$ since both of the components in the last expression have period $\pi$, the parameterization traces out the circular boundary twice over the interval $[0, 2 \pi]$; we only want to integrate over each direction $\theta$ once, so we should replace the interval over which we integrate $\theta$ by $[0, \pi]$.
We can see too why the triple integral you wrote evaluates to zero: The upper limit on the middle integral, is $2 \sin \theta$, and the symmetry of that function (together with the fact that $\theta$ appears nowhere else in the integral) ensures that the integral over $[\pi, 2\pi]$ has the same magnitude but the opposite sign as the desired integral over $[0, \pi]$.