Compute the volume of the solid bounded by the cone $z = 3\sqrt{x^2 + y^2}$, the plane $z=0$, and the cylinder $x^2 + (y-1)^2 = 1$.
So I tried parametrizing with $x = r\cos\theta$, $y = r\sin\theta + 1$, $z = z$. But then I'm not sure how to find the bounds for the triple integral. Can anybody walk me through it?
Thanks
First you need to find the place where cone completely cuts the cylinder. For that put $y=2, x=0$ which gives $z=6$. Changing $y \to y-1$ will not affect much to our calculations.
$$6\pi - \int_0^1 r \, dr \int_0^{2 \pi} d\theta \int_{3\sqrt{(r\cos \theta)^2 + (r \sin \theta - 1)^2}}^6 dz \tag{1}$$
The first term in $(1)$ is the volume of the cylinder between $z=0 \to z=6$. The second in $(1)$ is the volume trapped by the cylinder and the cone and the $z=6$ plane. By shifting $y\to y-1$ the volume of the system didn't change so it does not affect out calculations.