Confusing Triple Integral - Volume

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

• can we do it in rectangular coordinates? – Santosh Linkha May 3 '14 at 19:34
• Yea, that was my plan originally. But I get stuck pretty quickly trying to find the bounds for z, r , and $\theta$ – user127778 May 3 '14 at 19:36

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.
• why wouldn't it be $rsin\theta + 1$? – user127778 May 3 '14 at 21:31
• @user127778 I shifted $y \to y - 1$ that makes cylinder at center and $x = r \cos\theta$ and $y=r\sin\theta$ – Santosh Linkha May 4 '14 at 6:18