Triple integrals-finding the volume of cylinder.

Question

Find the volume of cylinder with base as the disk of unit radius in the $xy$ plane centered at $(1,1,0)$ and the top being the surface $$z=((x-1)^2+(y-1)^2)^{3/2}.$$ I just knew that this problem uses triple integral concept but dont know how to start. I just need someone to suggest an idea to start. I will proceed then. Thank you.

Answer 1

The volume of the cylinder $C$ is given by the following triple integral $$V=\iiint_{C}dzdydx=\iint_{(x-1)^2+(y-1)^2\leq 1}((x-1)^2+(y-1)^2)^{3/2} dx dy\\ =\iint_{X^2+Y^2\leq 1}(X^2+Y^2)^{3/2} dX dY.$$ Now use the polar coordinates $X=\rho\cos(\theta)$, $Y=\rho\sin(\theta)$. Then $X^2+Y^2=\rho^2$ and $dX dY=\rho d\rho d\theta$ where $\rho$ is the jacobian of the transformation. Thus we obtain $$V=\int_{\theta=0}^{2\pi}\int_{\rho=0}^1 \rho^{3} (\rho d\rho d\theta)=2\pi[\rho^5/5]_0^1=\frac{2\pi}{5}.$$