double integral integrand and region

Question

Find the volume of the cylinder with base as the disk of unit radius i the $xy$ plane centred at $(1,1,0)$ and the top being the surface $z=\left[(x-1)^2+(y-1)^2\right]^{3/2}$. I set up the integral as $$\int\int_C\int_{0}^{\left[(x-1)^2+(y-1)^2\right]^{3/2}}dzdydz$$ $$\implies \int\int_C[(x-1)^2+(y-1)^2]^{3/2}dydx$$ $C=\{(x,y):(x-1)^2+(y-1)^2=1\}$ Hence the integral becomes $$\int_0^{2\pi}\int_0^1r^4drd\theta$$ Is this correct?

Answer 1

This looks quite complicated, but you can make it easier. The equation of the cylinder's height is given by $z = ((x - 1)^2 + (y - 1)^2)^{\frac{3}{2}}$ and its base is given by $(x-1)^2 + (y-1)^2 = 1$. Simply shifting this over to origin will not change anything and will make it easier to work with. Therefore your new equations for height and base are $z = (x^2 + y^2)^{\frac{3}{2}}$ and $x^2 + y^2 = 1$, respectively. Setting up your integral you have $\int_C\int (x^2 + y^2)^{\frac{3}{2}}dydx$ Transforming this into polar coordinates is pretty easy so you get $\int_0^{2\pi}\int_0^1 (r^2)^{\frac{3}{2}}rdrd\theta = \int_0^{2\pi}\int_0^1 r^4 drd\theta$