Compute for each $R >0$ the following: $$ \int\int\int (x^2 + y^2+z) dV\;\text{ over the region: }\;x^2 + y^2+z^2 ≤ 2Rz $$ I decided to use polar:
Starting with $R=1$
rearranging the region I get a new region $x^2 + y^2+ (z-1)^2 ≤ 1 $ the sphere centered at $(0,0,1)$
Then I set up the integral $$ \int_{\theta=}^{\theta=}\int_{r=}^{r=}\int_{z=}^{z=}r^3+rz \,dzdrd\theta. $$
This is how far I got, I do not know how to define the bounds or whether this is the correct approach. Also after computing this integral for R=1 how do I find the general answer for $R > 0$ ?
That region is$$\{(x,y,z)\in\mathbb R^3\mid x^2+y^2+(z-R)^2=R^2\}.$$A first change of coordinates could be $x\mapsto X$, $y\mapsto Y$, and $z\mapsto Z+R$. Then your function becomes$$(X,Y,Z)\mapsto X^2+Y^2+Z+R.$$If you compute the integral of this function on the sphere centered at $(0,0,0)$ with radius $R$, you will get$$\int_0^R\int_0^{2\pi}\int_0^\pi\rho^2\sin(\varphi)\bigl(\rho^2\sin^2(\varphi)+\rho\cos(\varphi)+R\bigr)\,\mathrm d\varphi\,\mathrm d\theta\,\mathrm dr=\frac4{15}\pi R^4(2R+5).$$