Calculate the volume under the paraboloid $x^2+y^2=z$ and inside the cylinder $(x-1)^2+(y-1)^2=2$ above the plane z=0
the way I did it was with polar coordinates $x=1+rcos\theta$ and $y=1+rsin\theta$ , so according to my previos question here , if the shape is moved we will consider the center given as an origin in our case $(1,1)$ would be as $(0,0)$ so $0 \leq \theta \leq 2\pi$ and here is my mistake according to the solution , I did $ 0 \leq r \leq \sqrt{2}$ while it is supposed to be $ 0 \leq r \leq 1$ and $J=r$
my integral is $\iint 2r+2r^2cos\theta +r^3 +2r^2sin\theta drd\theta$ but that way I get $6\pi$ instead of $2.5\pi$
what they did in the solution is change of variables $u=x-1$ and $v=y-1$ and then the jacobian is $J=1$ all that is ok but the problem is I could not understand why the radius suddenly changed to $1$? could not understand how they got there with the change of variables or how I could get that the radius is to $1$ without changing variables
How can I get to that radius in both ways? with and without changing variables? thank you appreciate the help
I did the same change of coordinates and I also got $6\pi$, as I wrote in the comments. Then I decided to use the standard cylindrical coordinates. In that case,$$0\leqslant z\leqslant r^2\quad\text{and}\quad(x-1)^2+(y-1)^2\leqslant2\iff r\leqslant2(\cos(\theta)+\sin(\theta)).$$On the other hand, $\theta$ lies in $\left[-\frac\pi4,\frac{3\pi}4\right]$. So, the integral is$$\int_{-\pi/4}^{3\pi/4}\int_0^{2(\cos(\theta)+\sin(\theta))}\int_0^{r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta,$$which, again, is equal to $6\pi$.