Problem with finding the volume with cylindrical coordinates.

Question

I have the following function : $$Z=x^2+y^2, \ Z=x+y$$ that i want to solve (to find the volume) with cylindrical coordinates. I am evaluating the integral to get : $$V = \iiint dV=\iint\limits_{R} \hspace{-5pt} \int_{{x^2+y^2}}^{x+y} \, dz \, dA = \iint\limits_{R} x+y - (x^2+y^2) \, dA.$$ and from here i am trying to get the bounds for $r$ by intersecting the two functions and i get that $r=0 \ or \ r=1$ therefore i tought that $0\le r \le1$ but it dosen't seems right. But i get the following result : $$2\pi \int_{0}^{1}(r-r^2)rdrd\theta=2\pi\biggr(\frac{r^3}{3}-\frac{r^4}{4} \biggr ) \bigg|_0^1= \frac{\pi}{6}$$ and this is not the corrent answer, i should get $\frac{\pi}{8}.$ I know this is too specific to a given problem question i apologize for that but can someone tells me where i made a mistake ? Thank you in advance.

Answer 1

In cylindrical coordinates, you're after the volume of the region between $z=r^2$ and $z=r\bigl(\cos(\theta)+\sin(\theta)\bigr)$. Note that$$r^2\leqslant r\bigl(\cos(\theta)+\sin(\theta)\bigr)\iff r\leqslant\cos(\theta)+\sin(\theta).$$So, you want to have $\cos(\theta)+\sin(\theta)\geqslant 0$. That means that you have to compute$$\int_{-\pi/4}^{3\pi/4}\int_0^{\cos(\theta)+\sin(\theta)}\int_{r^2}^{r(\cos(\theta)+\sin(\theta))}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta,$$which is indeed equal to $\dfrac\pi8$.