I need to find the volume between two functions $ z=x^2+y^2 $ and $ z=x + y $. I converted the functions into polar coordinates and ended up with next integral:
\begin{cases} z=z \newline x=r\cos(\phi) \newline y=r\sin(\phi) \newline I=r \end{cases} $$ \int^{2\pi}_0\int^{1}_0\int^{2}_0rdzdrd\phi=...=2\pi $$
It does not seem correct, because I estimated that rectangular with height $2$ and sides $\sqrt2$ around the shape has volume of $4$, which is less than $2\pi$. Where is the mistake?
The volume is bounded between $z=x^{2}+y^{2}$ and $z=x+y$.
So $$\iint_{S}\int_{x^{2}+y^{2}}^{x+y}\,dz\,dy\,dx=\iint_{S}x+y-x^{2}-y^{2}\,dxdy$$
Where $S$ is the region enclosed by the curve $x+y=x^{2}+y^{2}$ which is the same as $(x-0.5)^{2}+(y-0.5)^{2}=0.5$
Thus you have $$\iint_{S}(\frac{1}{2}-(x-\frac{1}{2})^{2}-(y-\frac{1}{2})^{2})\,dxdy$$
Now you substitute $x=u+\frac{1}{2} , y=v+\frac{1}{2}$ to have
$$\iint_{T}(\frac{1}{2}-u^{2}-v^{2})\,dudv$$ where $T$ is the circular region $u^{2}+v^{2}\leq\frac{1}{2}$ .
Now you use polar coordinates to have
$$\int_{0}^{\frac{1}{\sqrt{2}}}\int_{0}^{2\pi}((\frac{1}{2}-r^{2})r\,d\theta dr $$
