I need to find the volume bounded by the following surfaces:
$$x^2+y^2=2z$$ $$x^2+y^2=3-z$$
I don't know how to proceed in solving this exercise, all I could think of was trying to make a system of these 2 and get $z=1$. I just wanna understand how am I supposed to proceed in finding the boundaries, no need to evaluate the integral.
You have $x^2+y^2=2(3-x^2-y^2)\implies x^2+y^2=6-2x^2-2y^2\implies 3x^2+3y^2=6$, so $x^2+y^2=2$, which gives you the circle of center $(0,0)$ and radius $\sqrt 2$. We can now compute the integral in cylindrical coordinates doing the following calculations $$\iiint_Vdxdydz=\int_0^{2\pi}\int_0^{\sqrt 2}\int_{\frac{\rho^2}{2}}^{3-\rho^2}\rho dzd\rho d\theta=\int_0^{2\pi}d\theta\int_0^{\sqrt 2} \rho\bigg( 3-\frac{3\rho^2}{2}\bigg)d\rho=$$ $$=2\pi\int_0^{\sqrt 2}3\rho-\frac{3\rho^3}{2} d\rho=2\pi\cdot\frac{3\rho^2}{2}-\frac{3\rho^4}{8}\Bigg\vert_0^{\sqrt 2}=3\pi.$$