I want to find the volume of the solid bounded between paraboloid $z=4-x^2-y^2$ and the plane $z=0$. I first tried with cylindrical coordinates:
$$V=\int_{0}^{2\pi} \int_0^2 \int_0^{4-x^2-y^2} dz \rho d\rho d\theta=\ldots =8\pi.$$
But then I tried to verify the result with cartesian coordinates:
$$V=\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^\sqrt{4-x^2} \int_0^{4} dz dy dx =\ldots=16\pi.$$
I have checked in Wolfram Alpha and both triple integrals are correct. So, in at least one of the approaches there is a mistake in the limits of integration. Any insight?
So thanks to Andreas my mistake was that z is bounded from above by $4-x^2-y^2$. Now the integral is:
$$V=\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^\sqrt{4-x^2} \int_0^{4-x^2-y^2} dz dy dx =\ldots=8\pi.$$