# Different volume with cartesian and cylindrical coordinates.

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?

• In your second integral, you have $z$ ranging from $0$ to $4$, regardlesss of $x$ and $y$. So you're getting the volume of the cylinder with the right base but a flat top at $z=4$ instead of the paraboloid top. Sep 27, 2020 at 23:49
• So I found the volume of a cylinder with radius equal to 2 and height equal to 4! Sep 27, 2020 at 23:52

So thanks to Andreas my mistake was that z is bounded from above by $$4-x^2-y^2$$. Now the integral is:
$$V=\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^\sqrt{4-x^2} \int_0^{4-x^2-y^2} dz dy dx =\ldots=8\pi.$$