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?

  • 2
    $\begingroup$ 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. $\endgroup$ Sep 27 '20 at 23:49
  • $\begingroup$ So I found the volume of a cylinder with radius equal to 2 and height equal to 4! $\endgroup$
    – Laxuist
    Sep 27 '20 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:

$$V=\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^\sqrt{4-x^2} \int_0^{4-x^2-y^2} dz dy dx =\ldots=8\pi.$$


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