Calculate the volume of the region bounded by the coordinate planes and the surface $z=4-x^2-y^2$

Then I think its $x=0, y=0$ and $z=0$ and the surface $z=4-x^2-y^2$

I think the integral is:


is right?


2 Answers 2


Your integral approach is correct. An easy way to do it is using polar coordinates. @Rezha Adrian Tanuharja said cylindrical coordinates, but you use them for triple integrals which is not the case (you can also solve this problem with $\displaystyle\iiint_E dV$).

Let $x=rcos(\theta)$, $y=rsin(\theta)$.

Your integration region is the circle in the first quadrant:

Region D

Then, we can see that $\theta$ should go from 0 (starting from point $C$) to B, and that angle is $90$ degrees. Then $0\leq\theta\leq \frac{\pi}{2}$. It is also easy to see that the radius of the cirlce goes from $0$ to $2$ (because $2^2=x^2+y^2$).

You should also remember that the jacobian when using polar coordinates is $r$, and we set up the following integral:

$\displaystyle\int_0^{{\pi/}{2}}\int_0^2(4-r^2)rdrd\theta=2\pi$ (this is an easy to solve integral, just a polynomial).


Use cylindrical coordinate: $z=4-r^{2}\implies r^{2}=4-z$

$$ \begin{aligned} V&=\frac{1}{4}\int_{0}^{4}\pi r^{2}dz\\ &=\frac{\pi}{4}\int_{0}^{4}(4-z)dz\\ &=2\pi \end{aligned} $$

  • $\begingroup$ @JohnWaylandBales you mean when I substituted $r^{2}=4-z$? $\endgroup$
    – acat3
    Commented Nov 7, 2020 at 12:59
  • $\begingroup$ I think the OP wanted to use double integrals, polar coordinates should be an option. $\endgroup$
    – Fabrizio G
    Commented Nov 8, 2020 at 16:35

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