I need to setup an iterated triple integral in cylindrical coordinates to solve the volume of the surface bounded by region inside the cylinder $x^2+y^2=12$, outside the hyperboloid $x^2+y^2-z^2=3$ and the first octant. Now, I tried to graph the cylinder and hyperboloid in polar coordinates. This is the graph in polar coordinates.

enter image description here

The bounds for z is $0 \leq z \leq \sqrt{r^2-3}$ since the region is also bounded by the 1st octant.

There are two radii r = $\sqrt{12}$ and r = $\sqrt{3}$ whenever z = 0. The bounds for $\theta$ is from 0 to $\frac{\pi}{2}$.

The iterated integral here is split.

$$ \int^{\frac{\pi}{2}}_{0} \int^{\sqrt{12}}_{0} \int^{\sqrt{r^2-3}}_{0} r dzdrd\theta - \int^{\frac{\pi}{2}}_{0} \int^{\sqrt{3}}_{0} \int^{\sqrt{r^2-3}}_{0} r dzdrd\theta$$

My idea here is to subtract the area of the circle with radius $\sqrt{3}$ from the area of the circle with radius $\sqrt{12}$. Is my process right?

  • $\begingroup$ Hyperboloid surface only forms for $x^2+y^2 \geq 3$ so subtracting for $0 \leq r \leq 3$ does not work as $z$ is not defined in this interval of $r$. You have the same mistake in the other question which I did not notice earlier. I did respond back on that again. I hope this clarifies. If you are not clear, let me know. $\endgroup$
    – Math Lover
    Jun 17, 2021 at 6:32

1 Answer 1


The integral setup you have is not correct. When you are taking the decision to subtract, you are only looking at the picture in XY-plane but you should also check for $z$. $z$ is bound above by the surface $z^2 = x^2+y^2-3$. The min value of $z^2$ is zero and so we must have $x^2+y^2 \geq 3$. In other words, for $r \lt \sqrt3, $ $z$ is not defined.

In any case, you do not need to split the integral in polar coordinates. That was needed in the last question as you were using cartesian coordinates. So the correct integral should be,

$\displaystyle \int^{\frac{\pi}{2}}_{0} \int^{\sqrt{12}}_{\sqrt3} \int^{\sqrt{r^2-3}}_{0} r \ dz \ dr \ d\theta$


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