# Setting up the triple integral of the volume using cylindrical coordinates

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.

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?

• 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. Jun 17 at 6:32

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