# Convert a triple integral to cylindrical coordinates?

Find the volume determined by $$z \le 6-x^2-y^2$$ and $$z \ge \sqrt{x^2+y^2}$$

I used cylindrical coordinates to change the bound for $z$ to $r \le z \le 6-r^2$. However, I am not sure how to find the bounds for $r$ and $\theta$.

I tried setting $r = 6 - r^2$ to find the intersection. This gives $r = -3$ and $r = 2$. Are those the bounds for $r$? What about theta?

Thank you!!

• now assume r is constant and see what is range of theta(r). – Panda Oct 27 '14 at 19:48

## 2 Answers

The key point is that $z \geq 0$. So the region is the rotation of a triangle-like shape, with curved top side ($6-r^2$.

The intersection of that curve with $z=r$ is at $r=2$.

The intergral is much more easily visualized using the "cylindar/shell" method than the "washer/disk" method; that is, we want to ingegrate over $\theta$ and $z$ first.

$$\int_{r=0}^2 dr \int_{z=r}^{6-r^2} dz \int_{\theta = 0}^{2\pi} r\, d\theta = 2\pi \int_{r=0}^2 dr \int_{z=r}^{6-r^2} r\, dz = 2\pi \int_{r=0}^2 (6r-r^3-r^2)dr$$ which is $$2\pi \left(( 3\cdot 2^2 - \frac{1}{4} \cdot 2^4 - \frac{1}{4} \cdot 2^3 \right) = \frac{32\pi}{3}$$

Notice that $6-u=\sqrt{u}$ only has the solution $u=4$, so the cone and paraboloid only intersect when $r=2$.

Therefore the limits for $r$ are $0\le r\le 2$ and the limits for $\theta$ are $0\le \theta\le 2\pi$ since the projection of the solid on the xy-plane is the region bounded by the circle $r=2$.