# Triple integral using cylindrical coordinates with constraints

The prompt is to find an iterated triple integral in cylindrical coordinate system where the surface S is defined as the solid cut out from the ball $x^2+y^2+z^2\leq 1$ by the half cone $z = \sqrt{(x^2 + y^2)}$ which represents the volume of the solid S.

So I made the integral $$V = \iiint r.f(x, y, z) dzdrd\theta$$ we know that $x = r\cos\theta$, $y = r\sin\theta$ and z = z.

using equations, $x^2+y^2+z^2\leq 1$ and $z = \sqrt{(x^2 + y^2)}$,

$$r^2 + z^2 = 1$$ $$z = \sqrt{r}$$ $$r^2 + r = 1$$ $$r(r + 1) = 1$$ this gives us $r = -\frac{1}{2}\pm\frac{\sqrt5}{2}$

$$V = \int_0^{2\pi}\int_{-\frac{1}{2}+\frac{\sqrt5}{2}}^{-\frac{1}{2}-\frac{\sqrt5}{2}}\int (r)dzdrd\theta$$

I'm not surehow to find limits for dz

• Spherical coordinates can be maybe better. Jun 20 '17 at 9:14

We do not want to integrate by $z$ last, as this would require two separate integrals. Noting that the volume we want has rotational symmetry (since both surfaces do) about the $z$-axis, the region on the $x$-$y$ or $r$-$\theta$ plane we want is a circle. So: $$\theta: 0 \text{ to } 2\pi$$ Now $r$ goes from $0$ to the value of $r$ where the cone and sphere touch. The cone is represented as $z = r$ and the sphere is represented as $r^2 + z^2 = 1$. Therefore, the value we want is $r^2 = 1/2$, or equivalently, our limits of integration are: $$r: 0 \text{ to } \frac{1}{\sqrt{2}}$$ Finally, we will integrate along the $z$ dimension first, integrating from the bottom of the shape, which is the cone so $z = r$, to the top of shape, which is the sphere so $z = \sqrt{1-r^2}$.
Recall that: $$dV = dx \, dy \, dz = r \, dr \, d\theta \, dz$$
The volume of the solid region $R$ we have described is: \begin{align*} V &= \iiint_R dV \\ &= \int_0^{2\pi} \int_0^\frac{1}{\sqrt{2}} \int_r^\sqrt{1-r^2} r \, dz \, dr \, d\theta \\ &= 2\pi \int_0^\frac{1}{\sqrt{2}} \int_r^\sqrt{1-r^2} r \, dz \, dr \\ &= 2\pi \int_0^\frac{1}{\sqrt{2}} (r\sqrt{1-r^2} - r^2) \, dr \\ &= \pi \int_0^\frac{1}{\sqrt{2}} (2r\sqrt{1-r^2} - 2r^2) \, dr \\ &= \pi \left[-\frac{2}{3}(1-r^2)^{3/2} - \frac{2}{3}r^3 \right]_0^{\frac{1}{\sqrt{2}}} \\ &= \frac{2\pi}{3} \left[(1-r^2)^{3/2} + r^3 \right]_{\frac{1}{\sqrt{2}}}^0 \\ &= \frac{2\pi}{3} \left(1 - \frac{2}{2\sqrt{2}} \right) \\ &= \frac{\pi}{3}(2-\sqrt{2}) \end{align*}