# Setting up Triple Integral

Evaluate the integral

$$\int x^2+y^2+z^2 \, dV$$

over the region within the cone $z^2=x^2+y^2$ and the sphere $x^2+y^2+z^2=z$.

I started to convert everything to cylindrical coordinates but it turned out to be a bigger mess to evaluate so I converted instead to spherical coordinates. I know theta goes from $0$ to $2\pi$ and $\phi$ ranges from $0$ to $\pi/4$. I'm struggling to find my values for $\rho$. I solve the equations for the cone and the sphere to get $\rho = \sqrt{z}$ and when I was solving the cone equation the $\rho$ cancelled.

Can someone help me figure out my $\rho$ integral? I can evaluate the integral itself.

• The surface $z=x^2+y^2$ is a paraboloid, not a cone. A cone would be $z^2=x^2+y^2$. Which did you intend? – David H Mar 29 '14 at 2:07
• It is the cone. Sorry for that error. There should be a square root in the equation. – Ayoshna Mar 29 '14 at 2:08 The enclosed volume is that of a cone with its apex at the origin and its symmetry axis on the $\ z-$ axis, having a height of $\ \frac{1}{2} \$ and a base radius of $\ \frac{1}{2} \$ , surmounted by a hemisphere of radius $\ \frac{1}{2} \$ with its center at $\ ( \ 0, 0, \frac{1}{2} \ ) \$ . The graph shows a "vertical" cross-section through this volume.

While the axial symmetry of the volume, and the form of the function $\ x^2 \ + \ y^2 \ + \ z^2 \$ make it tempting to use spherical coordinates, the cone and the hemisphere cannot both be easily described in such a system. In standard spherical coordinates, the hemisphere has a somewhat awkward description.

Cylindrical coordinates make describing the integration limits for the volume a bit more manageable, as shown by ketan, but carrying out the integration itself is not entirely convenient. This problem seems designed to thwart a direct approach.

The integration can be made easier by working with the cone and the hemisphere separately. Since the "slant edges" of the cone are described by $\ z \ = \ \pm r \$ , we can set up the volume integral as

$$\ \int_0^{2 \pi} d\theta \ \ \int_0^{1/2} \int_{r}^{1/2} \ (r^2 \ + \ z^2) \ \ dz \ r \ dr \$$

$$= \ \ 2 \pi \ \int_0^{1/2} \ \left( \ r^3 z \ + \ \frac{1}{3}rz^3 \ \right) \vert_{z=r}^{z=1/2} \ \ dr$$

$$= \ \ 2 \pi \ \int_0^{1/2} \ \left( \ \frac{1}{2}r^3 \ + \ \frac{1}{24}r \ - \ r^4 \ - \ \frac{1}{3}r^4 \ \right) \ \ dr$$

$$= \ \ 2 \pi \ \cdot \ \left( \ \frac{1}{8}r^4 \ + \ \frac{1}{48}r^2 \ - \ \frac{4}{15}r^5 \ \right) \vert_{0}^{1/2} \ \ = \ \ 2 \pi \ \cdot \ \left( \ \frac{1}{8} \cdot \frac{1}{16} \ + \ \frac{1}{48}\cdot \frac{1}{4} \ - \ \frac{4}{15} \cdot \frac{1}{32} \ \right)$$

$$= \ \ \pi \ \cdot \ \left( \ \frac{15 \ + \ 10 \ - \ 16 }{960} \ \right) \ = \ \frac{3 \pi }{320} \ \ .$$

To carry out the volume integration over the hemisphere, we will introduce the "shifted" coordinate $\ \zeta \ = \ z \ - \ \frac{1}{2} \$ , for which $\ d\zeta \ = \ dz \$ . The applicable equation for the hemisphere is now $\ x^2 \ + \ y^2 \ + \ \zeta^2 \ = \ \frac{1}{4} \$ , making the volume integral

$$\ \int_0^{2 \pi} d\theta \ \ \int_0^{1/2} \int_{0}^{\sqrt{\frac{1}{4} - r^2}} \ ( \ r^2 \ + \ [\zeta + \frac{1}{2}]^2 \ ) \ \ d\zeta \ \ r \ dr \$$

$$= \ \ 2 \pi \ \int_0^{1/2} \int_{0}^{\sqrt{\frac{1}{4} - r^2}} \ ( \ r^3 \ + \ r\zeta^2 \ + \ r\zeta \ + \ \frac{1}{4}r \ ) \ \ d\zeta \ \ \ dr \$$

$$= \ \ 2 \pi \ \int_0^{1/2} \ \left( \ r^3 \zeta \ + \ \frac{1}{3} r\zeta^3 \ + \ \frac{1}{2} r\zeta^2 \ + \ \frac{1}{4}r \zeta \ \ \right) \vert_{\zeta=0}^{\zeta=\sqrt{\frac{1}{4} - r^2}} \ \ \ dr \$$

$$= \ \ 2 \pi \ \cdot \ \int_0^{1/2} \ \left[ \ r^3 \ (\frac{1}{4} - r^2)^{1/2} \ + \ \frac{1}{3} r \ (\frac{1}{4} - r^2)^{3/2} \ + \ \frac{1}{2} r \ (\frac{1}{4} - r^2) \ + \ \frac{1}{4}r \ (\frac{1}{4} - r^2)^{1/2} \ \ \right] \ \ \ dr$$

[at this point, we'll make the substitution $\ u \ = \ \frac{1}{4} - r^2 \ , \ du \ = \ -2r \ dr \ \Rightarrow \ r^2 \ = \ \frac{1}{4} - u \$ ]

$$\rightarrow \ \ 2 \pi \ \cdot \ \frac{1}{2} \ \int_0^{1/4} \ \left[ \ (\frac{1}{4} - u) \ u^{1/2} \ + \ \frac{1}{3} u^{3/2} \ + \ \frac{1}{4}u^{1/2} \ \ \right] \ \ \ du$$ $$+ \ \ 2 \pi \ \cdot \ \int_0^{1/2} \ \frac{1}{2} r \ (\frac{1}{4} - r^2) \ \ dr$$

$$= \ \ \pi \ \left( \ \frac{1}{2} \cdot \frac{2}{3} \ u^{3/2} \ - \ \frac{2}{3} \cdot \frac{2}{5} u^{5/2} \ \right) \vert_0^{1/4} \ \ + \ \ \pi \ \left( \ \frac{1}{4} \cdot \frac{1}{2} r ^2 \ - \ \frac{1}{4}r^4 \ \right) \vert_0^{1/2}$$

$$= \ \ \pi \ \left( \ \frac{1}{3} \cdot \frac{1}{8} \ - \ \frac{4}{15} \cdot \frac{1}{32} \ + \ \frac{1}{8} \cdot \frac{1}{4} \ - \ \frac{1}{4} \cdot \frac{1}{16} \ \right)$$

$$= \ \pi \ \left( \ \frac{40 \ - \ 8 \ + \ 30 \ - \ 15 }{960} \ \right) \ = \ \frac{47 \pi }{960} \ \ .$$

Thus, we find the volume integral over the cone and hemisphere to be

$$\iiint_D \ x^2\ + \ y^2 \ + \ z^2 \ \ dV \ = \ \left( \frac{47 \ + \ 3 \cdot 3}{960} \right) \pi \ = \ \frac{7 \pi}{120} \ \ .$$

$$\ \$$

The total volume of the region is

$$\frac{\pi}{3} \cdot \left(\frac{1}{2}\right)^2 \cdot \frac{1}{2} \ + \ \frac{2\pi}{3} \cdot \left(\frac{1}{2}\right)^3 \ = \ \frac{\pi}{24} \ + \ \frac{\pi}{12} \ = \ \frac{\pi}{8} \ = \ \frac{15 \pi}{120} \ \ .$$

Since half of the volume is within roughly $\ \frac{1}{5} \$ unit of the $\ z-$ axis and $\ \frac{3}{5} \$ unit of the $\ x-$ axis, the value of the integrand is typically $\ r^2 \ + \ z^2 \ \sim \ \frac{2}{5} \$ , so a reasonable estimate for our volume integral is $\ \frac{2}{5} \cdot \ \frac{\pi}{8} \ \sim \ \frac{\pi}{20} \$ .

• wow! That's a lot of LaTeX! I upvoted for the amount of patience you have in typing this :D – usukidoll May 2 '14 at 9:44
• "Copy-paste" is definitely one's friend for such derivations: one only needs to revise appropriately from one line to another. Checking a post (the math and the LaTeX) takes longer than the typing... – colormegone May 3 '14 at 19:06

this could be the setup $$\int_{0}^{2\pi} \int_{0}^{1/2} \int_{r}^{\frac{\sqrt{1-4r^2}+1}{2}} r^2+z^2dzrdrd\theta$$
in Spherical it could be :>

$$\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{\frac{1}{2sin\phi}} f(\rho,\phi,\theta)\rho^2sin\phi d\rho d\phi d\theta$$