Can Triple Integrals Be Solved in Multiple Ways?

[rado5]

Homework Statement

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

$x^2+ y^2+ z^2= 2az$ when $x^2+ y^2+ z^2- 2az= 0$. We can complete the square by adding $a^2$ to both sides: $x^2+ y^2+ z^2- 2az+ a^2= x^2+ y^2+ (z- a)^2= a^2$. That is a sphere with center at (0, 0, a) and radius a. $x^2+ y^2+ z^2= a^2$ is, of course, a sphere with center at (0 0, 0) and radius a. The two spheres intersect when $(z- a)^2 - z^2= z^2- 2az+ a^2- z^2= a^2- 2az= 0$ or $z= a/2$, in which case $x^2+ y^2+ z^2= x^2+ y^2+ a^2/4= a^2$ or $x^2+ y^2= 3a^2/4$. That is, the intersection is a circle in the z= a/2 plane with center at (0, 0, a/2) and radius $a\sqrt{3}/2$.

In terms of spherical coordinates, each point on that circle of intersection has $\rho= a$ and $\theta= arcsin(\sqrt{3}/2)= \pi/6$

The integral, then, should be done in two parts: For the first, $\theta$ goes from 0 to $\pi/6$ and, for each $\theta$, $\rho$ goes to the "upper sphere", [math]x^2+ y^2+ z^2= a^2[/math], from 0 to a.

For the second part, $\theta$ goes from $\pi/6$ to $\pi/2$ and, for each $\theta$, $\rho$ goes from 0 to the "lower" sphere $x^2+ y^2+ z^2= 2az$ which in polar coordinates is $\rho^2= 2a\rho cos(\theta)$. That is, for each $\theta$, $\rho$ goes from 0 to $2a cos(\theta)$. (Note that when $\theta= \pi/6$, $cos(\theta)= 1/2$ so that $\rho= a$.)

$\phi$ goes from 0 to $2\pi$ for both integrals.

(In order to be consistent with the original post, I have used "engineering" notation of spherical coordinates in which $\theta$ is the "co-latitude" and $\phi$ is the "longitude" rather than "mathematics" notation in which the two Greek letters are reversed. I have used $\rho$ rather than "r" because that is the convention in either system.)

 

[Quinzio]

Divide the integral in 2 parts:

$$0< \varphi < \pi/3$$ where $$\rho = a$$
and
$$\pi/3< \varphi < \pi$$ where $$\rho = a\:sen2\varphi$$

$$z = \rho cos\varphi$$

So you solve

$$\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{a} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta <br /> +<br /> \int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen2\varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta$$

 

[Quinzio]

Divide the integral in 2 parts:

$$0< \varphi < \pi/3$$ where $$\rho = a$$
and
$$\pi/3< \varphi < \pi$$ where $$\rho = a\:sen2\varphi$$

$$z = \rho cos\varphi$$

So you solve

$$\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{a} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta <br /> +<br /> \int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen2\varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta$$

Only the firt part, if I didn't make mistakes:

$$=\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{a} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta$$

$$=\int_{0}^{2\pi} \int_{0}^{\pi/3} \left [ \frac{\rho^5}{5} (cos\varphi)^2 sen \varphi \: \right ]_0^a \: d\varphi\:d\theta$$

$$=\frac{ a^5 }{5} \int_{0}^{2\pi} \int_{0}^{\pi/3}(cos\varphi)^2 sen \varphi \: \: d\varphi\:d\theta$$

$$=\frac{ a^5 }{5} \int_{0}^{2\pi} \left[ \frac{-(cos\varphi)^3}{3} \: \right]_0^{\pi/3} \: d\theta$$$$=\frac{ a^5 }{5} \int_{0}^{2\pi} \left[ -\frac{3\sqrt3}{24} +\frac{1}{3} \: \right] \: d\theta$$$$=\frac{ a^5 }{5} \left( \frac{8-3\sqrt3}{24} \right) \int_{0}^{2\pi}\: d\theta$$

$$=2\pi \frac{ a^5 }{5} \left( \frac{8-3\sqrt3}{24} \right)$$

 

[Quinzio]

Other part

$$=\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen \varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta$$

$$=\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left [ \frac{\rho^5}{5} (cos\varphi)^2 sen \varphi \: \right ]_{0}^{2asen \varphi} d\rho \: d\varphi\:d\theta$$

$$=\frac{32a^5}{5}\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left (cos\varphi)^2 ( sen \varphi)^6 \: d\varphi\:d\theta$$

$$=\frac{32a^5}{5}\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left (( sen \varphi)^6-( sen \varphi)^8) \: d\varphi\:d\theta$$

Now have fun solving sine power eight :)

 

[rado5]

Dear HallsofIvy,

Thank you very much for your help. Yes my solution was wrong. At the end of my book the answer is $$\frac{59 \pi a^{5}}{480}$$ and when I solved it with your method I got the correct answer. I was actually wrong about the ranges!

Dear Quinzio,

I used HallsofIvy's method and I could easily solve it!

 

[rado5]

 

[Quinzio]

The two ways are really the same.