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 <br />

 

[Quinzio]

Divide the integral in 2 parts:

0&lt; \varphi &lt; \pi/3 where \rho = a
and
\pi/3&lt; \varphi &lt; \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 <br />

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.