Question

Session 77 lecture question: How would you evaluate the triple integral in spherical coordinates in his example? (how to integrate sec^3(theta)?)

Answer 1

can you post the integral?

Answer 2

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi/4}\int\limits_{1/\sqrt{2}\sec{\phi}}^{1} {\rho}^2\sin{\phi} d{\rho}d{\phi}d{\theta}\]
It's the volume of the portion of the unit sphere above z = 1/sqrt(2)

Answer 3

the inner gives
\[ \frac{\rho ^3}{3} \bigg|_{\sqrt{2}\sec \phi}^1= \frac{1}{3} - \frac{1}{6\sqrt{2}}
\cos^{-3} \phi\]

the middle is
\[ \int_0^\frac{\pi}{4} \frac{1}{3} \sin \phi - \frac{1}{6\sqrt{2}}\cos^{-3} \sin \phi \ d\phi\]

the first term gives
\[ -\frac{1}{3} \cos \phi \bigg|_0^\frac{\pi}{4} = - \frac{1}{3\sqrt{2}} +\frac{1}{3}\]

the second term is an example of u^(-3) du
where u= cos x and du = - sin x dx
it integrates to
\[ \frac{1}{6\sqrt{2}} \frac{\cos^{-2}\phi}{-2} \bigg|_0^\frac{\pi}{4}
\\ = -\frac{1}{12\sqrt{2}}\left( \frac{1}{\frac{1}{2}} - 1\right)= -\frac{1}{12\sqrt{2}}
\]
the two terms add up to
\[ \frac{1}{3} - \frac{1}{3\sqrt{2}}-\frac{1}{12\sqrt{2}}\\
= \frac{1}{3} - \frac{5}{12\sqrt{2}}
\]

finally, the last integral
\[ \int_0^{2\pi} \ d \theta = 2 \pi \]
and the answer is
\[ \frac{2 \pi}{3} - \frac{5\pi}{6\sqrt{2}}\]

Answer 4

oops: I noticed that I show
\[ \sqrt{2} \sec \phi\] as the lower bound on the first integral. It shoud read
\[ \frac{1}{\sqrt{2}}\sec \phi\]
The calculations use this value, so the result is correct.

Answer 5

Phi, you are awesome.