Question

Evaluate the iterated integral: (see attached file for problem and answer)

Answer 1

\[\int\limits_{0}^{2pi}\int\limits_{0}^{pi/4}\int\limits_{0}^{\cos \phi}(p^2\sin \phi) dp d \phi d \theta\]

I was able to evaluate the second integral then I got stuck

Answer 2

Sorry I meant the first integral:
\[1/3\int\limits_{0}^{2pi}\int\limits_{0}^{pi/4}(\cos^3\phi \sin \phi) d \phi d \theta\]

Answer 3

When evaluating this integral:

\[-1/12\int\limits_{0}^{2pi}(\cos^4 \phi)_{0}^{pi/4}d \theta\]

How are they just getting \[\cos^4 \phi\] isn't the integration of sin phi -cos(phi)
which would make the function \[-\cos^4 \phi = - \cos^5 \phi/5 \] when evaluated?

Answer 4

Split up that integral like this:
\[{1 \over 3} \int_{0}^{2\pi} \mathrm{d}\theta \int_{0}^{\frac{\pi}{4}}\cos^3\phi \sin\phi\ \mathrm{d}\phi\]
Then for the second integral, let u=cos(phi).

Answer 5

Actually, they're right. They just probably omitted the substitution.

Answer 6

\[
\int \sin (\phi ) \cos ^3(\phi )
\, d\phi=-\frac{1}{4} \cos ^4(\phi )\\
\int_0^{\frac{\pi }{4}} \sin
(\phi ) \cos ^3(\phi ) \,
d\phi=\frac{3}{16}
\]

Answer 7

Multiply the above result by
\[ 2 \frac \pi 3
\]
and youget the answer
\[\frac \pi 8
\]

Answer 8

\[\int \cos^4x\ \mathrm{d}x \neq {\cos^5x \over 5}\]

Answer 9

oh yeah your right ^

Answer 10

We do not have to do any major integration for the second integral. The first inegral is constant.
\[
\frac 1 3 \int_0^{
2\pi} \frac 3 {16} d\theta= \frac 1 {16} \int_0^{
2\pi} d\theta =\frac 1 {16} (2\pi)=\frac \pi 8
\]

Answer 11

Oh okay I see how your getting -1/4cos^4phi, I had to use wolfram to refresh my memory on how to integrate that particular function.

Answer 12

Use substitution
\[
u=\cos(\phi)\\
du = -\sin(\phi) d\phi
\]

Answer 13

^ yeah i just realized that. I think I've been doing too many probems not using u-sub that I temporarily forgot about it.

But how are you getting 3/16 up above?