Integration (Spherical) Question

[yUNeeC]

Hey guyzerz,

I'm reviewing for an upcoming Calc. III test and have come across a review problem that is giving me fits:

Evaluate the integral where E lies between the spheres p = 3 and p = 6 and above the cone phi = pi/4.

The integral is TRIPLEINT[xyz]dV

So basically, after converting to spherical coordinates, I get to where the integral (before any integration) looks like this:

(p^5)(sin^3(phi)cos(phi))(sin(theta)cos(theta))

I can split these up into:

INT[p^5] from p=6 to p=3

INT[sin^3(phi)cos(phi)] from phi = pi/4 to phi = 0

INT[sin(theta)cos(theta)] from 2pi to 0pi

The first two of these integrals are easy as pie. But on the third one, I get the integral to equal (after integration) 0.5(sin^2(theta)) from 2pi to 0...which gives me an answer of 0. This also equals zero if you integrate via u-substitution the other way around. I've tried integrating from pi/4 to 0 and multiplying the result by 8 (=2) and this doesn't work either. I really feel like I need to get my answer in terms of pi to account for the spherical nature of the problem, but I'm at a loss as to how to go about doing this.

Any help? :(

Gracias for tu time-o

PS: Yes I left off the dpdthetadphi stuff because it looked like i passed out on my keyboard.

 

[phyzguy]

\int_0^{2\pi}sin(\theta)cos(\theta)d\theta
is in fact zero. Why don't you tell us what the function E that you are trying to evaluate in this region is so we can see if you've set it up correctly? Also, you might try using tex so we can actually read the integrals.

 

[yUNeeC]

I apologize...didn't realize the Latex thing was on top of the message box. Anyway, time to make this more clear...

Evaluate the integral where E lies between the spheres p = 3 and p = 6 and above the cone phi = pi/4:

\int{\int{\int_E{xyzdV}}}

So basically, I convert to spherical coordinates, and put in the constraints and I get:

\int_0^{2\pi}{\int_0^{\pi/4}{{\int_3^6{(\rho^3}sin^2(\phi)cos(\phi)}sin(\theta)cos(\theta))\rho^2sin(\phi)d\rho d\phi d\theta}

Consolidating terms:

\int_0^{2\pi}{\int_0^{\pi/4}{{\int_3^6{(\rho^5}sin^3(\phi)cos(\phi)}sin(\theta)cos(\theta))d\rho d\phi d\theta}

Splitting the integral apart(?):

\int_3^6{\rho^5 d\rho} \int_0^{\pi/4}{sin^3(\phi)cos(\phi) d\phi} \int_0^{2\pi}{cos(\theta)sin(\theta)d\theta}

So basically, I'm realizing splitting the integral up probably isn't the way to go...because I end up with that zero from the integral of theta. Where am I going wrong here?

 

[Mute]

I apologize...didn't realize the Latex thing was on top of the message box. Anyway, time to make this more clear...

Evaluate the integral where E lies between the spheres p = 3 and p = 6 and above the cone phi = pi/4:

\int{\int{\int_E{xyzdV}}}

So basically, I convert to spherical coordinates, and put in the constraints and I get:

\int_0^{2\pi}{\int_0^{\pi/4}{{\int_3^6{(\rho^3}sin^2(\phi)cos(\phi)}sin(\theta)cos(\theta))\rho^2sin(\phi)d\rho d\phi d\theta}

Consolidating terms:

\int_0^{2\pi}{\int_0^{\pi/4}{{\int_3^6{(\rho^5}sin^3(\phi)cos(\phi)}sin(\theta)cos(\theta))d\rho d\phi d\theta}

Splitting the integral apart(?):

\int_3^6{\rho^5 d\rho} \int_0^{\pi/4}{sin^3(\phi)cos(\phi) d\phi} \int_0^{2\pi}{cos(\theta)sin(\theta)d\theta}

So basically, I'm realizing splitting the integral up probably isn't the way to go...because I end up with that zero from the integral of theta. Where am I going wrong here?

The angle \theta doesn't run all the way to 2pi. Its range is [0,\pi].

 

[yUNeeC]

That still results in theta integrating to 0.

Also, I'm reasonably sure that the range of theta is [0,2pi] as "the cone pi/4" means the angle pi/4 with respect to the z axis. It's a different plane, so "above" has nothing to do with the position of the x and y axis. The angle drops down from the z-axis in all directions.

I need help on how to not end up with theta integrating to zero.

 

[Mute]

That still results in theta integrating to 0.

Also, I'm reasonably sure that the range of theta is [0,2pi] as "the cone pi/4" means the angle pi/4 with respect to the z axis. It's a different plane, so "above" has nothing to do with the position of the x and y axis. The angle drops down from the z-axis in all directions.

I need help on how to not end up with theta integrating to zero.

You're right that it still integrates to zero, my bad for not looking more closely. The actual problem is that you've mixed up the limits of \phi and \theta. The metric

dV = r^2 \sin^2\phi dr d\phi d\theta

assumes that \theta is the angle from the z axis and \phi is the angle in the x-y plane. Conventionally, \phi runs from 0 to 2 pi and theta runs from 0 to pi. Your integral should be

<br /> \int_3^6{\rho^5 d\rho} \int_0^{2\pi}{sin^3(\phi)cos(\phi) d\phi} \int_0^{\pi/4}{cos(\theta)sin(\theta)d\theta}<br />

 

[yUNeeC]

It's zero.

Thinking about xyz in the range of 3\leq\sqrt{x^2+y^2+z^2}\leq 6

z can never go less than \sqrt{12} due to \phi, while the max of the function should occur at each axis having a value of \sqrt{12}...so I guess that, and a combination of the symmetry = 0.

Conceptually clear now. Shouldn't have even asked the question...I just didn't see that it could equal 0.

I appreciate the help, Mute.