Integration (Spherical) Question

In summary, the conversation is about evaluating a triple integral in spherical coordinates for a Calc. III test. The problem involves finding the integral of xyz over a region between two spheres and above a cone. The integral is split into three parts, but there is confusion about the limits of integration for \phi and \theta. After clarification, it is determined that the integral evaluates to zero due to symmetry and the range of the variables. The conversation ends with gratitude for the help provided.
  • #1
yUNeeC
34
0
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.
 
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  • #2
[tex]\int_0^{2\pi}sin(\theta)cos(\theta)d\theta[/tex]
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.
 
  • #3
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:

[tex]\int{\int{\int_E{xyzdV}}}[/tex]

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

[tex]\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}[/tex]

Consolidating terms:

[tex]\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}[/tex]

Splitting the integral apart(?):

[tex]\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}[/tex]

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?
 
  • #4
yUNeeC said:
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:

[tex]\int{\int{\int_E{xyzdV}}}[/tex]

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

[tex]\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}[/tex]

Consolidating terms:

[tex]\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}[/tex]

Splitting the integral apart(?):

[tex]\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}[/tex]

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 [itex]\theta[/itex] doesn't run all the way to 2pi. Its range is [itex][0,\pi][/itex].
 
  • #5
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.
 
  • #6
yUNeeC said:
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 [itex]\phi[/itex] and [itex]\theta[/itex]. The metric

[tex]dV = r^2 \sin^2\phi dr d\phi d\theta[/tex]

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

[tex]
\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}
[/tex]
 
  • #7
It's zero.

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

z can never go less than [tex]\sqrt{12}[/tex] due to [tex]\phi[/tex], while the max of the function should occur at each axis having a value of [tex]\sqrt{12}[/tex]...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.
 

1. What is spherical integration?

Spherical integration is a mathematical technique used to find the volume of three-dimensional objects that have a spherical shape or components. It involves integrating over a spherical coordinate system rather than a Cartesian coordinate system.

2. How is spherical integration different from other types of integration?

Spherical integration differs from other types of integration, such as rectangular or cylindrical integration, because it uses a different coordinate system and requires a different approach to solving the integral. Spherical integration is typically used for objects with spherical symmetry, while other types of integration may be used for objects with different shapes.

3. What are some applications of spherical integration?

Spherical integration has many applications in physics and engineering, such as calculating the electric field of a charged sphere or finding the center of mass of a spherical object. It is also commonly used in the calculation of triple integrals in multivariable calculus.

4. What is the formula for spherical integration?

The formula for spherical integration is ∫∫∫f(ρ,θ,φ)ρ²sinθ dρ dθ dφ, where ρ is the distance from the origin, θ is the polar angle, and φ is the azimuthal angle. This formula can be modified for different applications, but the basic structure remains the same.

5. What are some tips for solving spherical integration problems?

When solving spherical integration problems, it is important to carefully choose the limits of integration and to properly set up the integral using the correct coordinate system. It may also be helpful to sketch the region of integration and to use symmetry to simplify the problem. Additionally, practicing with different examples can improve understanding and mastery of spherical integration.

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