Surface Integrals: Find Value w/Divergence Theorem

In summary: Use "\" in front of Greek letters: \theta, \phi, \rho. Equation: \int _0^{2\pi} \int _0^{\pi} \int _0^2 { (7\rho^3 \sin^2\phi \sin^2{\theta} \cos{\phi} } \rho^2 d \rho d \phi d \theta }2) \rho is "rho", not "roe".
  • #1
duki
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Homework Statement



Find the value of the surface integrals by using the divergence theorem
[tex]\vec{F} = (y^2z)\vec{i} + (y^3z)\vec{j} + (y^2z^2)\vec{z}[/tex]

S: [tex] x^2 + y^2 + z^2 [/tex]

Use spherical coordinates.

Homework Equations



The Attempt at a Solution



I've gotten the integral I think. I want to make sure before I go along with evaluating it.

[tex]\int _0^{2\pi} \int _0^{\pi} \int _0^2 { (7\rho^3 \sin^2{\phi} \sin^2{\theta} \cos{\phi} } \rho^2 d \rho d \phi d \theta[/tex]

My latex is all messed up... maybe a mod can fix it for me?
 
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  • #2
duki said:

Homework Statement



Find the value of the surface integrals by using the divergence theorem
[tex]\vec{F} = (y^2z)\vec{i} + (y^3z)\vec{j} + (y^2z^2)\vec{z}[/tex]

S: [tex] x^2 + y^2 + z^2 [/tex]

Use spherical coordinates.

Do you mean [itex]x^2+y^2+z^2=4[/itex]?


I've gotten the integral I think. I want to make sure before I go along with evaluating it.

[tex]\int _0^{2\pi} \int _0^{\pi} \int _0^2 { (7\rho^3 \sin^2\phi \sin^2{\theta} \cos{\phi} } \rho^2 d \rho d \phi d \theta[/tex]

My latex is all messed up... maybe a mod can fix it for me?

That doesn't look quite right...what do you get for the divergence of F (in Cartesians and Sphericals)?
 
  • #3
Yes, I meant = 4. Thanks.

I got [tex] 5y^2 2z [/tex]
 
  • #4
duki said:
Yes, I meant = 4. Thanks.

I got [tex] 5y^2 2z [/tex]

I assume you mean [itex]5y^2 z[/itex]?...If so, you're right. What is that in spherical coordinates? What are you using for [itex]dV[/itex] (infinitesimal volume element) in spherical coordinates?
 
  • #5
Ok, I have [tex]5y^2z[/tex] in my notes but I thought that was wrong. When I take the partial of [tex]y^2z^2[/tex] with respect to z, why does that come out to just z?
 
  • #6
[tex]\frac{\partial}{\partial z} (y^2 z^2)=y^2 \frac{\partial}{\partial z} (z^2)=2y^2 z[/tex]
 
  • #7
OooOOooOOooooohhh
 
  • #8
I fixed the Latex:
1) Use "\" in front of Greek letters: \theta, \phi, \rho.

2) [itex]\rho[/itex] is "rho", not "roe".
 

FAQ: Surface Integrals: Find Value w/Divergence Theorem

1. What is a surface integral?

A surface integral is a type of integral that is used to calculate the total value of a function over a given surface. It takes into account the magnitude and direction of the function at each point on the surface.

2. What is the Divergence Theorem?

The Divergence Theorem is a mathematical theorem that relates a surface integral to a volume integral. It states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of the vector field over the enclosed volume.

3. When is the Divergence Theorem used in surface integrals?

The Divergence Theorem is used in surface integrals when the surface is closed and the vector field is defined over the entire enclosed volume. It allows for the simplification of calculating surface integrals by converting them into volume integrals.

4. How is the value of a surface integral found using the Divergence Theorem?

The value of a surface integral using the Divergence Theorem is found by first finding the divergence of the vector field over the enclosed volume. This is then multiplied by the volume of the enclosed region to find the total flux through the surface.

5. What are some real-world applications of surface integrals and the Divergence Theorem?

Surface integrals and the Divergence Theorem have many practical applications in fields such as physics, engineering, and fluid mechanics. They are used to calculate quantities such as electric flux, fluid flow, and heat transfer over surfaces and volumes.

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