Problem integrating over a sphere.

[S. Moger]

Homework Statement

Compute [itex]\int_S \vec{F} \cdot d\vec{S}[/itex]

[itex]\vec{F} = (xz, yz, z^3/a)[/itex]

S: Sphere of radius [itex]a[/itex] centered at the origin.

Homework Equations

[itex]x = a \sin(\theta) \cos(\varphi)[/itex]
[itex]y = a \sin(\theta) \sin(\varphi)[/itex]
[itex]z = a \cos(\theta) [/itex]

Phi : 0->2 pi, Theta : 0->pi/2 .

The Attempt at a Solution

[itex]\vec{F} = a^2 \cos(\theta) \cdot \{ \sin(\theta) \cos(\varphi), \sin(\theta) \sin(\varphi), \cos(\theta)^2 \}[/itex]

[itex]d\vec{S} = \frac{ \partial{\vec{r} }} {\partial{\theta} } \times \frac{ \partial{\vec{r}}}{\partial{\varphi}} d\theta d\varphi = a^2 \sin(\theta) \cdot \{ \sin(\theta) \cos(\varphi), \sin(\theta) \sin(\varphi), \cos(\theta) \}[/itex]

[itex]\int_S \vec{F} \cdot d\vec{S}[/itex] = [itex]\int_\varphi d\varphi \int_\theta ... d\theta = 2 \pi a^4 \int_\theta ... d\theta = 9 \pi a^4 / 10[/itex]

While the correct answer is [itex]\frac{4}{5} \pi a^4[/itex] .I'm relatively sure this isn't a book-keeping issue, I double checked the computations manually and with mathematica. Maybe I'm missing something (or maybe there's an easier way to "see" the answer).

 

[vela]

What's the integrand you came up with?

 

[mfb]

Stokes' theorem might lead to an easier integral (didn't test it).

$$\vec{F} = a^2 \cos(\theta) \cdot \{ \sin(\theta) \cos(\varphi), \sin(\theta) \sin(\varphi), \cos(\theta)^2 \}$$

This is still in cartesian coordinates? Then it shouldn't have that prefactor.

 

[S. Moger]

I got this:
[itex] \int_S \vec{F} \cdot d\vec{S}= a^4 \int \int \sin(\theta)^3 \cos(\theta) + \sin(\theta) \cos(\theta)^4 d\varphi d\theta[/itex]

The phi variable "disappears" ( = 1) because of the trigonometric identity, so I separate its integral from the expression involving theta.

The expression for the F-vector should be valid as it stands (in cartesian coordinates). I inserted the expressions for x y z in theta, phi and a, and extracted z.

 

[vela]

When you integrate the ##\sin^3\theta \cos\theta## term, it'll vanish because the integrand is odd around ##\theta = \pi/2##. Is that what you found? The final answer pops out from the second integral.

 

[mfb]

Ah right, you have the z everywhere.

 

[S. Moger]

When you integrate the ##\sin^3\theta \cos\theta## term, it'll vanish because the integrand is odd around ##\theta = \pi/2##. Is that what you found? The final answer pops out from the second integral.

If I integrate it from 0 to pi/2 I should get 1/4. (It's just [itex]\frac{\sin^4{\theta}}{4}[/itex], with theta=0 vanishing, if I'm not mistaken). But makes me think about the range (integration limits, not sure about the word for it). But if I spanned -pi/2 to pi/2 I would get zero in any case

 

[vela]

Oh, I didn't notice you had written down the wrong limits. ##\theta## does indeed go from 0 to ##\pi##, not ##\pi/2##.

 

[S. Moger]

Yes, that's it! Thank you very much.