- #1
Living_Dog
- 100
- 0
Let [tex]\vec{v} = r^{2} \hat{r}[/tex]. Show that the divergence theorm is correct using [tex]0 <= r <= R , 0 <= \theta <= \pi , and 0 <= \phi <= 2\pi [/tex].
[tex]$ \int \nabla \cdot \vec{v} d \tau = \int \vec{v} \cdot d \vec{a} $[/tex]
First the divergence of [tex]\vec{v}[/tex].
[tex]\nabla \cdot \vec{v} = 2r[/tex].
Then the volume integral:
[tex]
$ \int \int \int (2r) r^{2}sin \theta dr d\theta d\phi = 2 \int r^{3} dr \int sin \theta d\theta \int d\phi = 2 \cdot \frac{1}{4} r^{4} \left|^{R}_{0} \cdot 2 \cdot 2 \pi = 2 \pi R^{4}$
[/tex].
Ok, fine. But now do the area integral on the right. Since it is over the surface only, then [tex] r = R [/tex] and the integral is only over [tex]\theta[/tex] and [tex]\phi[/tex].
[tex]$ \int \vec{v} \cdot d\vec{a} = \int r^{2}\left|_{r=R} \cdot R^2 \int sin \theta d \theta \int d \phi = R^{4} \cdot 2 \cdot 2 \pi $[/tex]
And the last time I checked, [tex]2 \pi R^{4} \not= 4 \pi R^{4}[/tex]
Please help since I am going nuts with this minute detail!
-LD
[tex]$ \int \nabla \cdot \vec{v} d \tau = \int \vec{v} \cdot d \vec{a} $[/tex]
First the divergence of [tex]\vec{v}[/tex].
[tex]\nabla \cdot \vec{v} = 2r[/tex].
Then the volume integral:
[tex]
$ \int \int \int (2r) r^{2}sin \theta dr d\theta d\phi = 2 \int r^{3} dr \int sin \theta d\theta \int d\phi = 2 \cdot \frac{1}{4} r^{4} \left|^{R}_{0} \cdot 2 \cdot 2 \pi = 2 \pi R^{4}$
[/tex].
Ok, fine. But now do the area integral on the right. Since it is over the surface only, then [tex] r = R [/tex] and the integral is only over [tex]\theta[/tex] and [tex]\phi[/tex].
[tex]$ \int \vec{v} \cdot d\vec{a} = \int r^{2}\left|_{r=R} \cdot R^2 \int sin \theta d \theta \int d \phi = R^{4} \cdot 2 \cdot 2 \pi $[/tex]
And the last time I checked, [tex]2 \pi R^{4} \not= 4 \pi R^{4}[/tex]
Please help since I am going nuts with this minute detail!
-LD
Last edited: