# How can I verify the Divergence Theorem for F=(2xz,y,−z^2)

1. Mar 15, 2017

### kelvin56484984

1. The problem statement, all variables and given/known data
Verify the Divergence Theorem for F=(2xz,y,−z^2) and D is the wedge cut from the first octant by the plane z =y and the elliptical cylinder x^2+4y^2=16

2. Relevant equations
$$\int \int F\cdot n dS=\int \int \int divF dv$$
3. The attempt at a solution
For the RHS
r(u,v)=(4cosu,2sinu,v) where u=[0,2pi] and v=[0,1]
$$\overrightarrow{r}u \times \overrightarrow{r}v =(2cosu,4sinu,0)$$

$$\left \| \overrightarrow{r}u \times \overrightarrow{r}v \right \|=\sqrt{4cos^{2}u+16sin^{2}v}$$
$$\int \int f(r(u,v))\cdot \left \| \overrightarrow{r}u \times \overrightarrow{r}v \right \| dudv$$
For the Div F ． dv
Div F=1
What is the limit for the triple integral?
How can I do the triple integration to verify the divergence theorem?

thanks

Last edited: Mar 15, 2017
2. Mar 15, 2017

### LCKurtz

That last integral should be$$\pm\int \int \vec F(r(u,v))\cdot \vec r_u \times \vec r_v~ dudv$$where the sign is chosen appropriate for the outer normal.
$u$ doesn't go from $0$ to $2\pi$ for the first octant, and $v$ doesn't go from $0$ to $1$. Also, I assume you are aware that the surface integral(s) must include all four surfaces.

Have you drawn a picture of the wedge? Since the divergence is $1$ you are just doing a volume integral. You might pick some order of dz, dy, dx and set up a triple integral. Start with a picture.

Last edited: Mar 15, 2017
3. Mar 16, 2017

### kelvin56484984

Should u and v be [0,pi/2],[0,4] respectively ?
thus, the integration is
$$\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} (4cosu,2sinu,v)\cdot (2cosu,4sinu,0) dudv$$
$$=\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} (8cos^{2}u+8sin^{2}u)dudv$$

For the second part,
If I use the shperical coordinate to do the integration,
θ should be [0,pi/2] , ϕ,should be [0,pi/4] ?
How can I find the limit of p ?
z=y : psinϕ=psinθcosϕ

4. Mar 16, 2017

### LCKurtz

Yes for $u$, no for $v$. $v$ is just a rename of $z$ and it never gets as large as $4$. And its range depends on what $u$ is, which determines $x$ and $y$. You need a picture.

No, per above. Also what I have highlighted in red doesn't look like $\vec F(\vec r(u,v))$.

Why in the world would you think of spherical coordinates? There is nothing "spherical" about this problem. You have apparently ignored my post where I suggested to draw a picture and set it up in rectangular coordinates. And what about the other three surfaces for part $1$?

Last edited: Mar 16, 2017