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

[kelvin56484984]

Homework Statement

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

Homework Equations

\int \int F\cdot n dS=\int \int \int divF dv

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)[/B]
\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

 

[LCKurtz]

Homework Statement

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

Homework Equations

\int \int F\cdot n dS=\int \int \int divF dv

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)[/B]
\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

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.

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

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.

 

[kelvin56484984]

thanks for the reply
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

 

[LCKurtz]

thanks for the reply
Should u and v be [0,pi/2],[0,4] respectively ?

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.

thus, the integration is
\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} \color{red}{(4cosu,2sinu,v)}\cdot (2cosu,4sinu,0) dudv
=\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} (8cos^{2}u+8sin^{2}u)dudv

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

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

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 $1$?