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

  • #1

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


[tex]\int \int F\cdot n dS=\int \int \int divF dv[/tex]

The Attempt at a Solution


For the RHS
r(u,v)=(4cosu,2sinu,v) where u=[0,2pi] and v=[0,1]
[tex]\overrightarrow{r}u \times \overrightarrow{r}v =(2cosu,4sinu,0) [/tex][/B]
[tex]\left \| \overrightarrow{r}u \times \overrightarrow{r}v \right \|=\sqrt{4cos^{2}u+16sin^{2}v} [/tex]
[tex]\int \int f(r(u,v))\cdot \left \| \overrightarrow{r}u \times \overrightarrow{r}v \right \| dudv [/tex]
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:

Answers and Replies

  • #2
LCKurtz
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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


[tex]\int \int F\cdot n dS=\int \int \int divF dv[/tex]

The Attempt at a Solution


For the RHS
r(u,v)=(4cosu,2sinu,v) where u=[0,2pi] and v=[0,1]
[tex]\overrightarrow{r}u \times \overrightarrow{r}v =(2cosu,4sinu,0) [/tex][/B]
[tex]\left \| \overrightarrow{r}u \times \overrightarrow{r}v \right \|=\sqrt{4cos^{2}u+16sin^{2}v} [/tex]
[tex]\int \int f(r(u,v))\cdot \left \| \overrightarrow{r}u \times \overrightarrow{r}v \right \| dudv [/tex]
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.
 
Last edited:
  • #3
thanks for the reply
Should u and v be [0,pi/2],[0,4] respectively ?
thus, the integration is
[tex]\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} (4cosu,2sinu,v)\cdot (2cosu,4sinu,0) dudv[/tex]
[tex]=\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} (8cos^{2}u+8sin^{2}u)dudv[/tex]

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
LCKurtz
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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
[tex]\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} \color{red}{(4cosu,2sinu,v)}\cdot (2cosu,4sinu,0) dudv[/tex]
[tex]=\int_{0}^{4}\int_{0}^{\frac{\pi}{2}} (8cos^{2}u+8sin^{2}u)dudv[/tex]

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##?
 
Last edited:

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