Divergence theorem with hyperboloid and tetrahedron regions

[Divergent13]

Hi!

We are nearing the end of our course --- culminating in Stokes and Divergence Theorems for surface integrals, and I am having some difficulty with the following

1. F(x,y,z) = $$<x^3y, -x^2y^2, -x^2yz>$$

where S is the solid bounded by the hyperboloid x^2 + y^2 - z^2 =1 and the planes z = -2 and z=2.

I computed div F properly...

Well I know what the z limits are in my Triple Integral, however what must I use as my radius? Theta should go from 0 to 2pi correct?

2. F(x,y,z) = $$<x^2y, xy^2, 2xyz>$$

where S is the surface of the tetrahedron bounded by the planes x=0, y=0, z=0, and x+2y+z = 2

Here must my triple integral be from 0 to 2 for the x limit, then 0 to (2-x)/2 for my y limits, and for z just 0 to 2-x-2y.

Those seem correct, but a confirmation would be nice! Thanks a lot!

 

[Divergent13]

Whoops -- it's probably implied but the questions ask to compute


$$\int\int_SFdS$$ = $$\int\int\int_EdivFdV$$

 

[arildno]

1. Did you get the divergence to be zero?
I did, so is this an exercise in verifying that computing the surface integral directly also yields zero?

 

[arildno]

The limits in 2 seems right.

 

[Divergent13]

Ahh so maybe i didnt compute div F properly-- yup it's 0, making the answer 0. Thanks.

 

[Divergent13]

Ahh here's another one that's a bit more challenging:


Its BEGGING spherical coordinates:

$$F(x,y,z) = <x^3+ysinz, y^3+zsinx, 3z>$$

S: Surface of the solid bounded by the hemispheres $$z=sqrt(4-x^2-y^2)$$ and $$z=sqrt(1-x^2-y^2)$$ and the plane $$z=0$$.

I set: r(r,phi,theta) = <rsin(phi)cos(theta), rsin(phi)sin(theta), rcos(phi) >
Are the following the correct limits of integration?

1 < r < 2
0 < phi < pi/2
0< theta < 2pi

 

[arildno]

Are you talking about the limits of integration for the surface here??
(Remember, if you're talking about the solid, r=0 is certainly included.)

In order to solve this, I suggest you split your region in two solids, the two hemishperes, both bounded by the plane z=0.

Then it is easy to apply spherical coordinates on these solids separetely.

 

[arildno]

I'm sorry, total mistak on my parte (I thought there was a minus sign somewhere )
Your approach is perfefctly correct

 

[Divergent13]

So it is from 1 to 2... okay thanks!