I'm not completely sure that this is right, but it seems like it should be.

[sgupta31]

Homework Statement

Let D be the region x^2 + y^2 + z^2 <=4a^2, x^2 + y^2 >= a^2, and S its boundary (with
outward orientation) which consists of the cylindrical part S1 and the spherical part
S2. Evaluate the
ux of F = (x + yz) i + (y - xz) j + (z -((e^x) sin y)) k through
(a) the whole surface S using the Divergence Theorem,
(b) the surface S1 by calculating the flux integral directly,
(c) the surface S2 by calculating the flux integral directly.

Homework Equations

x^2 + y^2 + z^2 <=4a^2, x^2 + y^2 >= a^2
F = (x + yz) i + (y - xz) j + (z -((e^x) sin y))

The Attempt at a Solution

I see this to be a figure where the cylinder (radius a) is in the sphere (radius 2a).

a^2 + z^2= 4a^2
z= sqrt(3) *a

0<z<sqrt(3)*a
a<r<2a
0<theta<2*pi
I am not sure if I am doing it the right way, please point to the right direction!
Integrating wt z r theta, I get 9*sqrt(3)* (a^3)* pi for a)

 

[vela]

Your limits for z are incorrect. For one thing, you're only including the volume above the z=0 plane. Also, the upper limit should depend on r.

 

[sgupta31]

Hi,
Thanks for your response, however I do not understand when you say "above the z=0 plane." Is my visualization of the diagram correct? In the sense its a cylinder in a sphere right?

 

[Char. Limit]

I believe that by "above the z=0 plane", he means that you're only considering the region above the plane z=0, or in other words, you're only considering z>0. z should never be negative.

But hey, I'm not entirely certain.

 

[vela]

Yes, that's what I mean. Both the sphere and cylinder extend above and below the z=0 plane, so setting the lower limit equal to 0 excludes half the volume.

 

[sgupta31]

Hi,

Thanks for both your replies. I tried working out the z limits again and get:( 4*(a^2)- (r^2)) and the second limit, I am guessing should be negative, if it has to cover the entire volume. The limits for r are a and 2a and for theta 0 and 2 pi. Is this right?

Please help as I think I am getting the diagram right, if not then could you provide me with hints?

Thanks.

 

[vela]

Yes, those limits are correct.

 

[sgupta31]

Thanks. So I took the lower limit of z to be -(4*(a^2)-(r^2)) and the integral I get using Divergence theorem ie. I multiply the integral by divF=3. I get an answer of 48*pi*(a^4).

Not sure if this seems right, as there is a ^4 value to a..?I am a bit confused here and want to make sure before jumping on to b) and c). Thanks a lot.

 

[vela]

Your instinct is correct. I missed that you didn't take the square root. The limits of z should be ±sqrt[(2a)²-r²].

 

[sgupta31]

Thank you so much Vela.

So I understand why the limits were taken to be negative, as we were calculating all the area (ie. below the z=0 plane as well).

I got an answer of 12*sqrt(3)*pi*(a^3). Since there is an (a^3) I feel its about right, for a volume.

Now for part b) Surface S_1 which is the cylinder I am considering the equation (x^2)+(y^2)>= (a^2).
When I parametrize it, since I have to use the long method ie. the direct flux integral F.ds, I found
r(z,theta)=(acostheta i+ asintheta j+ z k)
The limits of theta are from 0 to 2pi and z are -sqrt(3)*a to sqrt(3)*a. Is this right? I am a bit confused as to parametrization, have spent a lot of time on this question and my brain is fried. Any help or hints would be appreciated.

 

[sgupta31]

and for part c) sphere I am using (x^2)+ (y^2) + (z^2)<= (4*(a^2)) which means that the vector
r(theta, phi)=2asintheta cosphi i+2a sinphi sintheta j+ 2a cosphi k where 0<theta< 2 pi
and 0<phi<pi.

I am not sure but finding vector n and F(r(theta,phi)). r'(theta,phi) seems very ugly and complex. Is this the right way to do it?

 

[vela]

You're at the point where you just have to grind it out. It's not too bad, actually. You might be able to reason geometrically what the results should be for dS for both surfaces.

For the cylinder, you should find d\mathbf{S} = -a\,d\theta\,dz\,\hat{\mathbf{r}} (where \hat{\mathbf{r}} is for cylindrical coordinates). You may have to put the negative sign in by hand because you want the normal vector pointing out of the volume.

For the sphere, you should find d\mathbf{S} = (2a)^2\sin\phi\,d\theta\,d\phi\,\hat{\mathbf{r}} (where \hat{\mathbf{r}} is for spherical coordinates).