# Evaluate Flux integral

• Dec 6th 2009, 06:14 PM
Evaluate Flux integral
Let S be the portion of the sphere: x^2 + y^2 + z^2 = 9 which lies between the planes: z = 1 and z = 2 , oriented by the unit normal vector N which points away from the origin. Evaluate the flux integral:

$\displaystyle \int\int_S \mathbf{F}\cdot\mathbf{N}\,dS$ , where F is the vector field given by: F (x, y, z) = ( x + y z^(3) )i + x z^(3)j - z k
• Dec 6th 2009, 06:25 PM
TheEmptySet
Quote:

Let S be the portion of the sphere: x^2 + y^2 + z^2 = 9 which lies between the planes: z = 1 and z = 2 , oriented by the unit normal vector N which points away from the origin. Evaluate the flux integral:

$\displaystyle \int\int_S \mathbf{F}\cdot\mathbf{N}\,dS$ , where F is the vector field given by: F (x, y, z) = ( x + y z^(3) )i + x z^(3)j - z k

Two questions.

First do you mean the closed surface? i.e are both of the planes part of the surface

Second, do you know the divergence Theorem?
• Dec 6th 2009, 06:33 PM
Thanks EmptySet
I am pretty sure z=1 and z=2 are part of the surface, but I am not sure. I don't know the divergence theorem, I'm soo lost. I'm barely hanging on to my B in the course.
• Dec 6th 2009, 08:14 PM
TheEmptySet
Quote:

Let S be the portion of the sphere: x^2 + y^2 + z^2 = 9 which lies between the planes: z = 1 and z = 2 , oriented by the unit normal vector N which points away from the origin. Evaluate the flux integral:

$\displaystyle \int\int_S \mathbf{F}\cdot\mathbf{N}\,dS$ , where F is the vector field given by: F (x, y, z) = ( x + y z^(3) )i + x z^(3)j - z k

We can paramterize the surface by

$\displaystyle \vec r(u,v)=< 3\sin(v)\cos(u) ,3\sin(v)\sin(u),3\cos(v)>$

$\displaystyle \vec r_u(u,v)=<-3\sin(v)\sin(u),3\sin(v)\cos(u),0>$

$\displaystyle \vec r_v(u,v)=<3\cos(v)\cos(u),3\cos(v)\sin(u),-3\sin(v)>$

Note that u ranges from 0 to $\displaystyle 2\pi$

and v $\displaystyle 2=3\cos(v_0) \iff v_0 = \cos^{-1}\left( \frac{2}{3}\right)$

$\displaystyle 1=3\cos(v_1) \iff v_1 = \cos^{-1}\left( \frac{1}{3}\right)$

Now

$\displaystyle \iint_S \vec{F} \cdot d\vec{S}=\int_{0}^{2\pi} \int_{\cos^{-1}(1/3)}^{\cos^{-1}(2/3)}\vec{F}\cdot (r_u \times r_v)dudv$

This should get you started.
• Dec 6th 2009, 10:06 PM
Quote:

Originally Posted by TheEmptySet
We can paramterize the surface by

$\displaystyle \vec r(u,v)=< 3\sin(v)\cos(u) ,3\sin(v)\sin(u),3\cos(v)>$

$\displaystyle \vec r_u(u,v)=<-3\sin(v)\sin(u),3\sin(v)\cos(u),0>$

$\displaystyle \vec r_v(u,v)=<3\cos(v)\cos(u),3\cos(v)\sin(u),-3\sin(v)>$

Note that u ranges from 0 to $\displaystyle 2\pi$

and v $\displaystyle 2=3\cos(v_0) \iff v_0 = \cos^{-1}\left( \frac{2}{3}\right)$

$\displaystyle 1=3\cos(v_1) \iff v_1 = \cos^{-1}\left( \frac{1}{3}\right)$

Now

$\displaystyle \iint_S \vec{F} \cdot d\vec{S}=\int_{0}^{2\pi} \int_{\cos^{-1}(1/3)}^{\cos^{-1}(2/3)}\vec{F}\cdot (r_u \times r_v)dudv$

This should get you started.

I totally understand until the last part. So in order to solve for this part "$\displaystyle (r_u \times r_v)dudv$" , all I have to do is, solve the cross product and integrate? You do mean the cross product right? Thanks soo much in taking ur time to help me out, it means soo soo much to me :)
• Dec 6th 2009, 10:59 PM
scorpion007
Yes, cross product. $\displaystyle \mathbf{\hat n}dS=(r_u\times r_v) du dv$
• Dec 6th 2009, 11:04 PM
Quote:

Originally Posted by scorpion007
Yes, cross product. $\displaystyle \mathbf{\hat n}dS=(r_u\times r_v) du dv$

I see, I don't know how to properly notate the cross product but I got -9sin(v)sin(u) as the cross product. I'm pretty sure this is not correct. Can you please show me the correct way? This is <<<<HARD>>>> :(
• Dec 6th 2009, 11:06 PM
scorpion007
The cross product is a vector. Not a scalar.
• Dec 6th 2009, 11:23 PM
Quote:

Originally Posted by scorpion007
The cross product is a vector. Not a scalar.

Oh gosh, I'm really lost.....
• Dec 6th 2009, 11:28 PM
scorpion007
Presumably you would have studied cross products before vector calculus?

In any case:

$\displaystyle r_u \times r_v = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ -3\sin(v)\sin(u)&3\sin(v)\cos(u)&0 \\3\cos(v)\cos(u)&3\cos(v)\sin(u)&-3\sin(v) \end{vmatrix}$

Do you know how to take this determinant?
• Dec 6th 2009, 11:38 PM
Quote:

Originally Posted by scorpion007
Presumably you would have studied cross products before vector calculus?

In any case:

$\displaystyle r_u \times r_v = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ -3\sin(v)\sin(u)&3\sin(v)\cos(u)&0 \\3\cos(v)\cos(u)&3\cos(v)\sin(u)&-3\sin(v) \end{vmatrix}$

Do you know how to take this determinant?

We did study cross products but very lightly, our professor expected us to know it on our own. But I never really got the hang of them. I don't know how to take the determinant, I would be really grateful if you can show me :)
• Dec 7th 2009, 12:21 AM
scorpion007
Did you learn cofactor expansion to find determinants? I suggest you learn that soon. You will need it many times in vector calculus.

$\displaystyle \mathbf{n}dS = (r_u \times r_v)dudv = <-9\sin^2{v}\cos{u},-9\sin^2{v}\sin{u},-9\sin{v}\cos{v}>dudv$

Can you now find $\displaystyle \mathbf{F}\cdot \mathbf{n}dS$?
• Dec 7th 2009, 12:44 AM
Quote:

Originally Posted by scorpion007
Did you learn cofactor expansion to find determinants? I suggest you learn that soon. You will need it many times in vector calculus.

$\displaystyle \mathbf{n}dS = (r_u \times r_v)dudv = <-9\sin^2{v}\cos{u},-9\sin^2{v}\sin{u},-9\sin{v}\cos{v}>dudv$

Can you now find $\displaystyle \mathbf{F}\cdot \mathbf{n}dS$?

Thanks soo much Scorpion007, I'm studying the cross product right now. Can you show me how to find $\displaystyle \mathbf{F}\cdot \mathbf{n}dS$? Thanks for taking your time to help me, I really appreciate it
• Dec 8th 2009, 02:04 PM
$\displaystyle \mathbf{n}dS = (r_u \times r_v)dudv = <-9\sin^2{v}\cos{u},-9\sin^2{v}\sin{u},-9\sin{v}\cos{v}>dudv$
Can you now find $\displaystyle \mathbf{F}\cdot \mathbf{n}dS$?