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Math Help - Evaluate Flux integral

  1. #1
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    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:

    \int\int_S <br />
\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
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  2. #2
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    Quote Originally Posted by mohammadmurtaza View Post
    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:

    \int\int_S <br />
\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?
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    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.
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    Behold, the power of SARDINES!
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    Quote Originally Posted by mohammadmurtaza View Post
    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:

    \int\int_S <br />
\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

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

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

    \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 2\pi

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

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

    Now

    \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.
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    We can paramterize the surface by

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

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

    \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 2\pi

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

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

    Now

    \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 " (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
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    Yes, cross product. \mathbf{\hat n}dS=(r_u\times r_v) du dv
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    Quote Originally Posted by scorpion007 View Post
    Yes, cross product. \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>>>>
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  8. #8
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    The cross product is a vector. Not a scalar.
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    Quote Originally Posted by scorpion007 View Post
    The cross product is a vector. Not a scalar.
    Oh gosh, I'm really lost.....
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    Presumably you would have studied cross products before vector calculus?

    In any case:

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

    Do you know how to take this determinant?
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  11. #11
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    Quote Originally Posted by scorpion007 View Post
    Presumably you would have studied cross products before vector calculus?

    In any case:

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

    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
    Last edited by mohammadmurtaza; December 7th 2009 at 12:40 AM. Reason: add comments
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    Did you learn cofactor expansion to find determinants? I suggest you learn that soon. You will need it many times in vector calculus.

    <br />
\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<br />

    Can you now find \mathbf{F}\cdot \mathbf{n}dS?
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  13. #13
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    Quote Originally Posted by scorpion007 View Post
    Did you learn cofactor expansion to find determinants? I suggest you learn that soon. You will need it many times in vector calculus.

    <br />
\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<br />

    Can you now find \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 \mathbf{F}\cdot \mathbf{n}dS? Thanks for taking your time to help me, I really appreciate it
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  14. #14
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    Quote Originally Posted by scorpion007 View Post
    Did you learn cofactor expansion to find determinants? I suggest you learn that soon. You will need it many times in vector calculus.

    <br />
\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<br />

    Can you now find \mathbf{F}\cdot \mathbf{n}dS?
    I see, how can I find the limits of the double integral? Thanx alot
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