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Math Help - Divergence Theorem Help!

  1. #1
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    Divergence Theorem Help!

    Here is the promblem

     F(x,y,z) = xz i + yz j + 3z^2 k
     E\ is\ the\ solid\ bounded\ by\ the\ paraboloid\ z\ = x^2 + y^2\ and\ the\ plane\ z = 1
    \int\int F \cdot dS = ?

    Divergence Theorem which i really don't understand well is  \int\int F \cdot dS  = \int\int\int div\ F\ dV

    I went ahead and got the gradient vector which is <z,z,6z> and then the function i got when i set up the triple integral is  8\int\int\int z dV

    and i guess all that is left for me to do is to set up and integrate. Math experts am im right up until this part, and if i am help me set up the triple integral. I'm assuming i have to switch to polar coordinates, i hope i don't have to switch to spherical or cylindrical coordinates, because i really don't understand that stuff very well!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by purplerain View Post
    Here is the promblem

     F(x,y,z) = xz i + yz j + 3z^2 k
     E\ is\ the\ solid\ bounded\ by\ the\ paraboloid\ z\ = x^2 + y^2\ and\ the\ plane\ z = 1
    \int\int F \cdot dS = ?

    Divergence Theorem which i really don't understand well is  \int\int F \cdot dS  = \int\int\int div\ F\ dV

    I went ahead and got the gradient vector which is <z,z,6z> and then the function i got when i set up the triple integral is  8\int\int\int z dV

    and i guess all that is left for me to do is to set up and integrate. Math experts am im right up until this part, and if i am help me set up the triple integral. I'm assuming i have to switch to polar coordinates, i hope i don't have to switch to spherical or cylindrical coordinates, because i really don't understand that stuff very well!
    By the divergence theorem, the integral you seek is \int \int \int \text{div}\bold F ~dV = \int_{-1}^1 \int_{- \sqrt {1 - x^2}}^{\sqrt{1 - x^2}} \int_{x^2 + y^2}^1 (z + z + 6z)~dz~dy~dx = \int_0^{2 \pi} \int_0^1 \int_{r^2}^1 8z \cdot r ~dz~dr~d \theta
    Last edited by Jhevon; December 4th 2009 at 08:59 AM. Reason: Made a mistake
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    By the divergence theorem, the integral you seek is \int \int \int \text{div}\bold F ~dV = \int_{-1}^1 \int_{- \sqrt {1 - x^2}}^{\sqrt{1 - x^2}} \int_{x^2 + y^2}^1 (z + z + 6z)~dz~dy~dx = \int_0^{2 \pi} \int_0^1 \int_{r^2}^1 8r^2 \cdot r ~dz~dr~d \theta
    I tried evaluating it with the bounds you set up and i got 4.18879 which was marked wrong . Im kind of confused on how you got  8r^2 when you set it up.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by purplerain View Post
    I tried evaluating it with the bounds you set up and i got 4.18879 which was marked wrong . Im kind of confused on how you got  8r^2 when you set it up.
    I'm sorry. I put z = x^2 + y^2 = r^2, but you should leave it as z. My bad. Try it again

    and try to find the exact answer rather than a decimal approximation.
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    I'm sorry. I put z = x^2 + y^2 = r^2, but you should leave it as z. My bad. Try it again

    and try to find the exact answer rather than a decimal approximation.
    Thanks, it worked!
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