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anonymous

  • 5 years ago

Evaluate the surface integral ∫∫ x^2 * z dσ S where the S is the part of the cylinder x^2 + z^2 =1 that is above the xy plane and between the planes y=0 and y=2. Can anyone help me?

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  1. anonymous
    • 5 years ago
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    Most of the questions i have to work with as examples use the double integral over σ and the dσ is dS... thats the first thing i am confused with... the second thing is that i started and thought to write the equation of the cylinder in terms of x so i could let that equal to g(y,z) but since there is no y in either function I dont know how to integrate over dydz

  2. nikvist
    • 5 years ago
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    \[x=\cos\phi,z=\sin\phi,d\sigma=d\phi dy\] \[\iint\limits_Sx^2zd\sigma=\iint\limits_S\cos^2\phi\sin\phi d\phi dy= \int\limits_0^\pi\cos^2\phi\sin\phi d\phi\cdot\int\limits_0^2dy=\] \[=-2\int\limits_0^\pi\cos^2\phi\,d(\cos\phi)=-2\frac{cos^3\phi}{3}_{\phi=0}^{\phi=\pi}=\frac{4}{3}\]

  3. anonymous
    • 5 years ago
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    the only problem with what you've written here is that you havent taken the g(x,y) and the partial derivatives of g with respect to x and y. the surface integral is supposed to look something like\[∫∫ f(x,y,z)*dS = ∫∫ f(x,y,z)*\sqrt{1+(∂x/∂y)^2+(∂x/∂z)^2}dA\]

  4. anonymous
    • 5 years ago
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    Here is how I started..... let x=\[\sqrt{1-z^2}\y and ∂x/∂y = 0 and ∂x/∂z=\[∂x/∂z = -z/\sqrt{1-z^2}\] so that my surface integral becomes \[∫∫ x^2*z*dS\] = \[[∫∫(1-z^2)*z*(1/\sqrt{1-z^2) }] = \] \[∫∫(1-z^2)^0.5\] I'm stuck here though because I would think the dA becomes dz*dy and I would integrate first with respect to z from 0 to 1 but then how to i integrate with respect to y if i dont have anything to find the range from

  5. anonymous
    • 5 years ago
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    that first line that got cut off should say ∂x/∂z = \[(-z)/(1-z^2)^0.5\]

  6. nikvist
    • 5 years ago
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    Best way is \[z=\sqrt{1-x^2}\,,\frac{\partial z}{\partial x}=\frac{-x}{\sqrt{1-x^2}}\,,\,\frac{\partial z}{\partial y}=0\] \[\iint\limits_Sx^2zd\sigma=\iint x^2\sqrt{1-x^2}\sqrt{1+\frac{x^2}{1-x^2}}dxdy= =\iint x^2dxdy=\] \[=\int\limits_{-1}^{1}x^2dx\cdot\int\limits_{0}^{2}dy=\frac{2}{3}\cdot 2=\frac{4}{3}\]

  7. anonymous
    • 5 years ago
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    so i was on the right track but i should have done the x^2 + z^2 = 1 in terms of z instead?! is there a reason? also, for the range of y is just from the question since it said between the planes y=0 and y=2 right.? and for x is for when you set \[z=\sqrt{(1-x^2)} \] equal to 0 right and solve for x= +1 and -1

  8. nikvist
    • 5 years ago
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    Find projection of surface S on xy plane, you will get rectangle (-1,0,0) , (1,0,0) , (1,2,0) , (-1,2,0).

  9. anonymous
    • 5 years ago
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    thank you alot

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