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I know this is really basic but I'm horrible at vector calculus. Any help is greatly appreciated. Thanks

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- Thread starter mewmew
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- #1

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I know this is really basic but I'm horrible at vector calculus. Any help is greatly appreciated. Thanks

- #2

berkeman

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Why would the x and y components cancel out? Is this a special 3-D situation?

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- #4

HallsofIvy

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Yes, it's the upper unit hemisphere! For every vector xi+ yj+ zk there will be a vector -xi- yj+ zk.berkeman said:Why would the x and y components cancel out? Is this a special 3-D situation?

mewmew, by "the normal vector" do you mean the

- #5

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Sorry, my terminology isn't that great. Ultimately I am trying to find < n_z >, which I believe is n_z integrated over my surface, divided by the area. Also "n_z" is the unit normal vector in the z direction, I tried to do it in tex but was having problems.

Since the unit vectors come out of the integrals then the integrals should just give me the surface area, correct? So <n_z> would just be the unite vector in the z direction? I am probably way off basis but my intuition tells me that would be the case...although it is wrong quite a lot!

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- #6

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No, n_z should be the portion of the unit normal in the z direction, right? The unit normal at a specified point on the surface can be written as n = n_x i + n_y j + n_z k, where n_x^2 + n_y^2 + n_z^2 = 1 (it's a unit normal!).

Suppose we were just dealing with a circle in polar coordinates. At any point on the circle (r, theta), you can write the unit normal as n = i cos theta + j sin theta, where i is the unit vector in the x direction and j is the unit vector in the y direction. On your hemisphere, you should be able to write a similar formula the normal in terms of i, j, k, theta, and phi. Then, you toss *this* vector into a double integral with appropriate limits (and R^2 sin theta dtheta dphi). Since this is a sum, you can break it up into three integrals, factor out the unit vectors and find (as you expected) the integrals for i and j vanish and you are left with your value for k.

Make sense?

Suppose we were just dealing with a circle in polar coordinates. At any point on the circle (r, theta), you can write the unit normal as n = i cos theta + j sin theta, where i is the unit vector in the x direction and j is the unit vector in the y direction. On your hemisphere, you should be able to write a similar formula the normal in terms of i, j, k, theta, and phi. Then, you toss *this* vector into a double integral with appropriate limits (and R^2 sin theta dtheta dphi). Since this is a sum, you can break it up into three integrals, factor out the unit vectors and find (as you expected) the integrals for i and j vanish and you are left with your value for k.

Make sense?

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