# Integrating spherical normal components

I am trying to integrate the the normal vector to the surface of a hemisphere but am having some trouble. I am pretty sure that the x and y components will cancel out, and that the z component in spherical coordinates is R*Cos[theta] but for some reason I am having trouble really understanding how to do it.

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

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

Well, can you express the normal vector at (R,theta,phi) in terms of the unit vectors in the x, y, and z directions? Then, you can break up the integral into the sum of 3 integrals, and since the unit vectors are constant, you can factor them out of the integrals.

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

mewmew, by "the normal vector" do you mean the unit normal vector? And are you integrating each component of the vector independently or are you integrating $\vec{v}\cdot d\vec{S}$?

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.

Well, can you express the normal vector at (R,theta,phi) in terms of the unit vectors in the x, y, and z directions? Then, you can break up the integral into the sum of 3 integrals, and since the unit vectors are constant, you can factor them out of the integrals.
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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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?

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