Surface Integrals

Would someone mind working this out? I know how to do it but i keep getting the different answer to the answer sheet.

thanks.

anyone?

I guess we all do mind working it out. Why don't you post your working and what the answer on the answer sheet is?

Reply 3

Plutoniummatt

Here it is, textbook answer is on the last page:

Reply 4

MC REN

Plutoniummatt

...

Erm you have F in cartesians, so just use n = (1,1,1)

EDIT/ you might need n = (x, y, z) / a or something actually, but i'm being stupid can't think exactly what it will be

Reply 5

Phynance

n is actually the unit normal vector.. So it has to be
n/|n|

(where n is the normal vector and |n| the magnitude of n)

n should be the normal vector to the sphere in other words to:
x^2+y^2+z^2 = a^2

Also, when working in 3D spherical coords the integration is not just r^2dthetadphi but r^2sintheta dthetadphi

Those are the errors that can be seen directly.

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Anyway I think the error is that you did not use r^2sintheta as integration element.. (You did not do anything wrong with the normal unit vector - that was a reply to the previous poster - the edit of his message is correct though)

Reply 6

miggeros

I think that thinking about the normal vector a bit might help; as stated since is normal to sphere must be in direction of (x,y,z) as you should be able to see by geometry of situation. To make it normal you need to divide by something (which will be related in a hopefully pretty obvious fashion to the radius of the sphere you integrating over)

Also, this may or may not be helpful, but after you have fixed the errors it might be profitable to note that symmetry implies that

\int_{S} x^3 dS = \int_{S} y^3 dS = \int_{S} z^3 dS

allowing you to avoid some horrible trig algebra by just doing the easiest of the above (getting you to notice this may well have been the "point" of the question)