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Surface Integral Watch

1. The surface, S, is the part of the cone x^2+y^2=z^2, z>0 which lies below the plane x+2z=1

find where the integral is over S.

Not sure how to start this one. Normally I am given the integral as the double integral of something involving ds, where I would then use the definition of ds and parameterise the surface. What do I do in this case?
2. You'll be wanting to change the coordinate system - given that it's a cone, you probably want cylindrical polars. (Remember to multiply by the Jacobian!)
3. (Original post by Smaug123)
You'll be wanting to change the coordinate system - given that it's a cone, you probably want cylindrical polars. (Remember to multiply by the Jacobian!)
But cylindrical polars are in 3 variables when my integral only involves 2.
4. (Original post by james22)
But cylindrical polars are in 3 variables when my integral only involves 2.
Yep, in the same way as cartesians are in 3 variables when the integral only involves 2. The easiest way to do a surface integral is to find a way to parametrise the surface in only two variables, keeping the third constant; while it probably is possible in Cartesians, it's not at all obvious how.
For instance, to parametrise over the surface of a sphere, you'd use spherical polars with r constant; to parametrise over a plane, you'd use Cartesians with n, the normal vector, constant. In the same way, here we'd use cylindrical polars with r constant for a given z.
5. Is 0 correct? I don't know of any easy way of checking it.
6. (Original post by james22)
Is 0 correct? I don't know of any easy way of checking it.
Hmm, OK, sorry, I didn't actually read the question properly - I suspect it's not 0 because the truncated cone is asymmetrical, but I haven't calculated it properly. The only possible avenue of attack I can think of is to use: - but I've got to rush out now. I'll have another look when I get back, unless a higher mind than mine can fix it…

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