Hey, I tried converting co-ordinates and all but I am having difficulties.

Evaluate

where and is the part of the cone beneath the plane with downward orientation.

Thanks for any help.

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- May 13th 2011, 08:49 PMNguyenAnother Surface Integral!
Hey, I tried converting co-ordinates and all but I am having difficulties.

Evaluate

where and is the part of the cone beneath the plane with downward orientation.

Thanks for any help. - May 13th 2011, 09:07 PMTheEmptySet
- May 14th 2011, 06:08 AMHallsofIvy
Or, working with cylindrical coordinates from the start, we can write the position vector of any point on the cone as .

The derivatives give vectors in the tangent plane:

The cross product of those two vectors gives the normal vector

where the order of multiplication has been chose to make the component negative ("with downward orientation").

The "vector differential of surface area" is

Of course, so that just as TheEmptySet says (with the sign changed of course).