Find the area of the portion of the sphere of radius 2 (centered at the origin) that is in the cone

What I have so far is that:

=

=

Then I have But is wrong, what is wrong?

Thank you.

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- Jul 31st 2009, 08:24 AMtttcomraderSurface area of a sphere
Find the area of the portion of the sphere of radius 2 (centered at the origin) that is in the cone

What I have so far is that:

=

=

Then I have But is wrong, what is wrong?

Thank you. - Jul 31st 2009, 09:03 AMarbolisQuote:

Take it from here, I think all should go well. - Jul 31st 2009, 09:46 AMHaytham
maybe you could try to find the cone generating angle and then calculate the integral with .

no need include anything else in the integral besides - Jul 31st 2009, 11:10 AMJester
- Jul 31st 2009, 12:22 PMOpalg
As

**Danny**says, you should write the equation of the sphere as . To find the region over which to integrate, you need to know where the sphere meets the cone, namely where or in other words . So the integral will be over the region , and the surface area is given by the integral . This cries out for conversion to polar coordinates, and it then becomes . From there on, it's easy. - Aug 2nd 2009, 11:12 AMalan collinsWhat about a shpere with radius 5 in the same question?
What about if the sphere were radius 5? I seem to be goofing it up just a little somewhere when converting to polar/spherical coordinates. Also, it seems as though the Jacobian should be (p^2)sin(phi) instead of just r because this should be changed to polar and not cylindrical coordinates... right? Any feedback, insight or explanation of this integral problem would be fantastic as I am seriously stuck and puzzled by this problem. Thanks.

- Aug 2nd 2009, 04:02 PMHaytham
I think it would be easier to do the all thing in spherical coordinates.

Notice that if the height of the cone is*z*(a generic point z), and the radius of the base is , so the generating angle for the cone, , is given by

as , and

this makes the integral for the area

it seems much more simpler to me, and no need to change the coordinates in between. the R being 2 or 5 makes no difference you just replace it in the integral as a constant.