Question

Find the surface area of the part of the sphere x^2+y^2+z^2=64 that lies above the cone z=sqrt(x^2+y^2)


I converted it to polar form and got the double integral of 64/(sqrt(64-r^2) , however I am not sure if that is correct nor am I sure if my limits of integration are correct. Please help!

Answer 1

x^2+y^2 = 32

Answer 2

by above the cone, they mean the surface area above the intersection of the cone and sphere right?

Answer 3

Yes, the projection of the shape, that is correct.

Answer 4

And Darwin, I got that for my limits of integration for r. from 0-sqrt(32)

Answer 5

if so, then when x = 0, y = 8

to me that tells me that everything above x=8 and y=8 gets included ... just my thought process so far

Answer 6

id go about it as taking a slice from the xy plane from y = |x| and the graph of x^2 + y^2 = 64

Answer 7

wow this thing is slow ....

Answer 8

ok I'll try it. Thanks a lot! Hard to get a reply with these types of questions.

Answer 9

one thing though is my teacher told me it is very difficult to solve without converting to polar coordinates. What do you think?

Answer 10

polar is fine, just getting a good idea about how it looks

Answer 11

i think we could take this slice of the surface area, and spin it around the y axis to get the results

Answer 12

an area covered by the cone on top of the sphere x^2+y^2=32

Answer 13

polar would be the easiest sul