Find the surface area of the part of the sphere x2+y2+z2=81 that lies above the cone z=√(x2+y2)
4 answers
Find the surface area of the part of the sphere x^2+y^2+z^2=81 that lies above the cone z=√(x^2+y^2)
How do i set up the integral?
the cone intersects the sphere where
x^2+y^2+(x^2+y^2) = 81
x^2+y^2 = 81/2
that is a circle of radius 9/√2 in the plane z=9/√2
That means the spherical cap has a height of 9(1-1/√2). There are many sites which can give you the formula for that.
As for the integral, it looks like you want
4∫[0,9/√2] ∫[0,√(81/2-x^2) dS
Check for the surface area element.
You could also use spherical coordinates which would look less complicated.
x^2+y^2+(x^2+y^2) = 81
x^2+y^2 = 81/2
that is a circle of radius 9/√2 in the plane z=9/√2
That means the spherical cap has a height of 9(1-1/√2). There are many sites which can give you the formula for that.
As for the integral, it looks like you want
4∫[0,9/√2] ∫[0,√(81/2-x^2) dS
Check for the surface area element.
You could also use spherical coordinates which would look less complicated.
A more general solution is found below; you can plug in your numbers.
http://www.freemathhelp.com/forum/threads/70982-surface-integrals-spherical-cap
http://www.freemathhelp.com/forum/threads/70982-surface-integrals-spherical-cap
0 answers