Question

find the surface area of the part of the sphere x^2+y^2+z^2=9 that lies above the plane z=2
I found 21V21pi/6-pi/6, but I am not sure.

Answer 1

so thats a sphere with radius 3 centered at origin

Answer 2

yeah, is the answer right?

Answer 3

Probably easiest to do this in spherical coordinates.

Answer 4

or polar coordinates maybe

Answer 5

phi goes from 0 to arccos(2/3)
theta goes from 0 to 2pi

Answer 6

\[\large
\int_0^{\arccos(2/3)}\int_0^{2\pi}r^2\sin(\phi)d\phi d\theta
\]

Answer 7

here is polar if you want to compare

http://www.wolframalpha.com/input/?i=%5Cint%5Climits_%7B0%7D%5E%7B2%5Cpi%7D%5Cint%5Climits_%7B0%7D%5E%7B%5Csqrt%7B5%7D%7D%283r%2Fsqrt%289-r%5E2%29+%29+drd%5Ctheta

Answer 8

thank you so much aum and ganeshie8. the hardest part for me was to write an equation. I got it. I can do the rest.
thanks.

Answer 9

see
http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx

Answer 10

I am getting \(\large 6\pi\) as the surface area.

Answer 11

same! same !

Answer 12

by polar coordinates, i found different boundaries. dr, 0 to V3 and dØ, 0 to 2 pi.

Answer 13

d delta, 0 to 2 pi

Answer 14

I see where you're doing the mistake

Answer 15

you need to look at the shadow in xy plane

Answer 16

you need to look at the shadow in xy plane of the surface u want

Answer 17

is that x^2+y^2=5?

Answer 18

exactly !

Answer 19

it should be :

x^2 + y^2 <= 5

Answer 20

its a disk

Answer 21

thats the region over which you need to evaluate teh double integral

Answer 22

so, my answer was right? 21V21pi/6-pi/6

Answer 23

you should get 6pi
i don't understand V's in your answer

Answer 24

it is square root of 5 and I took the boundaries, dr, 0 to squared root of 5 and d(deta) 0 to 2pi. Am I doing something wrong?

Answer 25

I found where I made the mistake. I found 6pi, too.

Answer 26

thank you so much, ganeshie8. I appreciate you for helping me. Thank you.