Question

Find the surface area of the portion of the sphere x^2+y^2+z^2=4 which lies above the xy-plane and BELOW the plane z=1.
I just need help with the radius part, please! (i'm using polar coordinates)

Answer 1

Can you help please?

Answer 2

You have to use polar coordinates?

Answer 3

yup

Answer 4

Is radius from 0 to 1?
and theta from 0to 2pi?

Answer 5

No, project the part of the sphere over the plane z=1 onto the xy-plane => x^2+y^2=4-z^2=3. r is from -sqrt(3) to sqrt(3), your theta is correct.

Answer 6

polar coordinates... I figured because it's surface area, It uses double integrals.

Answer 7

Thank you so much!!

Answer 8

Welcome!

Answer 9

wait.. why wouldn't it be 1 to sqrt3 because z =1? or z=1 is just used to plug it in the given equation to find out +-sqrt3

Answer 10

oops, sorry, r would be from 0 to sqrt(3), my bad

Answer 11

oh, as for your question about the limits of integration, since the projection is a circle of radius sqrt(3) on the xy-plane, we integrate from r=0 to r=sqrt(3). The limits of integration you said is correct if the projection is annular with inner radius=1 and outer radius=sqrt(3).

Answer 12

OKay... So r is from 1 to sqrt3?

Answer 13

No

Answer 14

you just said my limits of integration are correct

Answer 15

'The limits of integration you said is correct IF the projection is annular with inner radius=1 and outer radius=sqrt(3). (In this case it's not.)

Answer 16

Oh alright... Thank you!