Question

@slaw could use your help again, its more theoretical this time. I have the parametrization but I need help with the domain again. Find the surface area of sphere x^2+y^2+z^2=b^2 above x^2 +y^2 = a^2 where 0 12 years ago

Answer 1

I got the parametrization
\[r(\phi,\theta) = <bsin \phi \cos \theta, bsin \phi \cos \theta, bcos \phi>\]

Answer 2

well he signed off, can anyone else help me out here?

Answer 3

Hold on. Be patient

Answer 4

I am, just wanted to let you guys know he signed off, thats all. My apologies.

Answer 5

\[
\int _0^{2 \pi }\int _0^ub^2 \sin (\phi )d\phi d\phi =-2 \pi b^2 (\cos
(u)-1)
\]
Where u is the angle that you obtain by joining one of the intersection point of the sphere and the cylinder to the center and the z-axis

Answer 6

\[
\cos (u)=\frac{\sqrt{b^2-a^2}}{a}
\]

Answer 7

thats with \[d \phi d \theta\]
right?

Answer 8

yes, sorry

Answer 9

Wait a minute.

Answer 10

no worries, I appreciate the help. I'll wait

Answer 11

That is fine the answer is still the same it should be
\[
\int _0^{2 \pi }\int _0^ub^2 \sin (\phi )d\phi d\theta =-2 \pi b^2 (\cos
(u)-1)
\]

Answer 12

cos(u) is computed above.

Answer 13

may I ask why it is from 0 to u and not from u to pi/2?, I thought it would be since it was above the cylinder.

Answer 14

Final Answer must look like
\[
\int _0^{2 \pi }\int _0^ub^2 \sin (\phi )d\phi d\theta =2 \pi b^2
\left(1-\frac{\sqrt{b^2-a^2}}{a}\right)
\]

Answer 15

If it is from 0 to pi, you get the whole sphere. To get the area above the cylinder, you should let \(\phi \) varies from 0 to the angle where the cylinder touches the sphere.

Answer 16

That is the angle u

Answer 17

Do you understand?

Answer 18

I'm just trying to figure out the way you find cos(u), what did you look at? the point?

Answer 19

I am drawing a picture for you. Hold on.

Answer 20

I think I had my clyinder around the wrong axis. Thanks.

Answer 21

I understand now

Answer 22

YW