Question

find the area of the surface. The part of the sphere x^2+y^2+z^2=64 that lies above the plane z=3

Answer 1

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

Answer 2

find the partials and plugin

Answer 3

\(x^2+y^2+z^2=64\)

\(f_x = \dfrac{\partial z}{ \partial x} = \dfrac{-x}{z}\)

\(f_y = \dfrac{\partial z}{ \partial y} = \dfrac{-y}{z}\)

Answer 4

surface area :

\[\iint \limits_R \sqrt{f_x^2 + f_y^2 + 1} ~dA\]

Answer 5

switching to polar may help..

Answer 6

\[\iint \limits_R \sqrt{\frac{x^2}{z^2} + \frac{y^2}{z^2} + 1} ~dA\]

\[\iint \limits_R \sqrt{\frac{x^2+y^2+z^2}{z^2} } ~dA\]

\[\iint \limits_R \sqrt{\frac{64}{z^2} } ~dA\]

\[8\iint \limits_R \dfrac{1}{\sqrt{64-x^2-y^2}} ~dA\]

Answer 7

fine, so far ?

Answer 8

So far I am understanding. Thanks

Answer 9

next setup the bounds

Answer 10

the region of interest is above z = 3 and below the sphere :

64-x^2-y^2<= 3^2

x^2 + y^2 <= 55

Answer 11

so the shadow in xy plane is just a disk of radius sqrt(55),

maybe sketch it roughly if u dont see it yet..

Answer 12

If you do not mind can you guide me all the way to the end of the problem. If you are not to busy please?

Answer 13

sure :) did u get why the shadow in xy plane is a disk of radius sqrt(55) ?

Answer 14

yes and thanks