Multivariable surface integral

[thiskidistite]

a) Find the area of the part of the surface S = {x^2+ y^2+ (z-1)^2 = 4, 0 ≤ z ≤ 1}.

Note that this is part of the sphere of radius 2 with center (0,0,1).

 

[Pseudo Statistic]

Oh noes, a multivariable surface integral!
Well.. the coolest way (in my opinion) to do a question like this is to use spherical co-ordinate parametrization..
So try the substitution:
z-1 = 2 cos phi
x = 2 cos theta sin phi
y = 2 sin theta sin phi
(Since r = 2 in these cases)
Think you can work from there?
This should be in the homework help forum, by the way.

 

[thiskidistite]

so what should the limits of phi be?? I did it by using polar coordinates but I still got the wrong answer
Thanks for your help though

 

[Pseudo Statistic]

You're dealing with a sphere-- so you should be using spherical co-ordinates! (Polar would be too time-consuming)
Obviously we're dealing with the top half, 0<z<1, thus we have the limits:
0<theta<2pi
0<phi<pi/2
Since you want the surface area of the thing, you want:
$$\int_{S} dS = \int_{0}^{\pi/2} \int_{0}^{2\pi} ||r_{\theta} \times r_{\phi}|| d\theta d\phi$$
Where r is the position vector corresponding to the parametrization I provided above and subscripts denote the respective partial derivatives.

 

[HallsofIvy]

a) Find the area of the part of the surface S = {x^2+ y^2+ (z-1)^2 = 4, 0 ≤ z ≤ 1}.

Note that this is part of the sphere of radius 2 with center (0,0,1).

The surface area of a sphere of radius r is $4\pi r^2$. What is the area of half of a sphere of radius 2?