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Line integral to determine area of sphere?

  • Thread starter cyt91
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Find the area of the surface consisting of the part of the sphere of radius 2 centered at
origin that lies above the horizontal plane z = 1. (Equation of this sphere is given by
x^2 + y^2 + z^2 = 2^2 .)

x^2+y^2+1=4
x^2+y^2=3

This is the base of the solid. But how do we find the required surface area of the sphere?
How do we use line integral to determine this area? It's not an area that is parallel to the z-axis.
 

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  • #2
LCKurtz
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Doesn't your text have a formula for the surface area element dS for z = f(x,y)?
[tex]dS=\sqrt{1 + f_x^2+f_y^2}\ dxdy[/tex]
or parametric versions for cylindrical and spherical coordinates?
 
  • #3
HallsofIvy
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I don't understand your remark about the base not being parallel to the z- axis. Of course, it isn't- it lies in the plane z= 1 and so is perpendicular to the z-axis.

I would do this by writing parametric equations for the sphere using spherical coordinates. The surface can be written as [itex]x= 2cos(\theta)sin(\phi)[/itex], [itex]y= 2sin(\theta)sin(\phi)[/itex], [itex]z= 2cos(\phi)[/itex]. [itex]z= 2cos(\phi)= 1[/itex] gives [itex]cos(\phi)= 1/2[/itex] so [itex]\phi= \pi/3[/itex].

The vector equation for that sphere would be [itex]\vec{r}(\theta, \phi)= 2cos(\theta)sin(\phi)\vec{i}+ 2sin(\theta)sin(\phi)\vec{j}+ 2cos(\phi)\vec{k}[/itex]. The derivative of that with respect to the two parameters gives two tangent vectors to the sphere. The cross product of those two vectors is then perpendicular to the sphere and its length, with [itex]d\theta d\phi[/itex] is the "differential of surface area". Integrate that with [itex]\theta[/itex] from 0 to [itex]2\pi[/itex] and [itex]\phi[/itex] from 0 to [itex]\pi/3[/itex].
 

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