Find the area of the upper hemisphere of x^2+y^2+z^2=1 above the interior of one loop of r=cos(2theta). ans. pi[sqrt(2)]-1?

I drawn it and even checked it on a graphing application. However, I'm kinda lost in where to start. I'm always bothered how to use "z." Can you offer some advice?

Update:

xory..ans. (pi/2)-1..typo error

2 Answers

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  • rotchm
    Lv 7
    4 months ago
    Favorite Answer

    Hints:

    Sketch it to get an idea of the figure.

    For one leaf, above z=0 we may take z = √( 1-x² - y²).

    What are then the ranges for r & θ?

    The element of area is √( 1 +  (dz/dx)² + (dz/dy)² ) dx dy

    Evaluate all this. Then transform into cylindrical coords. Everything will "cancel" and be simplified. Don't forget that the element of area is now r dr dθ

    The sought area is the ∫∫  of that. 

    Show ur steps here and we can then continue if need be. 

  • 4 months ago

     Given upper hemisphere x² + y² + z² =1

    z= √(1-  x² - y²) = √(1- r²)

                                  π/2  cos2θ

    in one loop area =    ∫   ∫          √(1 +(∂z/∂x)² + (∂z/∂y)² )dxdy

                                  -π/2  0

                                  π/2 cos2θ

                               

                   Area         =   ∫ ∫      1/√(1-r² )rdrdθ

                                  -π/2  0

                             

                                      π/2            cos2θ

                                  =     ∫.     (r²-1)  dθ    =     π/2−1   

                                      - π/2          0

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