Erreix asked in Science & MathematicsMathematics · 4 months ago

# 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

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• rotchm
Lv 7
4 months ago

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