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?
xory..ans. (pi/2)-1..typo error
- rotchmLv 74 months agoFavorite Answer
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.
- jacob sLv 74 months ago
Given upper hemisphere x² + y² + z² =1
z= √(1- x² - y²) = √(1- r²)
in one loop area = ∫ ∫ √(1 +(∂z/∂x)² + (∂z/∂y)² )dxdy
Area = ∫ ∫ 1/√(1-r² )rdrdθ
= ∫. (r²-1) dθ = π/2−1
- π/2 0