Surface integral - Wikipedia, the free encyclopedia
The surface in question is the upper half of an ellipsoid (only the positive part of the z axis).
Can somebody please explain how to do the following question, I am struggling to get started and also don't know why it is asking for an area as opposed to a volume.
1.(i). Calculate the area of the ellipsoid (x^2/4) +(y^2/4)+z^2=1 that lies above the xy plane using the parameterisation:
r(θ,φ)=(2cosφsinθ, 2sinφsinθ, cosθ).
Surface integral - Wikipedia, the free encyclopedia
The surface in question is the upper half of an ellipsoid (only the positive part of the z axis).
If my calculations are correct
4sin^4(t)+16cos^2(t)sin^2(t) = 4sin^2(t)[sin^2(t)+4cos^2(t)]
= 4sin^2(t)[1+3cos^2(t)]
taking sqrt roots
=2sin(t)[ 1 +3cos^2(t) ]^1/2
letting u = cos(t) you obtain
int(1+3u^2)^1/2du
Use the trig sub u = 1/[3^(1/2)]tan(a) 3^(1/2)du = sec^2(a)da
You get an integral of the form sec^3(a)
whose AD is 1/2[ln[sec(a)+tan(a)] +sec(a)tan(a)]