Please help with this

Question

Find the area of the surface of the cap, cut from the

hemisphere x ² + y² + z² = 5, z ≥ 0, by the cylinder x² + y² = 1.

Answer 1

Another tutor, Tom, answered using polar substitution and then I added an answer that used spherical substitution and differentiated inside the integral. My answer involves some advanced proofs. Let me try Tom's method and give some explanation to help you follow.

There is a proof, a bit involved involving cross products and scalar products of vectors which concludes

For a surface z=f(x,y), surface area over region R in xy plane

is SA=∫∫√(f_x²+f_y²+1)dxdy integrated with limits defining R

computing

SA=∫∫√(([x²+y²]/[5-x²-y²])+1)dxdy over the circle x²+y²+=1

the circle projected up is the cylinder

note the the squaring of f_x and f_y removes the square root in the denominator

make 1=5-x²-y²/5-x²-y²

then SA=∫∫√[5/(5-x²-y²]dxdy over the circle x²+y²+=1

to convert dxdy=dA to a polar dA

there is a simple proof ,,,,dA=r*drdθ

where as usual for polar substitution ,,,r²=x²+y²,,,,x=rcosθ,,,,y=rsinθ

now SA = ∫∫√[5/(5-r²)]r*drdθ,,,where the region in r,θ plane is r from 0 to 1, θ from 0 to 2π

SA=2π((-√5)/2)*2*(5-r²)^1/2 where r limits are r=0 to r =1,,,,need-1/2 since -2r is differential of -r²

need 2 since integratlon raised power to 1/2 must divide by 1/2

see integration with respect to θ was computed first with limits 0 to 2π

SA=same answer as given by Tom