1R) 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

Let z= (5-x² -y²)^1/2 so dz/dx - -x/ (5- x² - y²)^1/2 and dz/dy = -y/ (5-x² -y²)^1/2. So A(S) =∫∫((x² +y²)/(5-x² -y²)^1/2 +1)dxdy which becomes A(S)=∫∫(5/(5-x²-y²))^1/2dxdy. For x² + y²=1 r=1 use rdrdθ for the area element. The equation becomes A(S)=(5)^1/2∫₀¹∫₀^2π(1/(5-r²))^1/2rdrdθ =2√5π∫₀¹rdr/(5-r²)^1/2. Let u=5-r² so du=-2rdr and rdr = -du/2 this simplifies integral to A(S)= 2√5π( -1/2) ∫₀¹du/u^1/2. Doing the integration gives A(S)= 2√5π(√5 -2) which gives π(10 -4√5) for the surface area of the cap.

Answer 2

If done with spherical coordinates

should compute to same answer found with rectangular mixed with polar substitution.

spherical is especially nice for spheres, change in radius*Suface Area=

d(rho)*Surface Area is directly proportional to dV

V=∫∫∫dV,,,,,,,,,,,,,,,,,,,,,,,,in Spherical dV=(rho)²sinΦd(rho)dθdΦ

dV/d(rho) at rho,,,,fundamental theorem, = ∫∫(rho)²dθdΦ

dV/d(rho)=Surface Area

where θ is from 0 to 2π and Φ is from 0 to cos^-1(2/√5) and (rho)²=5

note z=(rho)*cos(Φ) and the cylinder and sphere intersect at z=2

working out integral wind up with same answer