Find the surface area of x = y^2 + z^2 inside y^2 + z^2 = 9?

Answer 1

From the equations x= 9 and r²= 9 hence x= r² so the vector s = < r², rcosθ, rsinθ> taking the partials wrt r and θ gives s_r = < 2r, cosθ, sinθ> and s_θ =< 0,-rsinθ,rcosθ> and s_r x s_θ = <r, -2r²cosθ, -2r²sinθ>. So now the area is the integral ∫₀^2π∫₀³ || s_r xs_θ|| drdθ which equals ∫₀^2π∫₀³ √(r² +4r⁴cos²θ + 4r⁴sin²θ)drdθ. This simplifies to the integral ∫₀^2π∫₀³ √(r² +4r⁴)drdθ which becomes ∫₀^2π∫₀³ r√(1+4r²)drdθ. Doing the integration gives 2π∫₀³ √(1+4r²)dr which when using a u substitution gives the surface area as π/6((37)^3/2 -1).