Find the surface area of the boundary surface?

Question

Find the surface area of the boundary surface dR of the solid region R consisting of all points (x, y, z) satisfying x^2+z^2<=1 and y^2+z^2<=1.

Answer 1

There are four sides in this figure and all are congruent.

One side is y = √(1-z²) restricted to the domain D = { x^2+z^2 <= 1 }

Compute two partial derivatives of y:

∂y/∂x = 0

∂y/∂z = -z/√(1-z²)

Using four fold symmetry of the side, the area of the side is

∫∫_D √(1+(∂y/∂x)²+(∂y/∂z)²) dz dx

= 4∫₀¹ ∫₀^{√(1-x^2)} √(1 + 0² + z²/(1-z²)) dz dx

= 4∫₀¹ ∫₀^{√(1-x^2)} 1/√(1-z²) dz dx

= 4∫₀¹ sin^-1 √(1-x²) dx

= 4∫₀¹ cos^-1 x dx

= 4∫₀^π/2 cos x dx  ← you can get this by drawing a picture.

= 4sin x |₀^π/2

= 4

So the total area of the surface is four times that, which is 16.