Find the area of the portion of the surface

Question

Find the area of the portion of the surface 𝑥𝑦 = 𝑎𝑧 + 𝑏𝑧 in the first octant that lies in the cylinder 𝑥^2 + 𝑦^2 − 𝑏^2 = 𝑎^2. Assume that 𝑎, 𝑏 > 0.

Answer 1

Hi Ayaka,

Thanks for your question!

To solve this problem, we need to use the formula for the surface area of a function z = f(x,y) defined over a region R of the xy-plane:

S = ∫∫_R(f_x²+f_y²+1)^1/2dA.

In the formula above, subscripts indicate partial derivatives and dA indicates a small element of area of the region R.

For this problem, we are told f(x,y) = xy/(a+b). Therefore, f_x = y/(a+b) and f_y = x/(a+b), so our surface area formula becomes

S = ∫∫_R((y/(a+b))²+(x/(a+b))²+1)^1/2dA,

= 1/(a+b)∫∫_R((x²+y²+(a+b)²)^1/2dA.

We are told that our surface f must remain in the first octant of 3D space and that it must remain within the cylinder x²+y² = a²+b². Consequently, the region R over which f is defined must remain in the first quadrant of the xy-plane and remain inside the circle x²+y² = a²+b².

Our region R is best described in polar coordinates (r,θ) instead of rectangular coordinates (x,y). In particular, this region is described by (r,θ) satisfying 0 < r < (a²+ b²)^1/2 and 0< θ <π/2. As a result, we express the integral representing our surface area in polar coordinates:

S = 1/(a+b)∫₀^π/2∫₀^{(a^2+b^2)^(1/2)}((r²+(a+b)²)^1/2rdrdθ,

where we have used that x²+y² = r² and that dA = rdrdθ.

We now can solve our iterated integral. The inner integral can be solved by u-substitution with u = r² + (a+b)². After some work, one finds

S = 1/(a+b)∫₀^π/21/3[(a²+b²+(a+b)²)^3/2 - (a+b)³]dθ.

The outer integral is even simpler, as the integrand is effectively constant with respect to θ. We find

S = 1/3[(a²+b²+(a+b)²)^3/2 - (a+b)³]/(a+b)∫₀^π/2dθ,

= π/6[(a²+b²+(a+b)²)^3/2 - (a+b)³]/(a+b),

= π/6[(a²+b²+(a+b)²)^3/2/(a+b) - (a+b)²].