Evaluate the surface integral RR S x dS, where S is the triangular region with vertices (1, 0, 0), (0, −2, 0), and (0, 0, 4). Sketch the surface S and the domain of integration D.

Question

Evaluate the surface integral RR S x dS, where S is the triangular region with vertices (1, 0, 0), (0, −2, 0), and (0, 0, 4). Sketch the surface S and the domain of integration D.

Answer 1

Equation of surface (S) is:x/1 - y/2 + z/4 = 1; From here z = 4 - 4x + 2y

∫∫(S)xdS = ∫∫(D)x(1 + (∂z/∂x)² + (∂z/∂y)²)^1/2dxdy.

Here (D) is projection of S) on xy coordinate plane; so (D) is triangle with vertices (1,0,0), (0, - 2, 0) and (0,0,0). ∂z/∂x = - 4, ∂z/∂y = 2. In (D) 0 ≤ x ≤ 1, - 2 ≤ y ≤ 0.

Now we have ∫∫(S)xdS = ∫∫(D)x(1 + (- 4)² + 2²)^1/2dxdy = √21∫_-2⁰dy∫₀¹xdx = √21y_-2⁰x²/2₀¹ = √21