Question

Find the area of the portion of the plane 2x+3y+4z=28 lying above the rectangle 1≤x≤4,2≤y≤6 in the xy -plane.

Answer 1

Call this piece of the plane S. Parameterize S by the vector function

s(u, v) = u i + v j + (28 - 2u - 3v)/4 k

with 1 ≤ u ≤ 4 and 2 ≤ v ≤ 6. Take the normal vector to S to be

∂s/∂u x ∂s/∂v = 1/2 i + 3/4 j + k

which has norm √((1/2)^2 + (3/4)^2 + 1^2) = √29/4

The area of S is then

$\displaystyle\iint_S\mathrm dS=\int_1^4\int_2^6\left\|\frac{\partial\mathbf s}{\partial u}\times\frac{\partial\mathbf s}{\partial v}\right\|\,\mathrm dv\,\mathrm du=\frac{\sqrt{29}}4(4-1)(6-2)=\boxed{3\sqrt{29}}$