Question

Evaluate the surface integral. Sxy dSS is the triangular region with vertices (1, 0, 0), (0, 6, 0), (0, 0, 6)

Answer 1

Parameterize S by

`\vec r(s, t) = (1-s)(1-t)\,\vec\imath + 6s(1-t)\,\vec\jmath + 6t\,\vec k`

with 0 ≤ s ≤ 1 and 0 ≤ t ≤ 1.

Take the normal vector to S to be

`\vec n = (\partial\vec r)/(\partial s)*(\partial\vec r)/(\partial t) = (36-36t)\,\vec\imath+(6-6t)\,\vec\jmath+(6-6t)\,\vec k`

with magnitude

`\|\vec n\| = 6√(38)(1-t)`

Then the surface integral is

`\displaystyle \iint_S xy\,\mathrm dS = 6√(38) \int_0^1\int_0^1 (1-t)\,\mathrm dt\,\mathrm ds = \boxed{3\sqrt{\frac{19}2}}`