surface integrals help?

Evaluate the surface integral

Answer 1

1) Let r = .

Differentiating,
r_u = <1, 1, 2>
r_v = <1, -1, 1>.

So, r_u x r_v = <3, 1, -2>.
==> n = <-3, -1, 2> so that the normal points out of S in an upward orientation.
—-
So, ∫∫s F · dS
= ∫∫s · dS
= ∫∫s <(1+2u+v) e^(u^2 - v^2), -3(1+2u+v)e^(u^2 - v^2), u^2 - v^2> · <-3, -1, 2> dA
= ∫(v = 0 to 1) ∫(u = 0 to 2) (2u^2 – 2v^2) du dv
= ∫(v = 0 to 1) (2u^3/3 – 2uv^2) {for u = 0 to 2} dv
= ∫(v = 0 to 1) (16/3 – 4v^2) dv
= (16v/3 – 4v^3/3) {for v = 0 to 1}
= 4.
—————
2) ∫∫s (x + y + z) dS
= ∫∫s (x + y + z) ||r_u x r_v|| dA
= ∫(v = 0 to 1) ∫(u = 0 to 2) [(u+v) + (u-v) + (1+2u+v)] ||<3, 1, -2>|| du dv
= √14 * ∫(v = 0 to 1) ∫(u = 0 to 2) (4u + v + 1) du dv
= √14 * ∫(v = 0 to 1) (2u^2 + uv + u) {for u = 0 to 2} dv
= √14 * ∫(v = 0 to 1) (2v + 10) dv
= √14 * (v^2 + 10v) {for v = 0 to 1}
= 11√14.
—————-
I hope this helps!

Leave a Comment