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.

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So, ∫∫s F · dS

= ∫∫s

= ∫∫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.

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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.

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I hope this helps!