Question

Let D = D(R), where Þ(u, v) = (u², u + v) and

R = [1, 8] × [0, 6].
Calculate
y dA.
Note: It is not necessary to describe D.
Joyda
=

Answer 1

The region
`D` is essentially parameterized in three dimensions by

`\Phi(u,v) = (u^2, u+v, 0)`

with
`1\le u\le8` and
`0\le v\le6`.

The normal vector to
`D` is

`\vec n = (\partial\Phi)/(\partial u) * (\partial\Phi)/(\partial v) = (0,0,2u)`

with norm
`\|\vec n\| = 2u`.

Then the surface integral is

`\displaystyle \iint_D y \, dA = 2 \int_0^6 \int_1^8 (u+v)u \, du \, dv \\\\ = 2 \int_0^6 \left(\frac{1022}3 + 63 v\right) \, dv = \boxed{3178}`