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

$\vec n = \dfrac{\partial\Phi}{\partial u} \times \dfrac{\partial\Phi}{\partial v} = (0,0,2u)$

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

Then the surface integral is

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