Calculate $\int \int \vec{F} \cdot d\vec{S}$ over the section of surface surface is $x = u^2$, $y = uv$, $z = \frac{1}{2} v^2$ bounded by the curves $u = 0$, $u = 1$, $v = 0$, and $v = 3$, for the vector field $\vec{F} = y \vec{i} - x \vec{j} + xy \vec{k}$.
Looks like a single curved plane.
So now we need a unit normal vector to this surface.
$\vec{n} = \frac{\frac{\partial R}{\partial u} \times \frac{\partial R}{\partial v}}{\left | \frac{\partial R}{\partial u} \times \frac{\partial R}{\partial v} \right |}$
$\frac{\partial R}{\partial u} = \left < v , -2u, 3 u^2 v \right >$
$\frac{\partial R}{\partial v} = \left < u , 0, u^3 \right >$
$\frac{\partial R}{\partial u} \times \frac{\partial R}{\partial v} = \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\\ v & -2u & 3u^2v \\\ u & 0 & u^3 \end{pmatrix} = -2u^4 \vec{i} - (u^3v - 3u^3v) \vec{j} + 2u^2 \vec{k} = -2u^4 \vec{i} + 2u^3v \vec{j} + 2u^2 \vec{k}$
$\vec{n} = \frac{\left < -2 u^4, 2u^3 v, 2u^2 \right >}{\left | \frac{\partial R}{\partial u} \times \frac{\partial R}{\partial v} \right | }$
$\int_S (\vec{F} \cdot \vec{n}) dS = \int_0^3 \int_0^1 \left (\vec{F} \cdot \frac{\left < -2 u^4, 2u^3 v, 2u^2 \right >}{\left | \frac{\partial R}{\partial u} \times \frac{\partial R}{\partial v} \right | } \right ) \left | \frac{\partial R}{\partial u} \times \frac{\partial R}{\partial v} \right | du \, dv$
$ = \int_0^3 \int_0^1 \left ( \left < uv, -u^2, u^3 v\right > \cdot \left < -2 u^4, 2u^3 v, 2u^2 \right > \right ) du \, dv = \int_0^3 \int_0^1 ( -2 u^5 v - 2u^5 v + 2u^5v) du \, dv= \int_0^3 \int_0^1 -2 u^5 v du \, dv$
$ = \int_0^3 \left [ - \frac{1}{3} u^6 v \right ]_0^1 dv $
$ = \int_0^3 - \frac{1}{3} v \, dv = - \frac{1}{6} v^2 \bigr |_0^3 = - \frac{3}{2}$.
Looks like I picked the normal vector going the other way, so the answer is $\frac{3}{2}$.
Here is the problem: answer key says the answer is $13 \frac{7}{8}$! Where did I go wrong?
Oooops! You have done a serious mistake. :)
You have taken the derivatives for ${\bf{F}}$ not ${\bf{R}}$. As the question mentioned we have
$${\bf{R}} = {u^2}{\bf{i}} + uv{\bf{j}} + {1 \over 2}{v^2}{\bf{k}}$$
and hence
$$\eqalign{ & {{\partial {\bf{R}}} \over {\partial u}} = 2u{\bf{i}} + v{\bf{j}} \cr & {{\partial {\bf{R}}} \over {\partial v}} = u{\bf{j}} + v{\bf{k}} \cr} $$