Vector Calculus Surface Integral (Limits of Integration) I'm currently having trouble with the following problem. I believe that I have most of the problem set up, but I am having trouble finding what the limits of integration should be. 
$\int\limits_S \vec{F} \cdot d\vec{S}$
When, $\vec{F}(x,y) = <3x\sin y, 4\sin y,0>$
and $S$ is the graph of $f(x,y) = \frac{1}{2}(1+x^{2}-y^{2})$, $0\le x,y\le\pi$.
Now I don't know how important the picture is, but all that might be useful is that it is in the positive octant. 
I believe that the solution to this problem would be to calculate $ \vec{F} \cdot d\vec{S}$ for which I just took the gradient of our function $f(x,y)$ and obtained,
$ \vec{F} \cdot d\vec{S} = (3x^{2}\sin y-4y\sin y)dxdy$.
However, I don't know how badly I've butchered this problem. 
Please offer some assistance, and thank you!
 A: You almost got it!
The surface $S$ corresponds to:
$$
(x,y) \rightarrow (x,y,f(x,y))
$$
for $x$ and $y$ in $[0,\pi]$
So let's calculate at first $\mathbf{dS}$:
$$
\begin{array}{lcr}
\mathbf{dS}& = &(1,0,\frac{\partial f}{\partial x}) \times (0,1,\frac{\partial f}{\partial y})\\
& = & (-\frac{\partial f}{\partial x},-\frac{\partial f}{\partial x},1)
\end{array}
$$
Once you have that, you simply take the dot product of $\mathbf{F}$ and $\mathbf{dS}$ and integrate over $[0,\pi]\times[0,\pi]$:
$$
\int_0^\pi{\int_0^\pi{-3x^2\sin y + 4y\sin y~dxdy}}
$$
Hence you just have a problem of sign! But you can figure it out quickly by considering a simpler surface. (Try for example with $f(x,y) = x$ and $\mathbf{F}=(-1,1,0))$).
When your surface is of the form:
$$(x,y)\rightarrow(f_1(x,y),f_2(x,y),f_3(x,y))$$
you can compute $\mathbf{dS}$ at the point $P= (f_1(x,y),f_2(x,y),f_3(x,y))$ by looking for a small displacement in the x direction on the surface, which gives you the vector:
$$\mathbf{u}=(\frac{\partial f_1}{\partial x},\frac{\partial f_2}{\partial x},\frac{\partial f_3}{\partial x})$$
then you look for a small displacement in the y direction on the surface, which gives you the vector:
$$\mathbf{v}=(\frac{\partial f_1}{\partial y},\frac{\partial f_2}{\partial y},\frac{\partial f_3}{\partial y})$$.
These two vectors and the local point $P$ define the local tangent plane of the surface. $\mathbf{dS}$ is nothing more than the normal vector of this local tangent plane, and whose norm is the area of the surface defined by the two small displacement in the x and y direction. Hence you have the simple relationship:
$$\mathbf{dS} = \mathbf{u}\times\mathbf{v}$$
The cross product gives you that $\mathbf{dS}$ has the norm $u\times v$, and that it is orthogonal with $\mathbf{u}$ and $\mathbf{v}$, so it will be orthogonal to the tangent plane. Now you take $\mathbf{u}\times\mathbf{v}$ and not $\mathbf{v}\times\mathbf{u}$ because I assume you are working in the standard right-handed coordinate system.
