Problem
Let $S=\{(x,y,z)\mid x\geq0, y\geq0, z\geq0,2x+2y+z=2\}$, and $a=(2,2,1)$ be a normal vector oriented outside. Evaluate the surface integral $$\iint_{S}xz\,dy\,dz$$
I am having a hard time interpreting this question since every surface integral I have ever seen was in a form $\iint_{S}f(x,y,z)\,dS$. I don't even know what this integral even means.
If it was in a form $\iint_{S}f(x,y,z)\,dS$ then I can calculate using the method written here, but I am not even sure what is the first step.
What would be the first step to solve this question? Any help is appreciated.
By the way, the normal vector $a$ may be unnecessary to the question. If it is not required to solve it, you can ignore it.
Best Answer
All it means is you have a scalar field $f(x,y,z) = xz$ instead of a vector field.
Your surface is a plane given by $2x+2y+z=2 \,$ where $(x\geq0, y\geq0, z\geq0)$.
Given the integral is wrt $y, z$, rewrite your plane in the form $g(y,z) = x = 1 - y - \frac{z}{2} \,$.
$\displaystyle \frac{\partial g}{\partial y} = - 1, \frac{\partial g}{\partial z} = - \frac{1}{2}$
Based on your parametrization, your projected triangle in YZ plane $(x = 0)$ gives you upper bound of $y$ as $(1-\frac{z}{2})$ and upper bound of $z$ as $2$.
Your integral becomes $\displaystyle \iint_S (1 - y - \frac{z}{2}) z \, \sqrt{1 + (\frac{\partial g}{\partial y})^2 + (\frac{\partial g}{\partial z})^2} \, dy \, dz$