The question is:
Evaluate the surface integral:
$$
\iint\limits_S \, x^2yz\ \mathrm{d} S
$$
Where S is part of the plane z = 1 + 2x + 3y that lies above the rectangle [0,3] X [0,2]
I literally just don't understand the notation of this "rectangle". How is [0,3] X [0,2] a rectangle? Normally we are given vertices of some sort of shape, or instead just told something like "the part that lies in the first octant". So, I'm having trouble parameterizing the function in terms of u and v.
Basically, what I've done so far, in terms of parameterizing is:
u = x
v = y
So, z = 1 + 2u + 3v
$$
\vec{r} \ = u \vec{i} \ + v \vec{j} \ + (1 + 2u + 3v) \vec{k} \
$$
$$
\vec{r}_u \ = \vec{i} \ + 2 \vec{k} \
$$
$$
\vec{r}_v \ = \vec{j} \ + 3 \vec{k} \
$$
So:
$$
\vec{r}_u \times \vec{r}_v = -2\vec{i} \ – 3\vec{j} + \vec{k} \
$$
Then I get stuck around this step:
$$
\iint\limits_S \, u^2v(1 + 2u + 3v)\sqrt{14}\ \mathrm{d} v \, \mathrm{d} u
$$
I need limits of integration.
Sorry for any formatting mistakes.
Thank you for the help.
Best Answer
To make everything explicit: Knowing now that $[0,3]\times[0,2]$ is a rectangle in the $xy$-plane, we also have our limits of integration. We write $x^2yx=x^2y(1+2x+3y)=x^2y+2x^3y+3x^2y^2$ and integrate over the region $\{(x,y)|0\leq x\leq 3, 0 \leq y \leq 2\}$. Thus our integral is