# Help me understand a surface integral question?

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.

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[0,3]x[0,2] is the set of all ordered pairs (x,y) such that 0 \leq x \leq 3 and 0 \leq y \leq 2. If you draw this out in the plane it's a rectangle with vertices (0,0), (3,0), (0,2), (3,2). – Gyu Eun Lee Dec 10 '12 at 8:18
Thank you very much. So, the limits of integration would literally just be 0 <= u <= 3 and 0 <= v <= 2. I've just never seen this notation, and didn't want to assume [0,3] was x and [0,2] was y. Thank you! – Clark Henry Dec 10 '12 at 8:22

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
$\displaystyle\int_{x=0}^3\int_{y=0}^2(x^2y+2x^3y+3x^2y^2)dydx$.
The ${\rm d}\omega=|{\bf r}_u\times{\bf r}_v|\ {\rm d}(u,v)=\sqrt{14}\ {\rm d}(u,v)$ went lost here. – Christian Blatter Dec 10 '12 at 9:48
I just found the official answer to this question in the book - Stewart Calculus, 7E, question 16.7.009. Which stated that the answer is $= 171 \sqrt{14}$. How is this so, considering that $\sqrt{14}$ is no longer included in your answer? – Tarius Dec 7 '15 at 1:48