Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
1  
[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
up vote 0 down vote accepted

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

$\displaystyle\int_{x=0}^3\int_{y=0}^2(x^2y+2x^3y+3x^2y^2)dydx$.

share|cite|improve this answer
    
Where did the root(14) go in your integral? I'm sorry. I'm not familiar with integrating a surface integral without reparameterizing in terms of vectors first. I'm just learning these things now. – Clark Henry Dec 10 '12 at 8:34
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.