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The following are the problems that I have been working on. It involves change in variables with 2,3 variables respectively.

(1)Let $R$ be the trapezoid with vertices at $(0,1),(1,0),(0,2)$ and $(2,0)$. Using the substitutions $u = y-x$ and $v = y +x$, evaluate $$\int\int_R e^{{y-x}\over{y+x}} dA$$.

(2)Evaluate the following integral $$\int\int\int_D (x^2y + 3xyz)dV$$ where D is the region in 3-space defined by $1 \le x \le 2, 0 \le xy \le 2, 0 \le z \le 1$ using the substitution $u =x , v = xy, w=3z$.

Here are what I tried.

1), I understand the part where the parallel lines $x+y = 1$ and $x+y=2$ is used to find the limit of integration of $v$, but what am I supposed to do with that of $u$ ? One side of the trapezoid is vertical and the other side is horizontal and it looks nothing like the equation $v$. I was thinking about trying to doubling the trapezoid so that it becomes a parallelogram, but I am not sure if that even works.

I understand the idea and the fact that $$\int\int_R f(x,y) dA = \int\int_S f(g(u,v),h(u,v))|{ \partial{(x,y)}\over{\partial(u,v)}}|dvdu$$, but this is more of an algebraic thing that I am stuck with.

2), Again, I understand the theory, but I am stuck with the algebraic part. The approach I made was $$\int_0^1 \int_0^2 \int_1^2 (uv+3vw)|{1\over{3u}}|dudvdw$$ but I ended up having to integrate an improper integral with the result being $-\infty$... Can someone help out ?

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1 Answer 1

up vote 5 down vote accepted

I am adding some points for the first one so you can find the limits easily. Try to do the rest by yourself. :)

If we consider the 4 points above and mark them on $xy$ plane, we have the area as you see below: enter image description here

Now, as you probably know, we find $x$ and $y$ respect to $u$ and $v$ as we are indicated. So $$u=y-x,v=x+y,~~~\to y=\frac{u+v}2,~~x=\frac{v-u}2~~~~~~~(I)$$ For finding the limits for $u$ and $v$ in $uv$ plane do as follows and by $I$: $$(1,0)\longrightarrow u=-1,v=1\\\\\ (2,0)\longrightarrow u=-2,v=2\\\\\ (0,1)\longrightarrow u=1,v=1\\\\\ (0,2)\longrightarrow u=2,v=2\\\\\ $$ This means that you have a new region like:

enter image description here

Obviously $$-1\le v\le +1, ~~~-v\le u\le v$$

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The last part is exactly what I needed to know, I really appreciate your help @Babak S. –  hyg17 May 21 '13 at 7:17
    
It actually helped me with the second one, too. Thanks ! –  hyg17 May 21 '13 at 7:47
    
@hyg17: As $0\le x\le 2$ and $u=x$ so $0\le u\le 2$. As $0 \le xy \le 2$ so $0 \le v \le 2$ since $v=xy$. And so $w=3z, ~0 \le z \le 1$, we have $0\le w\leq 3$. –  B. S. May 21 '13 at 8:24
    
@Babak S.:☂ᖺᙓ ᗰᗝᔕ☂ ᑕᗢᙢᕈᒪᙓ☂ᙓ ᗩﬡᔕᗯᗴᖇ. $\Large{✓}$ –  Software May 21 '13 at 8:35
    
$+! \quad \ddot\smile$ –  amWhy May 22 '13 at 0:31

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