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I need to integrate $\int_ \! \int \sin \frac{1}{2}(x+y) \cos\frac{1}{2}(x-y)\,dx\,dy$ over region $R$:{triangle with vertices $(0,0),(0,2),(1,1)$}. They ask to use $u=\frac{1}{2}(x+y)$ and $v=\frac{1}{2}(x-y)$.

Attempt:First, I transformed $(x,y)$ to $(x=x(u,v),y=y(u,v))$. Namely, I solved for x and y: $$\begin{cases}u=\frac{1}{2}(x+y)\\v=\frac{1}{2}(x-y)\end{cases}$$

The Jacobian I found is $J(u,v)=\frac{\partial (x,y)}{\partial (u,v)}=-1$. I am having hard time founding the limits of integration. In xy-plane $R$ looks like that: enter image description here

So, the region R is bounded by $\begin{cases} y=0\\y=x\\y=-x+2 \end{cases}$

In uv-plane it looks like:

enter image description here

The region S is bounded by $\begin{cases} u=1\\ u=v\end{cases}$

Now the double integral looks like: $$\int_0^1 \! \int_0^v \sin u \cos v\,du\,dv$$

When, I solve it I get

$$\int_0^1 \! \int_0^v \sin u \cos v\,du\,dv=\frac{1}{2} (\frac{1}{2} \sin2 -1)$$

But in the answer key the answer is $1-\frac{1}{2} \sin2 $

Can you please tell me what I am doing wrong. Hints please.

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The signed Jacobian is $-2$, not $-1$. –  copper.hat Jun 12 '12 at 7:42
    
Also, I get the $S$ region to be slightly different, $S = \mathbb{co} \{(0,0), (1,-1), (1,0) \}$ (ie, your $S$ flipped through the $u$ axis). –  copper.hat Jun 12 '12 at 7:50
    
Thank you guys. –  Koba Jun 12 '12 at 23:26

2 Answers 2

up vote 1 down vote accepted

Two issues: (1) $|J| = 2$, as mentioned above. (2) Your $S$ region should be flipped about the $u$ axis. Check by computing the $u$-$v$ coordinates of $(0,0), (0,2), (1,1)$.

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The Jacobian :
You have $x=u+v$ and $y=u-v$ then : $|J|=2$

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