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I am trying to figure out the answer to this problem: Evaluate $$\iint sin(\frac{x+y}{2}) cos(\frac{x-y}{2})dA$$ on $R$, where $R$ is the triangle with vertices $(0,0),(2,0)$ and $(1,1)$.

I arrived at an answer using the substitution $u=(x+y)/2$ and $v=(x-y)/2$ and the change of variables theorem using Jacobians.

My answer was $2sin1-sin2$. But the answer that I have in this textbook is that $1 - sin(2)/2$.

Did I do something wrong? If so, then what did I do wrong? Please answer with the detailed solution.

Meshal
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Eric Brown
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  • You could have just used the product of sines/cosines formula and computed it very straightforwardly. –  Jun 09 '15 at 00:52

2 Answers2

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It's $\frac{1}{2} \int \int (\sin (2x) + \sin (2y)) dA=\frac{1}{2} \int\limits_{0}^{2}\int\limits_{x}^{-x+2} (\sin(2x)+\sin(2y)) dydx $

mich95
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Let $u=\frac{x+y}{2}$ and $v=\frac{x-y}{2}$. Adding and subtracting these equations gives $x=u+v$ and $y=u-v$.

Let $A$ denotes the area to be integrated in the $(x,y)$ plane, and $A^*$ to be the one in the $(u,v)$ plane. From the change of variables theorem, we can write $$dA^*=\left | J \right |dA$$ where $J$ is the Jacobian given by $$\left | J \right |=\left | \frac{dx}{du}\frac{dy}{dv} -\frac{dx}{dv}\frac{dy}{du} \right|=\left | -1-1 \right |=2$$

Hence on the $(u,v)$ plane, our integral becomes $$\int_{-1}^{0}\int_{-v}^{1}2sin(u)cos(v)dudv=2\int_{-1}^{0}-cos(v)cos(1)+cos^2(v)dv=1-\frac{sin(2)}{2}$$

Meshal
  • 779
  • Thanks for the help. I found the answer. The intervals of the new integral were wrong. I should have used (v,1) for u and (0,1) for v. Then I get the same answer of 1-(sin2))/2 – Eric Brown Jun 09 '15 at 04:07