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Evaluate $ \displaystyle\iint\limits_R \sin{(\frac{x+y}{2})}\cos{(\frac{x-y}{2})}dA$ where R is the triangle with vertices (0,0), (2,0) and (1,1). (Hint: Use the change of variables u = (x+ y)/2, v = (x− y)/2.) My attempt to answer: But when we plot sin((x + y)/2)*cos((x-y)/2), we get the below graph:
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Idea said: The region \(\displaystyle R\) is two dimensional in the x-y plane Plot the image \(\displaystyle S\) in the u-v plane. \(\displaystyle \int _0^?\int _0^?2 \sin u \cos vdvdu\) Click to expand... I got the answer. $ \displaystyle\int_0^1\displaystyle\int_0^u 2* \sin{u}* \cos{v}dvdu = 1-\frac{sin{(2)}}{2}= 0.545351286587$. Thanks for your reply.
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The region \(\displaystyle R\) is two dimensional in the x-y plane Plot the image \(\displaystyle S\) in the u-v plane. \(\displaystyle \int _0^?\int _0^?2 \sin u \cos vdvdu\)
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