Solving Double Integral: Struggling with Variables

[Schwartz]

I need to solve a double integral and I have no idea what to change the variables to:
\iint_{R} \cos ( \frac{y-x}{y+x}) \ dA
R=\{(x,y) \mid \ -x+1 \leq y \leq -x+2, 1 \leq x \leq 2 \}<br />
I tried to set u=y-x and v=y+x, but I still can't solve the resulting integral. I also tried setting v + u= \frac{y-x}{y+x} and differentiating implicitly when finding the Jacobian, but the Jacobian turns out to be zero.

 

[Schwartz]

\iint_{R} \cos ( \frac{y-x}{y+x}) \ dA = \iint_{R} \cos ( 1 - \frac{2x}{y+x}) \ dA
R=\{(x,y) \mid \ -x+1 \leq y \leq -x+2, 1 \leq x \leq 2 \} \\
u = 2x
v = x+y \\
R=\{(u,v) \mid \ 1 \leq v \leq 2, \ 2 \leq u \leq 4 \}
\frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{2}
\frac{1}{2} \int_{1}^{2} \int_{2}^{4} \cos ( 1- \frac{u}{v}) \ dudv

After putting in these variables, I can't solve the integral. If I change what u and v are, then either the Jacobian is zero or I still can't solve the intergral.

 

[Schwartz]

u=2x
v= \frac{1}{y+x}

R=\{(u,v) \mid \ \frac{1}{2} \leq v \leq 1, \ 2 \leq u \leq 4 \}

\frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{2v^2}

\frac{1}{2} \int_{ \frac{1}{2} }^{1} \int_{2}^{4} \frac{1}{2v^2} \cos ( 1- uv) \ dudv

I can't do this one either.

 

[benorin]

There's a good reason for that!

The value of that integral is:

\frac{1}{4} \left[ -\sin 1\cos 4 + 4\sin 1 \cos ^{2} 2 -4\cos 1 \cos 4 -20\sin 1 \mbox{Ci}(2)+16\sin 1 \mbox{Ci}(4)-16\cos 1 \mbox{Si}(4) \right.
-4\cos 1\cos 2 \sin 2 -4\sin 1\sin 4 +10\sin 1\sin 2+20\cos 1\mbox{Si}(2)-4\cos ^{2} 1 +3\cos 1\sin 2
\left. +10\cos 1\cos 2-3\sin 1\cos 2 -4\cos 1 \mbox{Si}(1) + 4\sin 1\mbox{Ci}(1)-4\sin^{2} 1 + \cos 1 \sin 4 \right]

so says maple, where Ci and Si are the cosine and sine integral functions .

 

[Schwartz]

Well, the answer should be \frac{3}{2} \sin 1, so I think that this problem is probably an error in my book.

I'll see what my professor says about it. Thanks.

 

[benorin]

I get that problem from Stewart, Calculus 4th ed. ch15.9 promblem # 21 (I had that text for clalc I->III). The given answer is \frac{3}{2} \sin 1 how ever the given region of integration is the trapezoidal region with verticies (1,0), (2,0), (0,2), and (0,1). Your bounds should then be:

R=\{(x,y) \mid \ -x+1 \leq y \leq -x+2, 0 \leq x \leq 2 \}

 

[Schwartz]

yeah, i figured it out. I was just being dumb. Thanks.