# Double Integral

1. Nov 15, 2005

### Schwartz

I need to solve a double integral and I have no idea what to change the variables to:
$$\iint_{R} \cos ( \frac{y-x}{y+x}) \ dA$$
$$R=\{(x,y) \mid \ -x+1 \leq y \leq -x+2, 1 \leq x \leq 2 \}$$
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.

Last edited: Nov 15, 2005
2. Nov 15, 2005

### 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.

Last edited: Nov 15, 2005
3. Nov 15, 2005

### 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.

Last edited: Nov 16, 2005
4. Nov 16, 2005

### 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 .

Last edited: Nov 16, 2005
5. Nov 16, 2005

### 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.

6. Nov 16, 2005

### 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:

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

7. Nov 17, 2005

### Schwartz

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