Recent content by Schwartz

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

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.

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.

\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}...

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...