I am asked to compute the integral:
$$\oint_{\gamma}(xy^2+\frac{1}{x}+y)\mathrm{d}x+(y\ln{x}+x+\frac{2}{y})\mathrm{d}y$$
where $\gamma$ is the positively oriented boundary of $D=\{x,y\geq0, x\leq y\leq 2x, 1\leq xy\leq 4\}$.
My thoughts are using the Green's theorem since $\gamma$ is closed and are piece wise smooth, simple closed curve. The integral equals to:
$$\iint \left( \frac{y}{x}-2xy\right) \mathrm{d}x\mathrm{d}y$$
Then using substitution $u=\frac{y}{x}, v=xy, dxdy=-2u\mathrm{d}u\mathrm{d}v$. The integral equals to:
$$\iint_{1\leq u\leq 2, 1\leq v\leq 4} (u-2v)(-2u)\mathrm{d}u\mathrm{d}v$$
Which calculates to be $31$ if I am not mistaken, but the correct answer is $\frac{3}{2}(1-5\ln{2})$. Appreciate any help!
Best Answer
Your error is in computing $dx dy$. Also, note that it is clearly wrong because somehow, you found that the absolute value of the determinant of the Jacobian is $-2u < 0$.
The correct relationship is that \begin{align} dx dy &= \left| \det \begin{pmatrix} \partial x/ \partial u & \partial x/ \partial v \\ \partial y/ \partial u & \partial y/ \partial v \end{pmatrix} \right| du dv \end{align} If you invert things properly, you should find that $x = u^{-1/2} v^{1/2}$ and $y = u^{1/2} v^{1/2}$. So, the determinant of the matrix above is $-\frac{1}{2u}$. Taking absolute value (AND using the fact that $u \in [1,2]$) yields: \begin{align} dx dy &= \dfrac{1}{2u} \, dudv \end{align} Now you should get the correct answer.