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here's a question I'm working on.
We have this integral over a region and we wish to use Green's Theorem to evaluate it.

$\int_D x\ln(y) dx$

$D:1\leq x \leq 2, e^x \leq y \leq e^{x^2}$

Here's how I set it up, I'm not sure if I'm correct at setting it up, but if I am, then I may have made an arithmetic mistake somewhere. The answer given is $\frac{-17}{12}$

$\int_D x\ln(y)dx + 0dy$
Where I take $x\ln(y)$ as $P$ and $0$ as $Q$.

Is this correct?

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  • $\begingroup$ so far so good. $\endgroup$ – Kuifje Dec 14 '15 at 13:38
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As per comments, your start was correct.

Spoiler alert -- complete solution below!

I'm not sure of the $P$ and $Q$ in your notation, so I'll use notation I'm familiar with and let you join the dots. Write Green's theorem for the vector field $\mathbf{F}(x,y) = F_1(x,y)~\mathbf{i}+F_2(x,y)~\mathbf{j}$ as: $$\oint_C\mathbf{F}\cdot d\mathbf{r} = \oint_CF_1~dx+F_2~dy=\iint_R\left(\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y}\right)~dx~dy$$

where $R$ is the region enclosed by the curve $C$.

You have a region $D: 1\le x\le2,e^x\le y\le e^{x^2}$. Writing $C$ as the boundary, $$\begin{eqnarray} \oint_Cx\ln(y)~dx&=&\oint_Cx\ln(y)~dx+0~dy\\ &=&\int_D\left(\dfrac{\partial}{\partial x} 0 - \dfrac{\partial}{\partial y}x\ln(y)\right)~dx~dy\\ &=&\int_1^2\!\!\int_{e^x}^{e^{x^2}}0- \dfrac x y~dy~dx\\ &=&\int_1^2\left[x\ln(y)\strut\right]_{y=e^x}^{e^{x^2}}~dx\\ &=&\int_1^2x(x^2)-x(x)~dx\\ &=&\left[\dfrac14x^4-\dfrac13x^3\right]_1^2 = (4-\dfrac83)-(\dfrac14-\dfrac13)=\dfrac{48-32-3+4}{12}=\dfrac{17}{12} \end{eqnarray}$$

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