How do I find this line integral using Green's Theorem and Change of Variables?

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Let C be the boundary of the square with vertices (0,0),(2,1), (1,-2), and (3,-1), oriented clockwise.

Let F(x,y)=《xy-sinx, y^2+cosy》.

Find the integral over the curve C of ∫F·dr

How do you use change of variables and greens theorem to solve this?

Thanks a mucho.

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1 Answer

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Can we at least assume that you know what "Green's theorem" is? Green's theorem says that $\oint Ldx+ Mdy= \int_D\int \left(\frac{\partial M}{\partial x}- \frac{\partial L}{\partial x}\right) dxdys$ where D is the region bounded by the closed curve the integral on the left is over.

Here we have $L= xy- sin(x)$ and $M= y^2- cos(y)$ so $\frac{\partial M}{\partial x}= 0$ and $\frac{\partial L}{\partial y}= x$. By Green's theorem, we only have to integrate $\int\int x dxdy$ over the region inside that contour.

Of course that contour is a little "complicated". It is a rectangle but the vertices are (0, 0), (2, 1), (1, -2), and (3, -1). The edges are not parallel to the x and y axes. That's where the "change of variable" comes in. The line from (0, 0) to (2, 1) is x= 2y. If we make the change of variable $u= \frac{2\sqrt{5}}{5}x+ \frac{\sqrt{5}}{5}y$, $v= \frac{\sqrt{5}}{5}x- \frac{2\sqrt{5}}{4}y$, (those coefficients are the sine and cosine from a 2, 1, $\sqrt{5}$ right triangle. I an effectively rotating the coordinate system) when (x, y)= (0, 0) (u, v)= (0, 0), when (x, y)= (2, 1), (u, v)= ($\sqrt{5}$, 0). When (x, y)= (1, -2), (u, v)= (0, $\sqrt{5}$), and when (x, y)= (3, -1), (u, v)= ($\sqrt{5}$, $\sqrt{5}$). That is now the rectangle with vertices at (0, 0), ($\sqrt{5}$, (0, $\sqrt{5}$), and ($\sqrt{5}$, $\sqrt{5}$). Integrate, in the u,v coordinate system, taking u from 0 to $\sqrt{5}$ and v from 0 to $\sqrt{5}$.

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