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### Section 5-7 : Green's Theorem

4. Use Green’s Theorem to evaluate $$\displaystyle \int\limits_{C}{{\left( {{y^4} - 2y} \right)\,dx - \left( {6x - 4x{y^3}} \right)\,dy}}$$ where $$C$$ is shown below.

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Okay, first let’s notice that if we walk along the path in the direction indicated then our left hand will be over the enclosed area and so this path does have the positive orientation and we can use Green’s Theorem to evaluate the integral.

From the integral we have,

$P = {y^4} - 2y\hspace{0.25in}\hspace{0.25in}Q = - \left( {6x - 4x{y^3}} \right) = 4x{y^3} - 6x$

Remember that $$P$$ is multiplied by $$x$$ and $$Q$$ is multiplied by $$y$$ and don’t forget to pay attention to signs. It is easy to get in a hurry and miss a sign in front of one of the terms.

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Using Green’s Theorem the line integral becomes,

$\int\limits_{C}{{\left( {{y^4} - 2y} \right)\,dx - \left( {4x{y^3} - 6x} \right)\,dy}} = \iint\limits_{D}{{4{y^3} - 6 - \left( {4{y^3} - 2} \right)\,dA}} = \iint\limits_{D}{{ - 4\,dA}} = - 4\iint\limits_{D}{{\,dA}}$

$$D$$ is the region enclosed by the curve.

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Okay, we are now pretty much done with this problem. After factoring out the “-4” from the integral we can use the following fact to finish the evaluation of the integral.

\begin{align*}\int\limits_{C}{{\left( {{y^4} - 2y} \right)\,dx - \left( {4x{y^3} - 6x} \right)\,dy}} & = - 4\iint\limits_{D}{{\,dA}}\\ & = - 4\left( {{\mbox{Area of }}D} \right) = - 4\left[ {\left( 6 \right)\left( 4 \right)} \right] = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 96}}\end{align*}

Don’t forget the fact that,

$\iint\limits_{D}{{\,dA}} = \mbox{Area of }D$

This is a very useful fact. It doesn’t come up often but when it does it can reduce the amount of work required to finish out a problem.

Of course, this fact is really only useful if $$D$$ is a region that we can easily determine the area for and in this case $$D$$ was just a rectangle that we could easily get the width, 6, and the height, 4, from the graph in the problem statement.