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Section 16.7 : Green's Theorem

2. Use Green’s Theorem to evaluate \( \displaystyle \int\limits_{C}{{\left( {6y - 9x} \right)dy - \left( {yx - {x^3}} \right)\,dx}}\) 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,

\[P = - \left( {yx - {x^3}} \right) = {x^3} - yx\hspace{0.25in}\hspace{0.25in}Q = 6y - 9x\]

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. It is also easy to get in a hurry and just assume that \(P\) is the first term in the integral and \(Q\) is the second. That is clearly not the case here so be careful!

Show Step 2

Using Green’s Theorem the line integral becomes,

\[\int\limits_{C}{{\left( {6y - 9x} \right)dy - \left( {yx - {x^3}} \right)\,dx}} = \iint\limits_{D}{{ - 9 - \left( { - x} \right)\,dA}} = \iint\limits_{D}{{x - 9\,dA}}\]

\(D\) is the region enclosed by the curve.

Show Step 3

We’ll leave it to you to verify that the equation of the line along the top of the region is given by \(y = 3 - x\). Once we have this equation the region is then very easy to get limits for. They are,

\[\begin{array}{c} - 1 \le x \le 1\\ - 1 \le y \le 3 - x\end{array}\] Show Step 4

Now all we need to do is evaluate the double integral. Here is the evaluation work.

\[\begin{align*}\int\limits_{C}{{\left( {6y - 9x} \right)dy - \left( {yx - {x^3}} \right)\,dx}} & = \iint\limits_{D}{{x - 9\,dA}}\\ & = \int_{{ - 1}}^{1}{{\int_{{ - 1}}^{{3 - x}}{{x - 9\,dy}}\,dx}}\\ & = \int_{{ - 1}}^{1}{{\left. {\left( {x - 9} \right)y} \right|_{ - 1}^{3 - x}\,dx}}\\ & = \int_{{ - 1}}^{1}{{\left( {x - 9} \right)\left( {4 - x} \right)\,dx}}\\ & = \int_{{ - 1}}^{1}{{ - {x^2} + 13x - 36\,dx}}\\ & = \left. {\left[ { - \frac{1}{3}{x^3} + \frac{{13}}{2}{x^2} - 36x} \right]} \right|_{ - 1}^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{218}}{3}}}\end{align*}\]