Calculus III - Green's Theorem

Section 16.7 : Green's Theorem

5. Verify Green’s Theorem for $ \displaystyle \oint_{C}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}}$ where $C$ is shown below by (a) computing the line integral directly and (b) using Green’s Theorem to compute the line integral.

$The curve $C$. It is the closed triangle with corners at $\left( { - 3,0} \right)$, the origin and $\left( {0,3} \right)$. The arrows show that it is traveled in the counter clockwise direction, going right along the x-axis to the origin, up the y-axis to $\left( {0,3} \right)$ and then back down the slanted side to $\left( { - 3,0} \right)$.$

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a computing the line integral directly Show All Steps Hide All Steps

Start Solution

So, let’s start off the problem with labeling the curves as follows,

$The curve $C$ with each piece labeled. It is a closed triangle with corners at $\left( { - 3,0} \right)$, the origin and $\left( {0,3} \right)$. The vertical piece up the y-axis is labeled ${C_1}$, the slanted piece from $\left( {0,3} \right)$ down to $\left( { - 3,0} \right)$ is labeled ${C_2}$, and the horizontal piece back along the x-axis is labeled ${C_3}$, with arrows showing the direction of travel on each.$

Following the specified direction for each curve here are the parameterizations for each curve.

${C_1}:\,\,\vec r\left( t \right) = \left\langle {0,t} \right\rangle \hspace{0.25in}0 \le t \le 3$
${C_2}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0,3} \right\rangle + t\left\langle { - 3,0} \right\rangle = \left\langle { - 3t,3 - 3t} \right\rangle \hspace{0.25in}0 \le t \le 1$
${C_3}:\,\,\vec r\left( t \right) = \left\langle {t,0} \right\rangle \hspace{0.25in} - 3 \le t \le 0$

Show Step 2

Here is the line integral evaluated over each of these curves.

\[\begin{align*}\oint_{{{C_1}}}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} & = \int_{0}^{3}{{\left[ {\left( 0 \right){{\left( t \right)}^2} + {{\left( 0 \right)}^2}} \right]\left( 0 \right)\,dt}} + \int_{0}^{3}{{\left[ {4\left( 0 \right) - 1} \right]\left( 1 \right)\,dt}}\\ & = \int_{0}^{3}{{ - 1\,dt}} = \left. { - t} \right|_0^3 = \underline { - 3} \end{align*}\] \[\begin{align*}\oint_{{{C_2}}}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} & = \int_{0}^{1}{{\left[ {\left( { - 3t} \right){{\left( {3 - 3t} \right)}^2} + {{\left( { - 3t} \right)}^2}} \right]\left( { - 3} \right)\,dt}} + \int_{0}^{1}{{\left[ {4\left( { - 3t} \right) - 1} \right]\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{81{t^3} - 189{t^2} + 81t\,dt}} + \int_{0}^{1}{{36t + 3\,dt}}\\ & = \int_{0}^{1}{{81{t^3} - 189{t^2} + 117t + 3\,dt}}\\ & = \left. {\left( {\frac{{81}}{4}{t^4} - 63{t^3} + \frac{{117}}{2}{t^2} + 3t} \right)} \right|_0^1 = \underline {\frac{{75}}{4}} \end{align*}\] \[\begin{align*}\oint_{{{C_3}}}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} & = \int_{{ - 3}}^{0}{{\left[ {\left( t \right){{\left( 0 \right)}^2} + {{\left( t \right)}^2}} \right]\left( 1 \right)\,dt}} + \int_{{ - 3}}^{0}{{\left[ {4\left( t \right) - 1} \right]\left( 0 \right)\,dt}}\\ & = \int_{{ - 3}}^{0}{{{t^2}\,dt}} = \left. {\frac{1}{3}{t^3}} \right|_{ - 3}^0 = \underline 9 \end{align*}\]

Show Step 3

Now, all we need to do is add up the results from the previous step to get the value of the line integral over the full curve. This gives,

\[\oint_{C}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} = \left( { - 3} \right) + \left( {\frac{{75}}{4}} \right) + \left( 9 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{99}}{4}}}\]

b using Green’s Theorem to compute the line integral Show All Steps Hide All Steps

Start Solution

Note that as the circle on the integral implies the curve is in the positive direction and so we can use Green’s Theorem on this integral.

From the integral we have,

\[P = x{y^2} + {x^2}\hspace{0.25in}\hspace{0.25in}Q = 4x - 1\]

Remember that $P$ is multiplied by $x$ and $Q$ is multiplied by $y$.

Show Step 2

Using Green’s Theorem the line integral becomes,

\[\oint_{C}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} = \iint\limits_{D}{{4 - \left( {2xy} \right)\,dA}} = \iint\limits_{D}{{4 - 2xy\,dA}}\]

$D$ 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 hypotenuse of the region is given by $y = x + 3$. Once we have this equation the region is then very easy to get limits for. They are,

\[\begin{array}{c} - 3 \le x \le 0\\ 0 \le y \le x + 3\end{array}\]

Show Step 4

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

\[\begin{align*}\oint_{C}{{\left( {x{y^2} + {x^2}} \right)\,dx + \left( {4x - 1} \right)\,dy}} & = \iint\limits_{D}{{4 - 2xy\,dA}}\\ & = \int_{{ - 3}}^{0}{{\int_{0}^{{x + 3}}{{4 - 2xy\,dy}}\,dx}}\\ & = \int_{{ - 3}}^{0}{{\left. {\left( {4y - x{y^2}} \right)} \right|_0^{x + 3}\,dx}}\\ & = \int_{{ - 3}}^{0}{{12 - 5x - 6{x^2} - {x^3}\,dx}}\\ & = \left. {\left( {12x - \frac{5}{2}{x^2} - 2{x^3} - \frac{1}{4}{x^4}} \right)} \right|_{ - 3}^0\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{99}}{4}}}\end{align*}\]

So, we got the same answer after applying Green’s Theorem to the line integral as we got by integrating the line integral directly.