Use Green's Theorem to evaluate the line integral \int_C \overrightarrow{F} \cdot...


Use Green's Theorem to evaluate the line integral {eq}\int_C \vec{F} \cdot d\vec{r}, {/eq} where {eq}\vec{F} = 2y \mathbf{i} + xy \mathbf{j}, {/eq} and {eq}C {/eq} is the triangle consisting of the line segments from {eq}(0, 0) {/eq} to {eq}(1, 0) {/eq} to {eq}(1, 1) {/eq} then back to {eq}(0, 0). {/eq}

Answer and Explanation:

Let {eq}T {/eq} be the triangular region which is bounded by {eq}C {/eq}. Then {eq}T {/eq} consists of the region above the {eq}x {/eq}-axis and below the line {eq}y=x {/eq} for {eq}0 \le x \le 1 {/eq}. So, in Cartesian coordinates, {eq}T {/eq} corresponds to the bounds {eq}0 \le x \le 1 {/eq}, {eq}0 \le y \le x {/eq}.

Now, we compute the line integral as follows:

$$\begin{align*} \oint_C \vec{F} \cdot d\vec{r}&=\iint_T \left(\frac{\partial}{\partial x}(xy)-\frac{\partial}{\partial y}(2y)\right) \, dA&\text{(by Green's theorem)}\\ &=\iint_T (y-2) \, dA&\text{(evaluating the partial derivatives)}\\ &=\int_{x=0}^1\int_{y=0}^x (y-2) \, dy \, dx&\text{(writing the area integral as an iterated integral in Cartesian coordinates)}\\ &=\int_{x=0}^1 \left(\frac{1}{2}y^2-2y\right|_{y=0}^x \, dx&\text{(evaluating the }y\text{-integral)}\\ &=\int_{x=0}^1 \left(\frac{1}{2}x^2-2x-\frac{1}{2}(0)^2+2(0)\right)\, dx\\ &=\int_0^1 \left(\frac{1}{2}x^2-2x\right) \, dx\\ &=\left(\frac{1}{6}x^3-x^2\right|_0^1&\text{(evaluating the }x\text{-integral)}\\ &=\frac{1}{6}(1)^3-1^2-\frac{1}{6}(0)^3+0^2\\ &=\frac{1}{6}-1\\ &=\frac{5}{6} \, . \end{align*} $$

In summary, {eq}\boxed{\oint_C \vec{F} \cdot d\vec{r}=\frac{5}{6} \, .} {/eq}

Learn more about this topic:

Line Integrals: How to Integrate Functions Over Paths

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 2

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