The question I'm working on states:
Let $C$ be the curve in $\mathbb R$ consisting of line segments from $(4,1)$ to $(4,3)$ to $(1,3)$ to $(1,1)$. Let $F(x,y) = (x+y)i + (y-1)^3 * e^{\sin(y)}j$. Evaluate the line integral: $$\int_{c} F* \,dr$$
I've never computed the line integral with four separate points before. I'm thinking that I'll need to use Green's Theorem and use the points as boundaries for integration? However I'm not even sure how to start the process. Any help would be greatly appreciated!
$\endgroup$ 21 Answer
$\begingroup$METHOD 1: EXPLOIT GREEN"S THEOREM
Note that $\frac{\partial F_x}{\partial y}=1$ and $\frac{\partial F_y}{\partial x}=0$.
On the line integral from $(1,1)$ to $(4,1)$, $\vec F=\hat x(x+1)$. So, the integral over the three segments of interest can be computed as the integral over all four segments minus the integral over the segment from $(1,1)$ to $(4,1)$. And that integral is trivial. The integral over the four segments constitutes a closed curve and application of Green's Theorem is again trivial.
Let $C_4$ be the segment from $(1,1)$ to $(4,1)$. Then, we have
$$\begin{align} \int_C \vec F \cdot d\vec r&=\oint_{C+C_1}\vec F \cdot d\vec r-\int_{C_1}\vec F \cdot d\vec r\\\\ &=\int_1^3\int_1^4\,(-1)\,dx\,dy-\int_1^4(x+1)dx\\\\ &=-6-\frac{21}{2}\\\\ &=-\frac{33}{2} \end{align}$$
METHOD 2: DIRECT LINE INTEGRAL APPROACH
If we wish to compute the line integral over the original contour $C$ directly, we have
$$\int_C \vec F \cdot d\vec r=\int_1^3 (y-1)^3e^{\sin y}dy+\int_4^1(x+3)dx+\int_3^1(y-1)^3e^{\sin y}dy$$
The first and third integrals on the right-hand side cancel and we have
$$\begin{align} \int_C \vec F \cdot d\vec r&=\int_4^1(x+3)dx\\\\ &=\left.\frac{(x+3)^2}{2}\right|_4^1\\\\ &=\frac{16-49}{2}\\\\ &=-\frac{33}{2} \end{align}$$
as expected!
It might be interesting to note that this answer would not be impacted if we changed $F_y$ to any smooth function that is independent of $x$.
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