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I recently learnt about line integrals and Green's Theorem. But the lecturer gave us an assignment to answer how to calculate line integrals "directly without using Green's Theorem". I've looked at the notes, but I can't seem to see the difference between line integral with Green's Theorem and without. The question is as below:

Consider the vector field $F(x,y) = xy\textbf{i} + x^2\textbf{j}=F_1\textbf{i} + F_2\textbf{j}$, let C be the rectangle with vertices $(0,0), (3,0), (3,1)$ and $(0,1)$, let $T$ denote the unit tangent vector to $C$ directed anticlockwise around $C$, and let $n$ denote the unit normal vector to $C$ directed out of the region bounded by $C$. Let $D$ denote this region bounded by $C$.

(a) Calculate the line integral $\int{F\cdot T ds}$ directly without using Green's theorem.

(b) Calculate the double integral $\int\int\left(\frac{dF_2}{dx} - \frac{dF_1}{dy}\right)dA$ without using Green's theorem.

Joey Zou
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Hani Zahrul
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2 Answers2

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[Please don't up vote - just paraphrasing answers in comments so question not in unanswered state forever]

In (A) you have to evaluate the line integral along a piecewise smooth path. This means breaking the boundary of the rectangle up into 4 smooth curves (the sides), parameterising the curves, evaluating the line integral along each curve and summing the results.

In (B) you have to expand $\dfrac{dF_2}{dx}, \dfrac{dF_1}{dy}$ and $dA$ and evaluate the result.

Frentos
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a) $$\int_{x=0}^{x=3}(x^2\textbf{j})\cdot (\textbf{i})dx+\int_{y=0}^{y=1}(3y\textbf{i}+3^2\textbf{j})\cdot(\textbf{j})dy+\int_{x=3}^{x=0}(x\textbf{i}+x^2\textbf{j})\cdot (\textbf{-i})dx+\int_{y=1}^{y=0}(0\textbf{i}+0\textbf{j})\cdot(\textbf{-j})dy=0+9+\frac92+0=\frac{27}2$$ b) $$\iint(\frac{\partial}{\partial x}x^2-\frac{\partial}{\partial y}xy)dA=\int_0^1\int_0^3(2x-x)dxdy=\int_0^1\frac92 dy=\frac92$$

Bob Dobbs
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