Green's Theorem on Line Integral

I am asked to find the line integral for the following field: $$F = (e^{y^2}-2y)i + (2xye^{y^2}+\sin(y^2))j$$ On the line segment with points $(0,0),(1,2)$ and $(3,0)$. I have to do it with Greens theorem. This is the setup I have so far.

$$\int_C F \cdot dr = \iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\, dA$$ $$\frac{\partial Q}{\partial x} = ye^{y^2} \qquad \frac{\partial P}{\partial y}=2ye^{y^2}-2$$ Line from $(0,0)$ to $(1,2)$ is $y = 2x$. Line from $(1,2)$ to $(3,0)$ is $y = -x + 3$ $$\int_0^1\int_0^{2x}ye^{y^2}-2ye^{y^2}-2 + \int_1^3\int_0^{-x+3}ye^{y^2}-2ye^{y^2}-2$$ $$\int_0^1\int_0^{2x}-ye^{y^2}-2 \, dy\,dx+ \int_1^3\int_0^{-x+3}-ye^{y^2}-2 \,dy\,dx$$ $$\int_0^1\left[\frac{e^{y^2}}{2} - 2y\right]_0^{2x} + \int_1^3\left[-\frac{e^{y^2}}{2} - 2y\right]_0^{-x+3}$$ $$\int_0^1\frac{e^{4x^2}}{2}-\frac{1}{2}-4x \,dx+ \int_1^3-\frac{e^{(-x+3)^2}}{2}-2(-x+3)+\frac{1}{2}\,dx$$ Let $u = -x + 3$ and $du = -1$. $$\int_0^1\frac{e^{4x^2}}{2}-\frac{1}{2}-4x \,dx+ \int_1^3\frac{e^{(u)^2}}{2}+2(u)-\frac{1}{2}\,du$$

$$\left[\frac{e^{4x^2}}{8x}-\frac{x}{2}-2x^2 \right]_0^1+ \left[\frac{e^{(u)^2}}{4u}+u^2+\frac{u}{2}\right]_1^3$$ $$\left[\frac{e^{4x^2}}{8x}-\frac{x}{2}-2x^2 \right]_0^1+ \left[\frac{e^{(-x+3)^2}}{4(-x+3)}+(-x+3)^2+\frac{-x+3}{2}\right]_1^3$$ $$\frac{e^{4}}{8}-\frac{1}{2}-2 \,-\frac{e^{4}}{8}-4-1$$ $$\frac{1}{2}-2 \,-5$$

But the answer key says the answer is $-3$. Where have I gone wrong?

• wow this is one really tedious integral. – cgo Dec 16 '15 at 9:04
• Try using $dx$, $dy$, $du$ in every integral you define. Also, use align and &= to facilitate for the viewer. – Guilherme Thompson Dec 16 '15 at 10:07

The problem lies in two places: the first one in $$\frac{\partial Q}{\partial x} = 2ye^{y^2},$$ but this is only algebraic, even though it made your life much harder.

The second one is much more sensible. Look at Green's Theorem $$\oint_C (P dx + Q dy) = \iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\,dA,$$ where D is CLOSED domain and $C = \partial D$. Therefore, the integral you are suppose to have is

$$\int_C F \cdot dr = -\left[\iint_{\triangle} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\, dA + \int_3^0 F(t,0) \cdot (1,0) \, dt\right],$$ where, $\triangle$ is the triangular domain of the question, $((0,0),(1,2),(3,0))$, and the the last term is the integral to bound you domain. Note that we integrate from $3 \to 0$ in order to close the domain, and we need the negative of RHS, because the chosen parametrization, that needs to respect the right-hand rule.

So, let $$G(x,y) = \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 2$$

We have \begin{align} \int_C F \cdot dr &= -\left[ \int_0^1\int_0^{2x} G(x,y) \, dy\,dx + \int_1^3\int_0^{-x+3}G(x,y) \,dy\,dx + \int_3^0 F(t,0) \cdot (1,0) \, dt,\right]\\ &= -\left[ \int_0^1\int_0^{2x} 2 \, dA + \int_1^3\int_0^{-x+3} 2 \,dA + \int_3^0 1 \, dt,\right]\\ &= -\left[ 2\int_0^1\int_0^{2x} \, dA + 2\int_1^3\int_0^{-x+3} \,dA - \int_0^3 \, dt,\right]\\ &=-\left[2\dfrac{1\times2}{2} + 2\dfrac{2 \times 2}{2} - 3 \right]\\ &=-3 \end{align}

The partial derivative of $Q$ regarding $x$ lacks a factor $2$.

• This is not a real answer. It's more like a comment. – Guilherme Thompson Dec 16 '15 at 13:17
• @GuilhermeThompson The question was "Where have I gone wrong?", so it is not answering this and not useful? Hm.. – mvw Dec 16 '15 at 13:25
• Somehow yes, but even with the modification you pointed out, his method would still be wrong. As I stated, this was a much smaller mistaken, because it's algebraic... On the other hand, his notion about Green's Theorem was wrong, and this is much more relevant, don't you think, @mvw? – Guilherme Thompson Dec 16 '15 at 13:30
• Someone else might identify another issue, would this justify diminishing the given effort so far? – mvw Dec 16 '15 at 13:48
• Your answer was very incomplete, such that even adjusting the mistaken you spotted he would still get the wrong answer! That's all. – Guilherme Thompson Dec 16 '15 at 14:05