Going through the proof for Green's Theorem there is one step that I am not clear about.

$$ \begin{eqnarray} \int_C M dx+Ndy &=& \iint_R\bigg(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\bigg)dA\\[10pt] \Rightarrow\quad \int_C Mdx &=& \int_{C_1}Mdx + \int_{C_2}Mdx\\ &=& \int_a^b M(x,f_1(x)) dx + \int_b^aM(x,f_2(x)) dx\\ &=& \int_a^b\big[M(x,f_1(x))-M(x,f_2(x))]dx \end{eqnarray} $$

The last step I am unclear about as the fundamental theorem of line integrals states

$$ \begin{eqnarray} \int_C \mathbf{F}\cdot d\mathbf{r} &=& f(x(b),y(b)) - f(x(a),y(a))\\ \mathbf{r}(t) &=& x(t)\mathbf{i} + y(t) \mathbf{j} \qquad a \leq t \leq b \end{eqnarray} $$

Given the integrate is equal to $f_b - f_a$ why isn't the last line of the proof $f_2 - f_1$? Does the converse mean we are treating the two functions as separate graphs where one represents the top half and the other the bottom? Is this what the horizontal and vertical simplicity refers to?

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Image credits: Centrage Learning. Calculus 9th Edition by Larsson, Edwards.

Edit: I have changed the limits of integration along $ C_2 $ to $ \int_b^a $.

  • $\begingroup$ There is nothing to do with fundamental theorem of line integrals. Just by the definition of Riemann integral since the formula you deal with only involves Riemann integral, not line integral. $\endgroup$
    – Bach
    Apr 20, 2018 at 17:31
  • $\begingroup$ Also, I corrected the limits of the integratoin along $ C_2 $ to $ \int_b^a $ $\endgroup$
    – Bach
    Apr 20, 2018 at 17:32
  • $\begingroup$ isn't the first line here the conclusion? $\endgroup$ Apr 20, 2018 at 17:51

2 Answers 2


While since you have already got

$$ \begin{align} I_1&=\int_{C}M dx \\&= \int_a^b [M(x, f_1(x))-M(x, f_2(x))] dx \end{align} .$$

By the same method you obtain

$$ \begin{align} I_2&=\int_{C}N dy \\&= \int_c^d [N(x, g_2(x))-N(x, g_1(x))] dy \end{align} $$.

Now I suggest look at the RHS,

$$ \begin{align} J_1&=\iint_{R}\frac{\partial N}{\partial x} dA \\&= \int_c^d \left(\int_{g_1(y)}^{g_2(y)}\frac{\partial N}{\partial x}dx\right)dy \\&=\int_c^d [N(x, g_2(y))-N(x, g_1(y))]dy\\ &= I_2 \end{align}$$

The same reason you get

$$ \begin{align} J_2 &=-\iint_{R}\frac{\partial M}{\partial y} dA \\&= -\int_a^b \left(\int_{f_1(x)}^{f_2(x)}\frac{\partial M}{\partial x}dy\right)dx \\&=-\int_a^b [M(x, f_2(x))-M(x, f_1(x))]dy \\&= I_1 \end{align}$$

Now add them together we get $ I_1+I_2=J_2+J_1 $ which is the Green's Theorem.


I notice that the symbol $f$ is being overused. That may be why the OP is having difficulties. In the original post, they use $\mathbf{F}=\nabla f(x,y)$ where $f(x,y)$ is the scalar potential for $\mathbf{F}$. That's a different $f$ than in the line integrals where we parameterize $y_1=f_1(x)$ and $y_2=f_2(x)$. For clarity, I avoided the use of $f$ altogether in my answer.

The object of interest here is $$\oint_C \mathbf{F}\cdot d\mathbf{r} \qquad;\quad\mathbf{F}= M(x,y)\hat{x}+N(x,y)\hat{y}$$

If you assume that $\mathbf{F}$ is a conservative field such that $\mathbf{F}=\nabla \phi(x,y)$ is the gradient of a scalar function $\phi(x,y)$, then yes, the gradient theorem

\begin{eqnarray} \int_{C_i} \mathbf{F}\cdot d\mathbf{r} &=& \phi(x(b),y(b)) - \phi(x(a),y(a))\\ \end{eqnarray}

would apply and the integral would vanish. But Green's theorem is more general than that. For a general $\mathbf{F}$ (i.e. not necessarily conservative) the closed contour integral need not vanish. That's why $C$ is separated into two portions $C_1$ and $C_2$ which both start and end at points $a$ and $b$.

The proof then goes on to parameterize $M$ and $N$ on either half of the curve. There are two simple ways to go about that: either choose $C_1,C_2$ to be, crudely speaking, the bottom and top halves, or choose them to be the left and right halves of $C$. If you go with the former choice, then you can parameterize the values of $y=y_1(x)$ and $y=y_2(x)$ as functions of $x$, leading to

$$ \int_C M dx= \int_a^b\big[M(x,y_1(x))-M(x,y_2(x))]dx $$


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