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Let C be a smooth simple closed curve in the xy-plane with the property that lines parallel to the axes cut it at no more than two points. Let R be the region enclosed by C and suppose that M, N, and their first partial derivatives are continuous at every point of some open region containing C and R. We want to prove the circulation-curl form of Green’s Theorem:

$$\oint_{C} Mdx+Ndy = \iint\frac{ \partial N}{\partial x} - \frac{\partial M}{\partial y}dxdy$$ The boundary curve $C$ is made up of $C_1$, the graph of y = $ƒ_1(x)$, and $C_2$ , the graph of $y$ = $ƒ_2(x)$.

The figure: The boundary curve $C$ is made up of $C_1$, the graph of y = $ƒ_1(x)$, and $C_2$ , the graph of $y$ = $ƒ_2(x)$.

The link above shows C made up of two directed pairs, $C_1 : y=f_1(x), a \leq x \leq b$, $C_2:$ $ b \geq x \geq a $

For any x between a and b, we can integrate $ \frac {\partial M}{\partial y}$ with respect to y from $y=f_1(x)$ to $ y=f_2(x)$ and obtain

$$\int_{f{_1}(x)}^{f{_2}(x)} \frac{\partial M}{\partial y} = M(x,y) \Big|_{y=f_1{(x)}}^{y=f_2(x)} = M(x,f_2(x))-M(x,f_1(x))$$.

We can then integrate this with respect to $x$ from $a$ to $b$:

$$\int_a^b\int_{f_1(x)}^{f_2(x)}\frac{\partial M}{\partial y}dydx = \int_{a}^{b} (M(x,f_2(x))-M(x,f_1(x))dx$$ ...

Please open to link shaded into blue to see the picture.

Please help me understand why we're integrating $\frac{\partial M}{\partial{y}}$ with the bounds $y=f_1(x)$ and $y=f_2(x)$. One case is if $ x=a$ or $ x=b$, then $f_1(x)=f_2(x)$ and what else would happen? The other case is if x $\neq a $and $x\neq b$.

Does the value of $M(x,f_2(x))-M(x,f_1(x))$ doesn't depend on the bounds of the integral $f_2(x)$ and $f_1(x)$?. I assume the upper and lower bounds $y=y_2(x)$ and $y=y_1(x)$ respectively are real numbers.

enter image description here

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I really am having a hard time with how to answer you here. You are integrating $$\iint_R \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} dA = \iint_R\frac{\partial N}{\partial x}dA - \iint_R\frac{\partial M}{\partial y}dA$$ over the region $R$. Apparently right now you are only concentrating on the 2nd integral.

Now in general, double integrals can be calculated as iterated integrals. For each $x_0$, define the set $R_{x_0} = \{ y \mid (x_0, y) \in R\}$. and suppose that $x = a$ is the least value of $x$ of any point in $R$, while $x = b$ is the greatest such value. Then

$$\iint_R g(x,y) \,dA = \int_a^b\int_{R_x} g(x,y)\,dy\,dx$$

for integrable functions $g$. Now in this case, we are given that $R_x$ is the interval $[f_1(x), f_2(x)]$. That is what all that discussion about $C$ is saying. So $$\int_{R_x} g(x,y)\,dy = \int_{f_1(x)}^{f_2(x)} g(x,y)\,dy$$

That is why you are integrating $\frac{\partial M}{\partial y}$ with respect to $y$ from $f_1(x)$ to $f_2(x)$.

When $x = a$ or $x = b$, of couse the integral is $0$, since $f_1(a) = f_2(a)$ and $f_1(b) = f_2(b)$. But for other values of $x$, it will be be non-zero. Note that for the purposes of this integral with respect to $y$, $x$ is just a constant, and so are $f_1(x)$ and $f_2(x)$. So by the FTC, you get $$\int_{R_x}\frac{\partial M}{\partial y}\,dy = \int_{f_1(x)}^{f_2(x)} \frac{\partial M}{\partial y}\,dy=M(x,y) \Big|_{y=f_1{(x)}}^{y=f_2(x)} = M(x,f_2(x))-M(x,f_1(x))$$ In all of that we are holding $x$ constant. But we get a different result for each $x$, which is the function of $x$ that the outer integral is integrating. $$\iint_R \frac{\partial M}{\partial y} \,dA = \int_a^b\int_{R_x} \frac{\partial M}{\partial y}\,dy\,dx = \int_{a}^{b} (M(x,f_2(x))-M(x,f_1(x)))dx$$


Yes, the value of $M(x,f_2(x))-M(x,f_1(x))$ depends on $f_1(x)$ and $f_2(x)$, Quite obviously so, unless $M$ is constant with respect to $y$. Nothing has been done that depends on them not varying.

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