Chapter 36
By the end of this section, you should be able to answer the following questions:
The following diagram summarises the relationships between conservative vector fields, path independent line integrals and closed line integrals we have seen so far.
Suppose a function of two variables $f$ is defined on a disc $D$ that contains the point $(a,b)$. If the functions $$\displaystyle \frac{\partial^2 f}{\partial x\partial y}\;\; \text{ and }\;\; \displaystyle \frac{\partial^2 f}{\partial y\partial x}$$ are both continuous on $D$, then \[ \frac{\partial^2 f}{\partial x\partial y}(a,b) = \frac{\partial^2 f}{\partial y\partial x}(a,b). \]
Say we have a conservative vector field $\F = F_1~\i+F_2~\j$. This means that there exists an $f(x,y)$ such that \[ F_1 = \frac{\partial f}{\partial x}, \quad F_2 = \frac{\partial f}{\partial y}. \]
An immediate consequence of Clairaut's theorem is that \[ \frac{\partial F_1}{\partial y} = \frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x} = \frac{\partial F_2}{\partial x}. \]
In other words, we have the following:
If $\F = F_1~\i + F_2~\j$ is a conservative vector field, then \[ \frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}. \]
Let's add this to our diagram:
If we can reverse the new arrow, then we would have the criterion that we need!
That is, the condition \[ \frac{\partial F_1}{\partial y} = \frac{ \partial F_2}{\partial x} \] would be a test for a conservative vector field. To do this, we require one more piece of the puzzle. That is Green's theorem. |
Let $D$ be a region in the $xy$-plane bounded by a piecewise-smooth, simple closed curve $C,$ which is traversed with $D$ always on the left. Let $F_1(x,y),$ $F_2(x,y),$ $\displaystyle \frac{\partial F_1}{\partial y}$ and $\displaystyle \frac{\partial F_2}{\partial x}$ be continuous in $D$. Then
\[ \iint_D\left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) ~dx~dy = \oint_C\left(F_1 dx + F_2 dy\right) \]
\[ \iint_D\left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) ~dx~dy = \oint_C\left(F_1 dx + F_2 dy\right) \]
This theorem relates a double integral to a line integral over a closed curve. For example, we can use Green's theorem to evaluate complicated line integrals by treating them as double integrals, or vice versa.
Regarding our discussion on conservative vector fields, we have the following corollary to Green's theorem: \[ \text{If } \;\frac{\partial F_1}{\partial y}= \frac{\partial F_2}{\partial x}, \;\text{ then } \;\oint_C \F \pd d\r = 0. \]
Note that $\F = F_1~ \i + F_2 ~\j$.
If we add this to our diagram, we can now link any four statements via the arrows.
In other words all four statements are equivalent.
In particular, we now have a test to determine whether or not a given two dimensional vector field is conservative:
The vector fied $\F$ is conservative |
Let's use Green's theorem: \[ \oint_C\F \pd d\r = \iint_D\left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) ~dx~dy \] \[ \quad \;\;=\iint_D\left( y^2- x^2\right) ~dx~dy \] Here we can use polar coordinates: $x= r \cos \theta,$ $y = \sin \theta.$ So we have \[ D = \left\{ (r, \theta) ~|~ 0\leq r\leq a, 0 \leq \theta \leq 2 \pi \right \} \] and $y^2-x^2 $ $= r^2\left(\sin^2\theta -\cos^2\theta \right)$ $= - r^2 \cos \left(2\theta\right)$. |
Then
Notice that $\F$ is not conservative! |
\[ \oint_C\big(F_1 dx + F_2 dy\big) =\iint_D\left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) ~dx~dy \]
where $C$ is the
path from $(0,1)$ to $(1,0)$ along $y = (x-1)^2$
and then from
$(1,0)$ to $(2, 1)$ along $y = x - 1$.
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Can we use Green's theorem here? Yes, but we need a closed curve. For example consider $C_3$ the segment from $(2,1)$ to $(0,1)$. |
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Then by Green's theorem we have
\[ \text{Thus }\;\;\int_C \F \pd d\r = -\int_{C_3}\F \pd d\r. \] |
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\[ \int_{C_3}\F \pd d\r = \;? \] On $C_3$ we have that $y=1$ and $x$ varies. A parametrisation of $C_3$ is: \[ \r(t) = 2(1-t)~\i+\j,\quad t\in[0,1]. \] Then $ \F\left(\r(t)\right) = 4(1-t)~\i + \left(4\left(1-t\right)^2+3 \right)\j,\, $ $\,d\r = (-2~\i)~dt.$ |
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$ \F\left(\r(t)\right) = 4(1-t)~\i + \left(4\left(1-t\right)^2+3 \right)\j,\, $ $\,d\r = (-2~\i)~dt.$ So
$\text{Hence }\;\;\displaystyle \int_C \F \pd d\r = -\int_{C_3}\F \pd d\r $ $=4$. |
the curve given by $\r(t) = \big(1-\cos(\pi t)\big)\i + \left(1+\sin^3(\pi t)\right)\j$ for $0\leq t\leq 1/2$.
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Direct computation using $ \r(t) $ is complicated. But notice that \[ \frac{\partial F_1 }{\partial y} = 2x =\frac{\partial F_2 }{\partial x} \] This implies that $\F$ is conservative. In other words, $\displaystyle \int_C \F \pd d\r\;$ is path independent. $\displaystyle \int_C \F \pd d\r = \int_{C_1} \F \pd d\r $ |
Let's use a simpler path: \[ C_1:= \r(t) = t~\i + (t+1)~\j \] with $0\leq t\leq 1.$ Then
You should check this! 📝 |
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We can also use the Fundamental Theorem for Line Integrals. Note that $\F= \nabla f,$ where \[ f = 3x+x^2y-y^3 + \text{constant}. \] Then
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