Chapter 16.3, Problem 35E

To determine

(a)

To show:

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$

Solution:

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$

Explanation:

1) Concept:

Compare $F\left(x,y\right)=\frac{-yi+xj}{x^2+y^2}$ and $F\left(x,y\right)=Pi+Qj,$ get $P$ and $Q.$ Find $\frac{\partial P}{\partial y}$ and $\frac{\partial Q}{\partial x}$.

2) Given:

$F\left(x,y\right)=\frac{-yi+xj}{x^2+y^2}$

3) Calculations:

Let $F\left(x,y\right)$ be $\frac{-yi+xj}{x^2+y^2}$

Comparing $F\left(x,y\right)=\frac{-yi+xj}{x^2+y^2}$ and $F\left(x,y\right)=Pi+Qj,$ get

$P=-\frac{y}{{(x}^2+y^2)}$ and $Q=\frac{x}{{(x}^2+y^2)}$

Differentiating $P$ partially with respect to $y,$

$\frac{\partial P}{\partial y}=\frac{\partial }{\partial y}\left(-\frac{y}{{(x}^2+y^2)}\right)$

By using the quotient rule of differentiation,

$\frac{\partial P}{\partial y}=-\left[\frac{{(x}^2+y^2)-y(2y)}{{(x}^2+y^2)^2 }\right]=-\frac{x^2-y^2}{{\left(x^2+y^2\right)}^2}= \frac{y^2-x^2}{{\left(x^2+y^2\right)}^2}\dots ..\left(a\right)$

Differentiating $Q$ partially with respect to $x,$

$\frac{\partial Q}{\partial x}=\frac{\partial }{\partial x}\left(\frac{x}{{(x}^2+y^2)}\right)$

$\frac{\partial Q}{\partial x}=\frac{x^2+y^2-x\left(2x\right)}{{\left(x^2+y^2\right)}^2}=\frac{y^2-x^2}{{\left(x^2+y^2\right)}^2}\dots \dots \left(b\right)$

From $\left(a\right)$ and $\left(b\right),$

$\frac{\partial P}{\partial y}= \frac{\partial Q}{\partial x}$

Conclusion:

 $\frac{\partial P}{\partial y}= \frac{\partial Q}{\partial x}$

(b)

To show:

${\int }_C^{\square }F·dr$ is not independent of path. Check if this contradicts theorem 6 or not.

Solution:

i) The integral $\underset{C}{\overset{}{\int }}F·dr$ is not independent of path.

ii) This does not contradict theorem 6.

Explanation:

1) Concept:

Let $F$ be a continuous vector field defined on a smooth curve $C$ given by a vector function $r\left(t\right), a\le t\le b.$ The line integral of $F$ along $C$ is

${\int }_C^{\square }F·dr={\int }_a^{b}F\left(r\left(t\right)\right)·r\text{'}\left(t\right)dt ={\int }_C^{\square }F·Tds$

2) Given:

$F\left(x,y\right)=\frac{-yi+xj}{x^2+y^2}$

3) Calculation:

i)

By using the given hint,

Let $C_1$ be the upper half circle $x^2+y^2=1$ from $\left(\mathrm{1,0}\right)$ to $\left(-\mathrm{1,0}\right).$

The parametric equation for $C_1$ is

$x=\cos t, y=\sin t, 0\le t\le \pi$

$r\left(t\right)=\cos ti+\sin tj$

Let $C_2$ be the lower half circle $x^2+y^2=1$ from $\left(\mathrm{1,0}\right)$ to $\left(-\mathrm{1,0}\right).$

The parametric equation for $C_2$ is

$x=\cos t, y=\sin t, 2\pi \le t\le \pi$

$r\left(t\right)=\cos ti+\sin tj$

Substitute $x=\cos t, y=\sin t$ in $F\left(x,y\right)=\frac{-yi+xj}{x^2+y^2},$ get $F\left(r\left(t\right)\right)$

$F\left(r\left(t\right)\right)=\frac{-\sin ti+\cos tj}{{{\cos }^{2}t+\sin }^{2}t}$

Differentiating $r\left(t\right)$ with respect to $t,$ get

$r^{\text{'}}\left(t\right)=-\sin ti+\cos tj$

By using the concept,

${\int }_{C_1}^{\square }F·dr=\underset{0}{\overset{\pi }{\int }}\left[\frac{-\sin ti+\cos tj}{{{\cos }^{2}t+\sin }^{2}t}\right]\left[-\sin ti+\cos tj\right]dt$

$=\underset{0}{\overset{\pi }{\int }}\left[\frac{-\sin t·(-\sin t)+\cos t·\cos t}{{{\cos }^{2}t+\sin }^{2}t}\right]dt$

$=\underset{0}{\overset{\pi }{\int }}\left[\frac{{{\cos }^{2}t+\sin }^{2}t}{{{\cos }^{2}t+\sin }^{2}t}\right]dt$

$=\underset{0}{\overset{\pi }{\int }}dt$

By integrating,

$={\left[t\right]}_0^{\pi }$

By using the fundamental theorem of calculus,

$=\pi -0=\pi$

${\int }_{C_1}^{\square }F·dr=\pi$

Also, by using the concept,

${\int }_{C_2}^{\square }F·dr=\underset{2\pi }{\overset{\pi }{\int }}\left[\frac{-\sin ti+\cos tj}{{{\cos }^{2}t+\sin }^{2}t}\right]\left[-\sin ti+\cos tj\right]dt$

$=\underset{0}{\overset{\pi }{\int }}\left[\frac{-\sin t·(-\sin t)+\cos t·\cos t}{{{\cos }^{2}t+\sin }^{2}t}\right]dt$

$=\underset{0}{\overset{\pi }{\int }}\left[\frac{{{\cos }^{2}t+\sin }^{2}t}{{{\cos }^{2}t+\sin }^{2}t}\right]dt$

$=\underset{2\pi }{\overset{\pi }{\int }}dt$

By integrating,

$={\left[t\right]}_{2\pi }^{\pi }$

By using the fundamental theorem of calculus,

$=\pi -2\pi =-\pi$

${\int }_{C_2}^{\square }F·dr=-\pi$

Since, ${\int }_{C_1}^{\square }F·dr\ne {\int }_{C_2}^{\square }F·dr,$ the line integral ${\int }_C^{\square }F·dr$ is not independent of path.

ii)

The domain of $F$ is ${\mathbb{R}}^2-\left\{0\right\}.$

That is,

The domain of $F$ has hole at origin.

This doesn’t contradict theorem 6, since the domain of $F$ is not simply connected.

Conclusion:

i) The integral ${\int }_C^{\square }F·dr$ is not independent of path.

ii) This does not contradict theorem 6.