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Chapter 1: Vector Analysis Problem 1.1: Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are co-planar. b) in the general case.  Solution   Problem 1.2 Is the cross product associative? $$(\vec{A}\times \vec{B}) \times \vec{C} \overset{?}{=} \vec{A}\times (\vec{B} \times \vec{C})$$ If so, prove it; if not, provide a counterexample (the simpler the better). Solution   Problem 1.3 Find the angle between the body diagonals of a cube.  Solution Problem 1.4 Use the cross product to find the components of the unit vector $\hat{n}$ perpendicular to the shaded plane in Fig. 1.11. Solution   Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form. Solution Problem 1.6 Prove that $$[\vec{A}\times (\vec{B}\times \vec{C})]+[\vec{B}\times (\vec{C}\times \vec{A})]+[\vec{C}\times (\vec{A}\times \vec{B})] = 0$$ Under what conditions does $\vec{A}\times (\vec{B}\times \v
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Chapter 1 Vector Analysis: Problem 1.55

Problem 1.55: Check Stokes’ theorem using the function $v = ay\vec{x} + bx \vec{y}$ ($a$ and $b$ are constants) and the circular path of radius $R$, centered at the origin in the $xy$ plane.

 Solution:

Stokes Theorem States That for a vector $\vec{v}$,

$$\int_S (\vec{\nabla} \times \vec{v})\cdot d\vec{a} = \oint_P \vec{v}\cdot d\vec{l}$$ 

Given function is $$\vec{v} = ay\hat{x} + bx\hat{y} $$ where $a$ and $b$ are constants. 

The path given is a circular path of radius $R$ in the $xy$ plane.


 

 

 

 

 

 

 

 In Cartesian coordinate System $$\vec{da} = dxdy\hat{z}$$

 and 

$$\vec{dl} = dx\hat{x} + dy\hat{y}$$

$$ (\vec{\nabla} \times \vec{v})= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ ay & bx & 0 \end{vmatrix}$$ $$\qquad \quad \space = \hat{x}(\frac{\partial}{\partial{y}} (0)- \frac{\partial}{\partial{z}} (x)) - \hat{y}(\frac{\partial}{\partial{x}} (0) - \frac{\partial}{\partial{z}} (ay))+\hat{z}(\frac{\partial}{\partial{x}} (bx) - \frac{\partial}{\partial{y}} (ay))$$

$$\qquad \quad \space = \hat{z}(b-a)$$ 

And hence 

$$\int_S (\vec{\nabla} \times \vec{v})\cdot \vec{da}=\int_S (b-a)\hat{z}\cdot dxdy\hat{z}$$

Since $(b-a)$ is constant 

$$\int_S (\vec{\nabla} \times \vec{v})\cdot \vec{da}=(b-a)\int_S dxdy$$ 

For the given surface 

$$\int_S dxdy = \pi R^2$$ 

Hence, 

$$\int_S (\vec{\nabla} \times \vec{v})\cdot \vec{da}=\pi R^2 \qquad ...(i)$$

$$\oint_C \vec{v}.\vec{dl} = \oint_C (ay\hat{x} + bx\hat{y})\cdot (dx\hat{x}+dy\hat{y})$$ $$\oint_C \vec{v}.\vec{dl} = \oint_C (aydx + bxdy)$$ 

For computing the line integral, Let's divide the circle in upper half $C_1$ and lower half $C_2$. For upper half, $$y = \sqrt{R^2-x^2}$$ and for lower half $$y = -\sqrt{R^2-x^2}$$ 

Let's compute the line integral for upper half first, 

$$dy = \frac{1}{2\sqrt{R^2-x^2}}(-2x)dx = \frac{-x}{\sqrt{R^2-x^2}}dx$$ So, $$\int_{C_1}\vec{v}.\vec{dl}=\int_{C_1} (a\sqrt{R^2-x^2}dx+bx(\frac{-x}{\sqrt{R^2-x^2}}))dx$$ $$\int_{C_1}\vec{v}.\vec{dl}=\int_{C_1} \frac{(a(R^2-x^2)-{bx^2})}{\sqrt{R^2-x^2}}dx$$ $$\int_{C_1}\vec{v}.\vec{dl}=\int_{C_1} \frac{aR^2-(a+b)x^2}{\sqrt{R^2-x^2}}dx$$ Since the Counterclockwise direction is considered to be positive, 

$\therefore x $ varies from $R$ to $-R$. $$\int_{C_1}\vec{v}.\vec{dl}=\int_{R}^{-R} (\frac{aR^2}{\sqrt{R^2-x^2}} - \frac{(a+b)x^2}{\sqrt{R^2-x^2}} )dx$$ $$\int_{C_1}\vec{v}.\vec{dl}=\big\{ {aR^2}.sin^{-1}({x\over R}) - (a+b)\big [{-x\over 2} \sqrt{R^2-x^2}+{R^2\over 2} sin^{-1}({x\over R})\big ]\big \}_R^{-R}$$ $$\int_{C_1}\vec{v}.\vec{dl}=\big\{ \frac{1}{2}(a-b)R^2sin^{-1}({x\over R})\big \}_R^{-R} \qquad \because \int_R^{-R} \frac{-x}{2}\sqrt{R^2-x^2}=0$$ $$\int_{C_1}\vec{v}.\vec{dl}=\frac{1}{2}R^2(a-b)sin^{-1}({-R\over R}) - \frac{1}{2}R^2(a-b)sin^{-1}({R\over R})$$ $$\int_{C_1}\vec{v}.\vec{dl}=\frac{1}{2}R^2(a-b)({-\pi \over 2}-{\pi \over 2})$$ $$\int_{C_1}\vec{v}.\vec{dl}=\frac{1}{2}R^2(b-a)\pi$$ For the second integral, $$dy = -\frac{1}{2\sqrt{R^2-x^2}}(-2x)dx = \frac{x}{\sqrt{R^2-x^2}}dx$$ $$\int_{C_2}\vec{v}.\vec{dl}=\int_{C_2} (a(-\sqrt{R^2-x^2})dx+bx(\frac{x}{\sqrt{R^2-x^2}})dx)$$ $$\int_{C_2}\vec{v}.\vec{dl}=\int_{C_2} \frac{-aR^2+(a+b)x^2}{\sqrt{R^2-x^2}}dx$$ Here $x$ varies from $-R$ to $R$ and hence $$\int_{C_2}\vec{v}.\vec{dl}=\int_{-R}^{R} \frac{-aR^2+(a+b)x^2}{\sqrt{R^2-x^2}}dx$$ Reversing the limits we get, $$\int_{C_2}\vec{v}.\vec{dl}=\int_{R}^{-R} \frac{aR^2-(a+b)x^2}{\sqrt{R^2-x^2}}dx = \int_{C_1}\vec{v}.\vec{dl}$$ and hence $$\oint_C \vec{v}\cdot \vec{dl} =\frac{1}{2}R^2(b-a)\pi + \frac{1}{2}R^2(b-a)\pi $$ $$\oint_C \vec{v}\cdot \vec{dl} =\pi R^2(b-a)$$ Hence stokes theorem is proved. 

 

If you have any doubt regarding the solution or you want solution of some problem which is not posted please let me know by commenting. This encourages me to answer more question because sometime it feels like all I am doing is just a waste. If it helps someone I will be happy to do it.


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