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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}...
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Chapter 1 Vector Analysis: Problem 1.56

Problem 1.56: Compute the line integral of
$$v = 6 \hat{x} + yz^2 \hat{y} + (3y + z) \hat{z}$$
along the triangular path shown in Fig. 1.49. Check your answer using Stokes’
theorem.

 

Solution: 

$$\vec{v} = 6\hat{i} + yz^2\hat{j} + (3y+z)\hat{k}$$ 
$$\vec{dl} = dx\hat{i} + dy\hat{j} + dz \hat{k}$$ 
Break the line integral into 3 parts where $L_1$ denotes the line integral where $y$ changes from $0$ to $1$, $x=0, z=0$ 
$$\int_{L_1}\vec{v}\cdot \vec{dl} =\int_{L_1} (6\hat{i} + 3y\hat{k})\cdot (dy\hat{j}) = 0$$ $L_2$ denotes the line integral of the line $z = -2y+2, x=0$ $dz = -2dy$ 
$$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (6\hat{i} + yz^2\hat{j} + (3y+z)\hat{k})\cdot (dy\hat{j} + dz\hat{k})$$ 
$$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (yz^2dy + (3y+z)dz$$ 
$$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_{L_2} (y(-2y+2)^2dy + (3y+(-2y+2))(-2dy)$$ $$\int_{L_2}\vec{v}\cdot \vec{dl} =\int_1^0 (4y^3 - 8y^2+2y-4)dy$$ 
$$\int_{L_2}\vec{v}\cdot \vec{dl} =\big[y^4 - \frac{8y^3}{3}+y^2-4y \big]_1^0= \frac{8}{3} + 2$$ 
$L_3$ denotes the line integral where $z$ changes from $2$ to $0$, $x=0, y=0$ 
$$\int_{L_3}\vec{v}\cdot \vec{dl} =\int_{L_3} (6\hat{i} + z\hat{k})\cdot (dz\hat{k}) $$ $$\int_{L_3}\vec{v}\cdot \vec{dl} =\int_2^0 z dz = -2 $$ 
$$\oint_L \vec{v}\cdot \vec{dl} = \int_{L_1} \vec{v}\cdot \vec{dl} + \int_{L_2} \vec{v}\cdot \vec{dl} + \int_{L_3} \vec{v}\cdot \vec{dl} = \frac{8}{3}$$ 
$$\vec{\triangledown} \times \vec{v} = \begin{vmatrix} \hat{i} && \hat{j} && \hat{k}\\ {\partial \over \partial x} && {\partial \over \partial y} && {\partial \over \partial z}\\ 6 && yz^2 && (3y+z) \end{vmatrix} $$ 
$$\vec{\triangledown} \times \vec{v} = \hat{i}(3-2yz) -0\cdot\hat{j} + 0\cdot\hat{k} = \hat{i}(3-2yz)$$ 
$$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} =\oint_S (3-2yz)\hat{i}\cdot dydz\hat{i} = \oint_S (3-2yz) dy dz$$ 
Here when $y$ varies from $0$ to $1$, $z$ varies from $0$ to $-2y+2$ 
$$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \int_0^1 \int_0^{-2y+2}(3-2yz)dzdy$$ 
$$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \int_0^1\big[ \big[3z-yz^2\big]_0^{-2y+2}dz\big]dy$$ 
$$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \int_0^1 \big[3(-2y+2)-y(-2y+2)^2\big]dy$$ 
$$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \int_0^1 \big[-4y^3 + 8y^2 -10y+6\big]dy$$ 
$$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = \big[-y^4 + \frac{8}{3}y^3 -5y^2+6y\big]_0^1 $$ 
$$\oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da} = -1+\frac{8}{3}-5+6 = \frac{8}{3}$$ 
Hence, $$\oint_L \vec{v}\cdot \vec{dl} = \oint_S (\vec{\triangledown} \times \vec{v})\cdot \vec{da}= \frac{8}{3}$$ 
Hence, Stokes theorem is verified.











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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