Solution To MAT304 Test



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This is my first post on vector analysis, i will be solving questions involving the stokes theorem, greens theorem and divergence of a vector field.
Now to understand all this, you must have a strong knowledge of fundamental calculus. Below are the questions and solutions.


Please kindly use the comment box or contact form on the homepage of this blog for any comment, corrections and observations, because i cannot guarantee a $100\%$ accuracy due to typographic errors and others errors made during the process of solving.

Questions

NOTE: The initial solution to problem 1 and 2 had few mistakes, so i updated the solution.

(1) Evaluate the integral $\int_c{y^2dx+z^2dy+x^2dz}$, where $C$ is the triangular closed path joining the points $(0,0,0),(0,a,0),(0,0,a)$, by transforming the integral to surface integral using stoke's theorem.
(2)Verify the Green's theorem in the form $\oint_c(3x^3-8y^3)dx+(2y-3xy)dy$ where $C$ is the boundary of the region defined by $y=x,y=x^2$.
(3) Show that $div(\bigtriangledown{\phi_{1}}\times{\bigtriangledown{\phi_2}})=0$.

Solution 1

 We are to apply the stoke's theorem to evaluate the integral


graphics image below

The stokes theorem is given by $\oint{A.dr}=\iint_s(\bigtriangledown\times{A}).nds$.
where $n$ is the unit normal vector.
Now we are to obtain $A$ from the given integral by $\int_c{A.dr}=\int_c[y^2dx+z^2dy+x^2dz]$
to obtain $A$ factor out $dx,dy,dz$ so that we obtain an integral vector.
$=\int(y^2i+z^2j+x^2k)(idx+jdy+kdz)\Rightarrow{y^2i+z^2j+x^2k}$
Now we can easily let $A=y^2i+z^2j+x^2k$
$\therefore$ $curl.A=\bigtriangledown\times{A}$
\begin{equation}
\begin{vmatrix}
i    &    j     &    k\\
\frac{\mathfrak{d}}{\mathfrak{d}x}    &    \frac{\mathfrak{d}}{\mathfrak{d}x}        &    \frac{\mathfrak{d}}{\mathfrak{d}z}\\
y^2        &        z^2        &        x^2
\end{vmatrix}
\end{equation}

We take the determinant of the matrix
$i[\frac{d}{dy}(x^2)-\frac{d}{dz}(z^2)]-j[\frac{d}{dx}(x^2)-\frac{d}{dz}(y^2)]+k[\frac{d}{dx}(z^2)-\frac{d}{dy}(y^2)]$
by taking the partial derivatives we get
$=i(-2z)-j(2x)+k(-2y)$
$=-2(zi+xj+yk)$
Since the x-coordinate of each vertex $x$ of triangle is $0$ therefore the triangle is in the $y-z$ plane
$(\bigtriangledown\times{A}).n=(-2zi-2xj-2yk).i$
The equation of the line $AB$ is $z=y$
$\int_c{A}.dr=\iint_R(-2z)ds$
$=\int^a_0dy\int^0_y(-2z)dz$
$\int^a_0dy[-z^2]^0_y=\int^a_0y^2dy$
$=[\frac{y^3}{3}]^a_0$
$=\frac{a^3}{3}$

Solution 2

Since the boundary of the region is given by $y=x$ and
$y=x^2$, then we substitute $y=x^2, dy=2xdx$ into the integral
$=\oint(3x^3-8(x^2)^3)dx+(2(x^2)-3x(x^2))(2x)dx$
$=\oint(3x^3-8x^6)dx+(4x^3-6x^4)dx$
$=\oint(-8x^6-6x^4+7x^3)dx$
Along the curve $(1,0)$
$=\int^1_0(-8x^6-6x^4+7x^3)dx$
$=\frac{-8x^7}{7}-\frac{6x^5}{5}+\frac{7x^4}{4}|^1_0$
$=\frac{-8}{7}-\frac{6}{5}+\frac{7}{4}=\frac{-160-168+245}{140}$
$=\frac{-83}{140}$..............(1)
Now along the $y=x, dy=dx$ from $(1,1)$ to $(0,0)$
$\int^0_1(3x^3-8(x)^3)dx+(2(x)-3x(x))dx$
$=\int^0_1(3x^3-8x^3)dx+(2x-3x^2)dx$
$\int^0_1(-5x^3-3x^2+2x)dx=\frac{-5x^4}{4}-\frac{3x^3}{3}+\frac{2x^2}{2}|^0_1$
$0-(\frac{-5}{4}-1+1)=0-(\frac{-5-4+4}{4})=\frac{5}{4}$........(2)
The require line integral is thus:
=eqn(1)-eqn(2)
$\frac{-83}{140}-\frac{5}{4}=\frac{-83-175}{140}=\frac{-258}{140}$
Now we proceed to evaluate using Green's theorem
$\oint(M\mathfrak{d}x+N\mathfrak{d}y)=\iint_R(\frac{\mathfrak{d}N}{\mathfrak{d}x}-\frac{\mathfrak{d}M}{\mathfrak{d}y})dxdy$
Using the integral
$\oint_c(3x^3-8y^3)dx+(2y-3xy)dy$
Let $M=(3x^3-8y^3)dx$ and $N=(2y-3xy)dy$
so that
$\int^1_{x=0}\int^x_{y=x^2}(\frac{\mathfrak{d}}{\mathfrak{d}x}(2y-3xy)-(\frac{\mathfrak{d}}{\mathfrak{d}y}(3x^3-8y^3))dxdy$
$\int^1_{x=0}\int^x_{y=x^2}[(-3y)-(-24y^2)]dxdy$
$\int^1_{x=0}\int^x_{y=x^2}[-3y+24y^2]dydx$
$\int^1_{x=0}[\frac{-3y^2}{2}+\frac{24y^3}{3}]^x_{y=x^2}$
$\int^1_{x=0}[\frac{-3(x)^2}{2}+8x^3-(\frac{-3(x^2)^2}{2}+\frac{24(x^2)^3}{3})]$
$\int^1_{x=0}[\frac{-3x^2}{2}+8x^3+\frac{3x^4}{2}-8x^6]$
$=\frac{3x^3}{6}+\frac{8x^4}{4}+\frac{3x^5}{10}-\frac{8x^7}{7}|^1_0=\frac{-x^3}{2}+2x^4+\frac{3x^5}{10}-\frac{8x^7}{7}|^1_0$
$=(\frac{-1}{2}+2+\frac{3}{10}-\frac{8}{7})-0$
$\frac{-35+140+21-80}{70}=\frac{46}{70}=\frac{23}{35}$


Solution 3

$div(\bigtriangledown{\phi_{1}}\times{\bigtriangledown{\phi_2}})=0$.

We are to show that $\bigtriangledown(\bigtriangledown{\phi_{1}}\times{\bigtriangledown{\phi_2}})=0$.
\begin{equation}
=\bigtriangledown
\begin{vmatrix}
i    &    j    &    k\\
\frac{\mathfrak{d}\phi_1}{\mathfrak{d}x}&\frac{\mathfrak{d}\phi_1}{\mathfrak{d}y}&\frac{\mathfrak{d}\phi_1}{\mathfrak{d}z}\\
\frac{\mathfrak{d}\phi_2}{\mathfrak{d}x}    &  \frac{\mathfrak{d}\phi_2}{\mathfrak{d}y}    & \frac{\mathfrak{d}\phi_2}{\mathfrak{d}z}
\end{vmatrix}
\end{equation}

$=\bigtriangledown[i(\frac{\mathfrak{d}\phi_1\phi_2}{\mathfrak{d}y\mathfrak{d}z}-\frac{\mathfrak{d}\phi_1\phi_2}{\mathfrak{d}y\mathfrak{d}z})-j(\frac{\mathfrak{d}\phi_1\phi_2}{\mathfrak{d}x\mathfrak{d}z}-\frac{\mathfrak{d}\phi_1\phi_2}{\mathfrak{d}x\mathfrak{d}z})+k(\frac{\mathfrak{d}\phi_1\phi_2}{\mathfrak{d}x\mathfrak{d}y}-\frac{\mathfrak{d}\phi_1\phi_2}{\mathfrak{d}x\mathfrak{d}y})]$
$=\bigtriangledown(i(0)-j(0)+k(0))$
$=\frac{\mathfrak{d}(0)}{\mathfrak{d}x}+\frac{\mathfrak{d}(0)}{\mathfrak{d}y}+\frac{\mathfrak{d}(0)}{\mathfrak{d}z}$
$=0$, $\Box$