STEM Problem Detail

Problem Statement

Prompt

Provide the step by step solution of Ex. 1.

Solution Rendering...

**Solution. Part (i)**

We have
\[
\operatorname{curl}\,\mathbf{q}
=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\
-\dfrac{y}{x^{2}+y^{2}} & \dfrac{x}{x^{2}+y^{2}} & 0
\end{vmatrix}
=
\mathbf{k}
\left[
\frac{(y^{2}-x^{2})+(x^{2}-y^{2})}{(x^{2}+y^{2})^{2}}
\right]
=0.
\]

Hence,
\[
\operatorname{curl}\,\mathbf{q}=0
\]
everywhere except at the origin. Thus the flow is irrotational. It has a
singularity at the origin where the velocity becomes infinite.

**(a)** Draw a square in the Cartesian plane with vertices
\[
A(1,0),\quad B(2,0),\quad C(2,1),\quad D(1,1).
\]

The circulation around the square \(ABCD\) is
\[
\Gamma=\oint \mathbf{q}\cdot d\mathbf{r}
=
\int_A^B \mathbf{q}\cdot d\mathbf{r}
+
\int_B^C \mathbf{q}\cdot d\mathbf{r}
+
\int_C^D \mathbf{q}\cdot d\mathbf{r}
+
\int_D^A \mathbf{q}\cdot d\mathbf{r}.
\tag{1}
\]

Along \(AB\) (\(y=0\)),
\[
d\mathbf{r}=dx\,\mathbf{i},
\]
and
\[
\int_A^B \mathbf{q}\cdot d\mathbf{r}
=
\int_{1}^{2}
\left[
\frac{-iy+jx}{x^{2}+y^{2}}
\right]_{y=0}
\cdot (dx\,\mathbf{i})
=0,
\tag{2}
\]
since \(\mathbf{j}\cdot\mathbf{i}=0\).

Along \(CD\) (\(y=1\)),
\[
\int_C^D \mathbf{q}\cdot d\mathbf{r}
=
\int_{2}^{1}
\left[
\frac{-iy+jx}{x^{2}+y^{2}}
\right]_{y=1}
\cdot (dx\,\mathbf{i})
=
\int_{1}^{2}\frac{dx}{x^{2}+1}
=
\tan^{-1}2-\tan^{-1}1.
\tag{3}
\]

Along \(BC\) (\(x=2\)),
\[
d\mathbf{r}=dy\,\mathbf{j},
\]
so
\[
\int_B^C \mathbf{q}\cdot d\mathbf{r}
=
\int_{0}^{1}
\left[
\frac{-iy+jx}{x^{2}+y^{2}}
\right]_{x=2}
\cdot (dy\,\mathbf{j})
=
\int_{0}^{1}\frac{2\,dy}{y^{2}+4}
=
\tan^{-1}\!\left(\frac12\right).
\tag{4}
\]

Along \(DA\) (\(x=1\)),
\[
\int_D^A \mathbf{q}\cdot d\mathbf{r}
=
\int_{1}^{0}
\left[
\frac{-iy+jx}{x^{2}+y^{2}}
\right]_{x=1}
\cdot (dy\,\mathbf{j})
=
-\tan^{-1}1.
\tag{5}
\]

Using (2), (3), (4) and (5) in (1),
\[
\Gamma
=
\tan^{-1}\!\left(\frac12\right)
+\tan^{-1}2
-\tan^{-1}1
-\tan^{-1}1.
\]

Hence,
\[
\Gamma
=
\cot^{-1}2+\tan^{-1}2-\frac{\pi}{4}-\frac{\pi}{4}
=
\frac{\pi}{2}-\frac{\pi}{4}-\frac{\pi}{4}
=0,
\]
since
\[
\tan^{-1}2+\cot^{-1}2=\frac{\pi}{2}.
\]

Thus, the circulation around the square is zero, which also follows directly
from Stokes’s theorem since \(\operatorname{curl}\,\mathbf{q}=0\) inside the region.

**(b)** For circulation around the unit circle centred at the origin, use
polar coordinates:
\[
x=r\cos\theta,\qquad y=r\sin\theta.
\]

Then
\[
\mathbf{q}
=
\frac{-\sin\theta\,\mathbf{i}+\cos\theta\,\mathbf{j}}{r}
=
u\,\mathbf{i}+v\,\mathbf{j},
\]
where
\[
u=-\frac{\sin\theta}{r},
\qquad
v=\frac{\cos\theta}{r}.
\]

Hence,
\[
q_r=u\cos\theta+v\sin\theta=0,
\qquad
q_\theta=-u\sin\theta+v\cos\theta=\frac{1}{r}.
\]

Therefore,
\[
\Gamma=\oint \mathbf{q}\cdot d\mathbf{r}
=
\int_{0}^{2\pi}\left(\frac{1}{r}\cdot r\right)d\theta
=2\pi.
\]

**Note.**

1. Unlike part (a), the circulation is nonzero since the curve encloses the
origin where a singularity exists.
2. The circulation is independent of the radius; for any curve enclosing the
origin,
\[
\Gamma=2\pi.
\]

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Problem #35