Let $E$ denote the ellipse
$$E=\{(x, y):\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\}$$
The question is, by direct computation, showing that
$$\int \int_{\mbox{inside E}} div X dA = \int_E X n dl $$
where $l$ denotes length, and $n$ denotes the outward pointing normal vector
at point $(x, y)$ on the ellipse, and $X=(y, x)$. First of all,
since $div X=0$, the integral should be zero, but I'm not sure how to compute
$\int_E X n dl $ in this case. In particular, what should the expression for $n$ be?
For the ones involving three coordinates, $(x, y, z)$, I heard that you use cross product, but
what should be done for this case? Any help would be greatly appreciated. Once I figure out how to compute $n$ for this kind of cases, I think I will be set.
Answer
The integral is a line integral that can be evaluated by parameterization. The usual choice on an ellipse is
$$
\mathbf r(t) = \langle x(t),y(t) \rangle = \langle a \cos t, b \sin t \rangle, \quad 0 \le t \le 2\pi, \quad a,b > 0.
$$
The tangent vector is $$\mathbf r'(t) = \langle -a \sin t, b \cos t \rangle,$$
and the arclength element is
$$
dl = |\mathbf r'(t)| dt = \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} \, dt.$$ There are various ways to find an exterior normal vector, but in this case note that $\langle b \cos t, a \sin t \rangle$ is orthogonal to $\mathbf r'(t)$ and points away from the origin. The unit exterior normal is thus $$ \mathbf n(t) = \frac{1}{\sqrt{a^2 \sin^2 t + b^2 \cos^2 t}} \langle b \cos t, a \sin t \rangle.$$ Your vector field is $$\mathbf X(t) = \langle y(t),x(t) \rangle = \langle b \sin t, a \cos t \rangle$$ so that $$ \mathbf X(t) \cdot \mathbf n(t) = \frac{(a^2 + b^2) \sin t \cos t}{\sqrt{a^2 \sin^2 t + b^2 \cos^2 t}}.$$ Put it all together to get $$ \oint_E \mathbf X \cdot \mathbf n \, dl = \int_0^{2\pi} \mathbf X(t) \cdot \mathbf n(t) |\mathbf r'(t)| \, dt = (a^2 + b^2) \int_0^{2\pi} \sin t \cos t \, dt.$$
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