Line Integral of circle in counterclockwise direction

[songoku]

Homework Statement

Please see below

Relevant Equations

Line Integral
Parametric

My attempt:
Let $x=a \cos \theta$ and $y=a \sin \theta$

$$\int_{L} xy^2 dx-x^2ydy$$
$$=\int_{0}^{2\pi} \left( (a\cos \theta)(a\sin \theta)^2 (-a\sin \theta)-(a\cos \theta)^2 (a \sin \theta)(a\cos \theta)\right) d\theta$$
$$=-a^4 \int_{0}^{2\pi}\left( \sin^3 \theta \cos \theta+\cos^3 \theta \sin \theta \right) d\theta$$

I get zero as the result of the integration. Is it possible? Thanks

 

[Orodruin]

Yes, the integral is zero by symmetry.

 

[Orodruin]

Yes, the integral is zero by symmetry.

To qualify this:

Take the first term and consider the integral along the lower half-circle. You obtain $y^2 = a^2 - x^2$ and can parametrize it by $-a < x < a$. The integral along the lower half-circle is therefore
$$
\int_{-a}^a y^2 x \, dx = \int_{-a}^a (a^2 - x^2) x \, dx
$$
which is an integral of an odd function over an even interval and therefore zero. A similar argument applies to the upper half-circle and for the integral of the other term.

Alternatively, you can use Green's formula and conclude that
$$
\oint_\Gamma \left(y^2 x \, dx - x^2 y \, dy\right)
= \int_{S} \left(-\frac{\partial(x^2 y)}{\partial x} - \frac{\partial(y^2 x)}{\partial y}\right) dx\, dy
= - 4 \int_{S} xy \, dx\, dy,
$$
where $S$ is the disc enclosed by the circle. This integral is obviously zero since the integrand is odd in both $x$ and $y$ whereas $S$ is symmetric with respect to $x \to -x$ as well as $y \to -y$.

 

[songoku]

Thank you very much for the help and explanation Orodruin

 

[FactChecker]

FYI, there is a very large class of functions that will give zero integrals on closed curves. They are related to potentials and to analytic functions with no enclosed singularities.

 

[Orodruin]

FYI, there is a very large class of functions that will give zero integrals on closed curves. They are related to potentials and to analytic functions with no enclosed singularities.

Should be pointed out that this is not the case here though. The zero really comes from symmetry. This may be seen from the fact that the integrand of the area integral in Green’s formula above is not zero so there will exist areas such that the integral around them are non-zero.

(Yes, I tried that first before arguing symmetry )