Proof $\iint_{x^2+y^2\le1}x \cdot f(\cos(x),y)dxdy=0$

Question

I’m trying to proof :

$$\iint_{x^2+y^2\le1}x \cdot f(\cos(x),y)dxdy=0$$

For a continuous function $f$ in $\mathbb{R}^2$ .

My try was according to parametrization but I failed to get any help through it : We’ll assume : $x=\cos(t)$ and $y=\sin(t)$ Hence $dS=dxdy=rdt$

The integral will be : $\int_{0}^{2\pi} r\cos(t)\cdot f(t,\sin(t)) \,dt $ And I’m not exactly sure this integral gives me anything.. I’d appreciate some help on this, thanks!

Answer 1

This is a consequence of symmetry. Let's do a change of variables: $u=-x, v=y$, then the absolute value of the Jacobian of this change of variables is $1$, so we have $$I=\iint_{x^2+y^2\le1}x \cdot f(\cos(x),y)dxdy=\iint_{u^2+v^2\le1}(-u) \cdot f(\cos(-u),v)dudv.$$ Since $\cos(-u)=\cos(u)$, we have $$\iint_{u^2+v^2\le1}(-u) \cdot f(\cos(-u),v)dudv=-\iint_{u^2+v^2\le1}u \cdot f(\cos(u),v)dudv=-I,$$ i.e., $I=-I$. Therefore, $I=0$.

Answer 2

Notice that, once we fix $y$ the function is odd wrt to $x$, and the domain is symmetric wrt. to the $y$-axis. More in particular: $$ \int\int_{x^2+y^2\leq1}xf(cos(x),y)\:dxdy = \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}xf(cos(x),y)\:dxdy $$ Let's fix $y$. Then the function $xf(cos(x),y)$ is odd wrt. to $x$. Since the interval $[-\sqrt{1-y^2},\sqrt{1-y^2}]$ is symmetric wrt. to the $y$-axis, then its integral is $0$. Therefore:

$$ \int_{-1}^{1}0\:dy = 0 $$