Contour integral of $\int |z|^2 dz$

Question

I'm not really sure if I'm doing it right this integral, could you please tell me if there's something wrong, please?

$$ \int |z|^{2} dz $$ For the contourn: $|z+3| + |z-3| =10$ I already solve the contourn, it's an ellipse: $$ \frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 $$ So, I used $$ x = 5 \cos (\theta) ; y = 4\sin(\theta) ; dx = -5\sin(\theta) ; dy = 4\cos(\theta)$$

And when i do the integral from $0$ to $2\pi$, the result is $0$ (zero) Am I doing something wrong?

I'm doing

$$\int (x^{2} + y^{2}) dx + i(x^{2} + y^{2}) dy $$

Answer 1

You are doing the change of variable

$$z=5\cos\theta+i4\sin\theta\implies dz=(-5\sin\theta+i4\cos\theta) d\theta$$

And

$$|z|^2=25\cos^2\theta+16\sin^2\theta=16+9\cos^2\theta$$

Then if $E$ is the ellipse we have that

$$\int_E|z|^2 dz=\int_0^{2\pi}(16+9\cos^2\theta)(-5\sin\theta+i4\cos\theta) d\theta\\ =\int_0^{2\pi}(45\sin^3\theta+i36\cos^3\theta)d\theta=0$$

because odd powers of sine or cosine have integral zero in $[0,2\pi]$, what can be shown using arguments of symmetry.