Compute $\int_C ze^{\sqrt{x^2+y^2}} \mathrm ds$ 
Compute $\int_C ze^{\sqrt{x^2+y^2}} \mathrm ds$ where
$$C:x^2+y^2+z^2=a^2, x+ y=0, a \gt 0$$

At first I thought to parametrize this as: $x=a \cos t , y=a \sin t, z =0$, but then the integral will result in $0$ and this might not be true.

The curve $C$ is a circle in the plane $x+y=0$ centered at the origin with radius $a$, so it is symmetric with respect to the plane $z=0$. Moreover the integrand is odd with respect to $z$and therefore, by symmetry, the given integral $\int_C ze^{\sqrt{x^2+y^2}} \mathrm ds$ is zero.
BTW a convenient parametrization for the circle $C$ could be:
$$x(t)=-y(t)=\frac{a\cos(t)}{\sqrt{2}},\quad z=a\sin(t)\quad \text{with $t\in [0,2\pi]$}$$
