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.
 A: Ok, so lets start with $x + y = 0$ since it has been given to you. This implies that $x^2 = y^2$. From this result we can simplify the following:
\begin{align*}
C: 2x^2 + z^2 = a^2
\end{align*}
As you can see this implies more of an elliptical curve. Your initial parametrization assumed a circular curve. Also, it is never acceptable to just choose a value for a variable like $z = 0$. This would have to be given to you.
Now, we only have two variables, $x$ and $z$. I am assuming that $a$ is just a positive scalar since $x$, $y$ and $z$ cover the three spacial dimensions. Since the curve relation only has these two variables, we can write one of them in terms of the other as follows:
\begin{align*}
z^2 &= a^2 - 2x^2 \\
\end{align*}
Now, I did not write the relation as a function, because I want to keep it in this form for easier calculations. If I was to write it in terms of $z$, by itself, it would require piece-wise defined functions, which can be avoided in this scenario as I will soon demonstrate. Applying implicit differentiation on the above relation, we can find the following useful result:
\begin{align*}
\frac{d}{dx}\left(z^2\right) &= \frac{d}{dx}\left(a^2\right) - 2\frac{d}{dx}\left(x^2\right) \\
2z\frac{dz}{dx} &= -4x\frac{dx}{dx} \\
2z\frac{dz}{dx} &= -4x \\
\frac{dz}{dx} &= -\frac{2x}{z} \\
\\
\left(\frac{dz}{dx}\right)^2 &= \frac{4x^2}{z^2} \\
\left(\frac{dz}{dx}\right)^2 &= \frac{4x^2}{a^2 - x^2} \\
\end{align*}
In general we have the relation, for 3-dimensional space, of:
$$ds = (dx^2 + dy^2 + dz^2)^\frac{1}{2}$$
Since we have $x = -y$, we can deduce that $dx^2 = dy^2$ which allows us to simplify the above relation and reformulate it to a sensible form as follows:
\begin{align*}
ds &= (2dx^2 + dz^2)^\frac{1}{2} \\
ds &= \left(2 + \left(\frac{dz}{dx}\right)^2 \right)^\frac{1}{2}dx \\
ds &= \left(2 +  \frac{4x^2}{a^2 - x^2} \right)^\frac{1}{2}dx \\
\end{align*}
Let me know if I should give more information.
A: 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:
$$x(t)=-y(t)=\frac{a\cos(t)}{\sqrt{2}},\quad z=a\sin(t)\quad \text{with $t\in [0,2\pi]$}$$
A: $x=a \cos t , y=a \sin t, z =0$ is not a parametrisation of $C$ as $x + y \neq 0$. Try plugging $x+y=0$ into $x^2+y^2+z^2 = a^2$ and then use your approach with $\sin$ and $\cos$.
