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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:
$$x(t)=-y(t)=frac{acos(t)}{sqrt{2}},quad z=asin(t)quad text{with $tin [0,2pi]$}$$

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^2right) &= frac{d}{dx}left(a^2right) - 2frac{d}{dx}left(x^2right) \
2zfrac{dz}{dx} &= -4xfrac{dx}{dx} \
2zfrac{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.

$x=a cos t , y=a sin t, z =0$ 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 $cos$.