Compute $\int_C ze^{\sqrt{x^2+y^2}} \mathrm ds$

Question

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.

Answer 1

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: $$x(t)=-y(t)=\frac{a\cos(t)}{\sqrt{2}},\quad z=a\sin(t)\quad \text{with $t\in [0,2\pi]$}$$

Answer 2

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

Answer 3

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.