How can I parameterize the intersection of $x+y+z=0$ and $x^2 +y^2 + z^2 =a^2$?

Question

I'm doing exercises using stokes theorem.

And i'm looking for parameterization of the intersection:

$x+y+z=0$ and $x^2 +y^2 + z^2 =a^2$

in terms of x and y such that $r(x,y)$ .

This is because I want to find the normal of the surface created from the intersection.

So that I can use the cross-product. I.e:

$r_{x}$ x $ r_{y}$ $=n$

to find the normal.

What I tried to do was to write z interms of x and y.

Thus giving me $z= -x-y$

Which would give me:

$r(x,y) = xi + yj -(x+y)k$

Why does this not work?

Answer 1

If you actually want to give a parameterization for the intersection of the surfaces $$S_1\colon x+y+z=0$$ and $$S_2\colon x^2+y^2+z^2=a^2,$$ that is, the intersection of the sphere $S_2$ and a plane $S_1$ that goes through its center, then you are trying to parameterize a curve, not a surface, and so you need to use a function of one variable.

Since you need $z=-(x+y)$ and $x^2+y^2+(x+y)^2=a^2$, that is $$2x^2+2y^2+2xy=a^,2$$ try to find $x(t)$ and $y(t)$ that verify that last equation, and then use the parameterization $$r(t)=\big(x(t),y(t),-x(t)-y(t)\big).$$

Answer 2

Choosing a parameterization for $x^2+y^2+z^2 = a^2$

as

$$ x = a \cos\alpha\sin\beta\\ y = a \sin\alpha\sin\beta\\ z = a \cos\beta $$

we have

$$ \cos\alpha\sin\beta+\sin\alpha\sin\beta+\cos\beta = 0\Rightarrow \beta = \arctan\left(-\frac{\sin\alpha+\cos\alpha}{\sqrt{2+\sin(2\alpha)}},\frac{1}{\sqrt{2+\sin(2\alpha)}}\right) $$

now substituting back we have

$$ \left\{ \begin{array}{rcl} x & = & \frac{a \cos (\alpha )}{\sqrt{\sin (2 \alpha )+2}} \\ y & = & \frac{a \sin (\alpha )}{\sqrt{\sin (2 \alpha )+2}} \\ z & = & -\frac{a (\cos (\alpha )+\sin (\alpha ))}{\sqrt{\sin (2 \alpha )+2}} \\ \end{array} \right. $$