Intersection of a sphere and a plane.

Question

I have a curve: $\gamma =:[ x^2+y^2+z^2 = 1, x+y+z=1] $ How I can parametrize it? Or write it out in explicit form? I need this for compute the integral $\iint_{S}dS$, where $S$ is a surface of circle. Of course, I can compute the integral: $\iint_{D}dxdy$, where $D$ is projection of surface $S$ on $XoY$ plane. I can find $D$ solving the system of equations, which describe $\gamma$, but it is harder, than find equation, which describe the circle $\gamma$, I think.

Thank you in advance!

Answer 1

That curve is the intersection between a sphere and a plane and it has more than one point. Therefore, it is a circle. It is orthogonal to the line $l$ defined by $(0,0,0)$ and $(1,1,1)$ and its center belongs to it. Actually the center is $c=\frac13(1,1,1)$ (the intersection between $l$ and the plane). Since one of the points of the circle is $q=(1,0,0)$, the radius is$$r=\|q-c\|=\sqrt{\frac23}.$$Now, another point $p$ of the circle such that $p-c$ and $q-c$ are orthogonal is$$p=\left(\frac{1}{3},\frac{1}{3}-\frac{1}{\sqrt{3}},\frac{1}{3}+\frac{1}{\sqrt{3}}\right.)$$So, your circle is$$\left\{c+(p-c)\cos\theta+(q-c)\sin\theta\,\middle|\,\theta\in[0,2\pi]\right\}.$$

Answer 2

From $$x+y+z=1$$ you get $$z=1-x-y$$

plug in $$x^2+y^2+z^2 = 1$$ to get $$x^2+y^2+(1-x-y)^2 = 1$$

Let $x=t$, and solve the resulting quadratic for $ y=y(t)$.

The parametric equation then is $$ x=t, y=y(t), z=1-t-y(t) $$