Find the parametric equation of intersection between the two surfaces

x^2+y^2+z^2=1, x=y+1

Since the two surfaces intersect, we can plug in y + 1 for x in the equation of the sphere to get:
(y + 1)^2 + y^2 + z^2 = 1 ==> 2y^2 + 2y + z^2 = 0.

Now, we seek to parametrize 2y^2 + 2y + z^2 = 0.

If we complete the square here, we have:
2(y^2 + y + 1/4) + z^2 = 2(1/4) ==> 2(y + 1/2)^2 + z^2 = 1/2.

Note that the fact that the y and z-terms are squared suggest that we use something similar to polar coordinates, namely:
y + 1/2 = (r√2/2)cos(t), z = sin(t)
==> y = [r√2*cos(t) - 1]/2, z = sin(t).

Since x = y + 1:
x = [r√2*cos(t) - 1]/2 + 1 = [r√2*cos(t) + 1]/2.

So, the required parametrization is:
x = [r√2*cos(t) + 1]/2, y = [r√2*cos(t) - 1]/2, and z = sin(t).

I hope this helps!