I'm trying to solve the following problem, regarding Stokes Theorem:
$F = z i + xj + yk $; C the curve of intersection of the plane $x + y + z = 0$ and the sphere $x^2 + y^2 + z^2 = 1$ [Hint: Recall that the area of an ellipse $x^2/a^2 + y^2/b^2 = 1$.]
I know that whenever we have a plane that intersects with sphere, this intersection produces a curve that has the form of an ellipse. Knowing this how can you determine the radii of said ellipse?
If you want to find the intersection of these two surfaces.
Either say: $z = -x-y$ and plug this into the the equation for the sphere. $x^2 + y^2 + x^2 + 2xy + y^2 = 1\\ 2x^2 + 2y^2 + 2xy = 1$
Which is an elipse rotated 45 degrees off from standard, and rotate it back if you wish...This is the projection of the intersection on the xy plane.
Or say. $(\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2},0)$ and $(0,\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2})$ lie on the intersection of the two surfaces.
$(x,y,z) = (\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2},0) \cos t + (0,\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2}) \sin t$
or
$x = \frac{\sqrt 2}{2} \cos t\\ y = -\frac{\sqrt 2}{2} \cos t +\frac{\sqrt 2}{2} \sin t\\ z = -\frac{\sqrt 2}{2} \sin t$
Is a parametric equation of the curve.
To apply stokes theorem...
$\int (-\frac{\sqrt 2}{2} \sin t)(-\frac{\sqrt 2}{2} \sin t) + (\frac{\sqrt 2}{2} \cos t)(\frac{\sqrt 2}{2} \sin t +\frac{\sqrt 2}{2} \cos t)+(-\frac{\sqrt 2}{2} \cos t +\frac{\sqrt 2}{2} \sin t)(-\frac{\sqrt 2}{2} \cos t) dt = \iint curl F \,dA$
And you can use the hemisphere or the circle and get the same answer. I would use the circle.