area of a plane inside an sphere I'm trying to find the area of the plane $x+y+z = 1$ lying inside the unit sphere, i.e $x^2+y^2+z^2=1$
In order to compute the double integral, I have tried substituting $z = 1-x-y$ into the sphere and expressing $x$ in terms of $y$ by solving the quadratic and hence parametrize the intersection curve and then find the minimum and maximum $x$ in order to find the limits of integration for $x$. I find this procedure too tedious and I'm sure there must be an alternative approach using appropiate coordinates. Any ideas?
 A: $|\vec{r}-\vec{r_0}|=a $ is the vector equation of sphere where $a$ is radius and  $\vec{r_0}$ is origin. Now equation of plane $(\vec{r}-\vec{r_o}).\vec{n}=0$ here $\vec{r_o }$ is a point on plane and $\vec{n}$ is normal to plane. in your case 
Equation of sphere $|\vec{r}|=1 $
Equation of plane $(\vec{r}-\hat{i}).\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}=0$ 
Now distance of origin from plane,
$$|(-\hat{i}).\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}|=\frac{1}{\sqrt{3}}$$ 
Radius of intersection formed
$$R=\sqrt{1-(\frac{1}{\sqrt{3}})^2}=\sqrt{\frac{2}{3}}$$
Area of intersection
$$A = \pi R^2$$
A: The sphere is centered at the origin and has radius $1$.
The plane is orthogonal to the diagonal line 
$$
\left\{ \matrix{
  x = t \hfill \cr 
  y = t \hfill \cr 
  z = t \hfill \cr}  \right.
$$
that crosses it at the point $C=(1/3,1/3,1/3)$,
which is distant $1/ \sqrt{3}$ from the origin, and will be the center of the circle.
Then the radius of the circle will clearly be $r=\sqrt{(1-1/3)}=\sqrt{(2/3)}$.
And the area follows immediately.  
If you want to calculate it by an integral, there are plenty of ways to do that, in cartesian and spherical coordinates.
A possible starting point would be to express the plane in parametric coordinates, possibly starting from 
the point $C$.
Otherwise, as suggested, to rotate the reference and have the $z$ axis along the diagonal line.
Etc. ...
