Volume of sphere sliced by a plane.

Question

Given there is a simple plane

$x+y+z = 1$

and a sphere of radius 1 centred at the origin.

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

How to calculate the bigger volume bounded by them using integration ? The point I can not figured out is the range of $z$ in terms of $x, y$. I can see upper bound is $z = -(x +y)$ from the plane. However I am not sure if lower bound is $z =-\sqrt{1-x^2-y^2}$ since the negative root means the lower half of the sphere. The desired volume is bounded by the plane and some part of sphere which consists of some of upper and lower hemispheres.

I have tested whether $-\sqrt{1-x^2-y^2}$ can be a bound by letting the plane be

$x+y+z = 0$

so that it divides the sphere equally and volume should be $\frac{2}{3}\pi$. Integration based on my guess is

$V = \int_{0}^{2\pi}\int_0^{1}\int_{-\sqrt{1-r^2}}^{-r(cos(\theta)+sin(\theta))}dz\ rdrd\theta$

As a result I have got desired answer so I am suspecting the sign of root does not matter, yet this may work only for this case so I would like to confirm if my bound is correct.

Answer 1

Since the distance form the center of the sphere and the plane is equal to $\frac{\sqrt 3}3$ the problem is equivalent to the intersection between

• sphere $x^2+y^2+z^2=1$
• plane $z=\frac{\sqrt 3}3$

then we can use for example cylindrical coordinates to find volume.