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.

  • $\begingroup$ Rotate the plane to $x=1/\sqrt3$ say. Now it's finding the volume of a spherical cap, a solid of revolution. $\endgroup$ Oct 19, 2019 at 7:03
  • $\begingroup$ Those bounds you have are not correct. The lower bound is fine, but the upper bound changes depending on where in the circle you are. For an example in 2D, try setting up the double integral over a region $x^2+y^2\leq 1$ and $y\leq x$ $\endgroup$ Oct 19, 2019 at 7:45
  • $\begingroup$ I understood your hint. If I rotate the entire object, the upper bound will be constant z . Yet, doesn't the upper bound still consist of both plane and some parts of sphere even after the rotation? $\endgroup$
    – NMZ
    Oct 19, 2019 at 8:01

1 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.


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