So the problem that I'm trying to solve is as follows:

Assume 1/8 of a sphere with radius $r$ whose center is at the origin (for example the 1/8 which is in $R^{+}$). Now two parallel planes are intersecting with this portion where their distance is $h$ (Note:$h$ < $r$) and one of the planes passes through the origin.What is the area cut off by these two planes on the 1/8 of the sphere? You may assume anything that you think is required to calculate this area, as given, my suggestion would be the angles at which the parallel planes cross xyz planes.

Obviously I'm not interested in trivial cases for example when the plane which passes through origin is one of xy, xz or zy planes.

I hope I explained everything clearly. I wish I could draw a picture for this but I don't how. Let me know if you need more clarification. Any hint or help about how to find this area is highly appreciated.

enter image description here

  • $\begingroup$ It would help avoid misunderstanding if the problem stated that the surface area can be anywhere from 0 to some positive value and that it clearly is a function of 2(?) angles (obviously in addition to h) which define the planes' exact orientation. Or state that the 1/8 circle is defined to the (+,+,+) section of the (x,y,z) coordinate space. By saying that it is 1/8 but leaving it up in the air as to its precise location, it's too easy to assume we are talking about 2 planes slicing a whole circle and then taking 1/8 of that somehow. It would also help to clarify that the sphere has radius r. $\endgroup$
    – Jose_X
    Commented May 20, 2011 at 18:15
  • $\begingroup$ It is actually stated in the problem that as an example you can assume the 1/8 sphere which is in $R^{+}$ so it is defined in (+,+,+), although this does not have any effect on the final answer. $\endgroup$
    – mohak
    Commented May 20, 2011 at 21:44
  • $\begingroup$ I have also stated that I'm not interested in trivial cases so all the situation you have mentioned, are just trivial cases which needs no calculation. But I agree with you abut the case where h>=r I wanted to mention that somehow but I thought that would be clear, I'll change that in the problem. $\endgroup$
    – mohak
    Commented May 20, 2011 at 21:48
  • $\begingroup$ @Jose_X: please do not use answers to make comments. $\endgroup$ Commented May 20, 2011 at 21:56

3 Answers 3


I don't care to make the calculation for you, but I think(?) this is the region you intend:
Sphere sector

Addendum. Here is an idea to avoid computing a complex integral. Let me assume that the slice "fits" in one octant, as I drew above (as opposed, e.g., to exiting through the bottom of the octant). First, imagine that you wanted, not the area in an octant, but the area in a halfspace, demarcated by one of the coordinate planes. Then, from a side view, we have this:
Sphere slice
Your two planes are $p_1$ and $p_2$, with $p_1$ through the sphere center. The halfspace boundary is coordinate plane $q$. Now note that the difference between the half of the $2 \pi h$ area of the slice and the truncated area is just the area of two spherical triangles (one illustrated, one in back), whose angles are all known.

So I believe the area in your octant differs from a quarter of $2 \pi h$ by two spherical triangles, one at either end of the octant. The dimensions of both triangles are known from the orientation of the planes $p_1$ and $p_2$.

  • $\begingroup$ This figure is awesome. The area between two planes is exactly what I'm trying to calculate. Hope somebody can point me to the right direction for finding the solution. $\endgroup$
    – mohak
    Commented May 20, 2011 at 0:34
  • $\begingroup$ @Joseph O'Rourke: Sweet figure ; how did you generate it? CAD? $\endgroup$
    – Jacob
    Commented May 20, 2011 at 20:17
  • 1
    $\begingroup$ @user11128: Do you want us to insert this picture in your question? $\endgroup$ Commented May 20, 2011 at 22:40
  • 1
    $\begingroup$ @Jacob: Thanks! Mathematica. $\endgroup$ Commented May 21, 2011 at 0:10
  • $\begingroup$ @Américo, yes please. $\endgroup$
    – mohak
    Commented May 21, 2011 at 5:56

This first draft gives the main components of what might be a "near solution". As the edits take place later on, hopefully a convincing argument will (and can) be made. This attempt at a solution has not yet been reviewed that much, so it might end up not being that near to a solution.

Future edits might include a yet-to-be-written computer program that would use the results of this solution to calculate the required surface area given a set of input parameters.

Future edits might also come with graphs (if not a real-time picture generator) and use latex or whatever it is that is used at math.stackoverflow.com to improve appearance.

Critique is greatly welcomed. Proofs below are adhoc and can be improved.


Determine the area of surface, S, in 3 space: S overlaps a sphere of radius R centered at the origin. S is further limited to the intersection of this sphere with the following: the xy, xz, and yz Cartezian planes (so S lies entirely within the (+,+,+) octant), as well as two other planes, parallel to each other, separated by a distance h, and where one of the planes, P0, goes through the origin. Depending on the orientation of P0, the other parallel bounding plane, P1, might lie "below" or "above" it. P0 does not overlap the xy, xz, or yz planes. A full solution would include suitable constraints on h and on any other variables so that two such parallel planes touching S will exist.

The solution does not attempt to find a reduced form for various terms. Please comment if you notice simplifications that were not carried out.


Because we have to deal with both a spherical entity and a planar one, the chosen coordinate system is a right-hand Cartezian coordinate system in 3 dimensions. [This may not be ideal.]

P0 is defined with arbitrary constants, A, B, and C: Ax + By + Cz = 0

P1 is defined: A'x + B'y + C'z = D'

Since P1 is parallel to P0, A' = A, B' = B, C' = C. [not proven]

The first significant step:

To solve for D' [and there may be a much better way to do this], we observe that P1 is tangent to a sphere of radius h centered at the origin. [Proof of existence of this tangent point is left for a later edit.]

We solve for the tangent point, labeled (x,y,z) below in order to keep clutter down (instead of (x0, y0, z0)).

First, let's calculate some partial derivatives of the sphere (shown using d notation for now) dz/dy dz/dx and dy/dx [todo: use instead dz/dy, dy/dx, and dx/dz to (perhaps) preserve symmetry in solution of D']

For x^2 + y^2 + z^2 = h^2,

dy/dx = -x/(sqrt(h^2-x^2-z^2))

dz/dy = -y/(sqrt(h^2-x^2-y^2))

dz/dx = -x/(sqrt(h^2-x^2-y^2))

These linear changes in the given directions will match the corresponding "slopes" of P1.

For Ax + By + Cz + D',

dy/dx = -A/B

dz/dy = -B/C

dz/dx = -A/C

From this we can find (x,y,z). After equating corresponding pairs of these 6 equantions, removing the denominators, and squaring:

(B^2)(x^2) = A^2(h^2-x^2-z^2)

(C^2)(x^2) = A^2(h^2-x^2-y^2)

(C^2)(y^2) = B^2(h^2-x^2-y^2)

Moving terms around, it becomes clearer that we have a system of 3 equations linear in x^2, y^2, and z^2

(A^2+B^2)x^2 + 0 + (A^2)z^2 = (A^2)(h^2)

(A^2+C^2)x^2 + (A^2)y^2 + 0 = (A^2)(h^2)

(B^2)x^2 + (C^2+B^2)y^2 + 0 = (B^2)(h^2)

Solving using Cramer's rule [details in later edit].. and note the assymetry [todo: did I make a mistake above or is this simply a consequence of breaking symmetry in not using dx/dz?]

x^2 = ( (h^2)(A^2)(C^2) ) / ( (A^2+C^2)(C^2+B^2) )

y^2 = ( (h^2)(B^2) ) / (C^2+B^2)

z^2 = ( (h^2) ((A^2)(B^2)+(C^4)) ) / ( (A^2+C^2)(C^2+B^2) )

When we take the square roots, we will have to determine whether + or - is required. [todo: not sure yet precisely which + or - is taken for each of x, y, and z, but one set corresponds to a plane "below" P0 and another to one "above". We may or may not have to figure out + or - this way]

For emphasis, let's use x0, y0, and z0:

x0 = +- ( (h)(A)(C) ) / (sqrt( (A^2+C^2)(C^2+B^2) ))

y0 = +- ( (h)(B) ) / (sqrt( (C^2+B^2) ))

z0 = +- ( (h) sqrt((A^2)(B^2)+(C^4)) ) / (sqrt( (A^2+C^2)(C^2+B^2) ))

Knowing the tangent point, we can calculate D'

D' = A(x0) + B(y0) + C(z0)

From here on out, we'll just use D' (or D) but knowing that it has been resolved in terms of A, B, C, and h.

The second significant step:

We'll need to find where the parallel planes intersect the radius R (+,+,+) sphere.

But first we will create the strips and summation that will approximate the surface area.

We will choose the strips approximating S as follows. Each strip is approximately the intersection of a very thin ring (along a minor circle) of the sphere's intersection with S. These rings are parallel to the yz plane (so along the x axis). A ring's area can be seen as the difference in area of the side of two similar cones with one's height being H + delta x and the other's being H. Each such strip (ring) will correspond to a pair of cones that would be tangent at the center x value of that strip. Adjacent strips share a side. A similar construction can be done inside the sphere where the strips are sections of secant planes that lie entirely within the sphere.

Note, the sum of the external strips may not form an upper bound to the surface area of S. It is assumed for the moment that as the strip's width (delta x) goes to 0, the sum of their areas approach the area of S arbitrarily close and give us our answer. This construction and argument needs to be redone. A later edit may set everything up carefully and try to prove this final step as a Riemann integral. The integral (and the series) may not be reduced fully, but the series that comprises the Riemann sum can be directly encoded into a computer algorithm to produce an answer hopefully arbitrarily close to the exact result.

My very limited experience is that if we develop a Riemann sum that for infinitesimals has each component area approximate the exact area of (flat) planar region tangent to the surface piece being approximated, then the formalism that leads to the integral will be possible to carry out (whether we carry it out in our proof or not). Note the following problem.

continuing for the moment...

The endpoints of each strip are defined by where the two parallel planes intersect the sphere.

Rather than calculating the cones' areas and subtracting [todo: reconsider such a cone calculation and compare]. We will assume [without proof for the moment] that the area of each such flat and circular strip is equal to the area of a rectangular strip with

width = delta s

height = (r)(alpha1 - alpha0)

We see the latter because

height = minor circle arc length

= 2(pi)(r) ( (alpha1-alpha0)/(2(pi)) )

We also note that delta s does not equal delta x as the ring strips are slanted at an angle that matches the slope of the sphere at that minor circle corresponding to x. The precise slopes depend on y as well, but there is significant symmetry here (see reference to cone shape). So the delta s is the hypotenuse of a right triangle with delta x as one of the sides and delta r as the other.

These alphas, in radians, are the angle values taken from the center of the implied minor circle (ie, angle formed at x axis by rays that go towards the sphere surface). An alpha value of zero would correspond to the point where each ray overlap and lie on the xy plane. When one ray is on the xy plane and the other on the xz plane, we have pi/2. Remember, these points that define one of the rays that form a pair, one each for alpha1 and alpha0, for a given strip (ie, at a given x), come from the intersection, respectively, of P1 and P0 (at that x) with the sphere. [These two values of alpha will be calculated in the third section.]

Why do we select the strips this way (ie, slices of the sphere parallel to the yz plane)? Because the length and width of each such strip can easily be calculated. Each such length is a section of the perimeter of a minor circle of the sphere.. so we can avoid doing a second limit/sum to calculate their lengths since we know the lengths already for circles. We can also express both the width and the length of each strip in terms of x. This leaves the entire integral (ie, the limit of the sum of strips) expressing the area of S as a function of x only (not y or z).

The radius of the minor circle: r = sqrt(R^2-x^2)

The width: (delta s)^2 = (delta r)^2+(delta x)^2

delta r (a constant for each ring which equals the radial distance from the x axis to the longer end of the strip - the distance from x axis to the shorter end) is equal to delta z when we look at where those rings intersect the xz plane.

(delta s)^2 = (delta z)^2+(delta x)^2

On this xz plane, we see that

(delta z) / (delta x) = dz/dx

From the equation of the sphere, with y=0, z=sqrt(R^2-x^2)

dz/dx = x / (sqrt(R^2-x^2))

delta z = x (delta x) / (sqrt(R^2-x^2))


(delta s)^2 = (x^2(delta x)^2)/(R^2-x^2) + (delta x)^2

= (delta x)^2 ( (x^2+1)/(R^2-x^2) )

delta s = (delta x) (sqrt( (x^2+1)/(R^2-x^2) ))

Thus the area of the rectangle that equals the area of the corresponding circular ring strip:

strip area = [delta s] [r] [alpha1 - alpha0]

= [ (delta x) (sqrt( (x^2+1)/(R^2-x^2) )) ] [sqrt(R^2-x^2)] [alpha1 - alpha0]

The total area of all strips is SUM x=0 to R of all of these strips above.

This moves to the integral, x=0 to x=R, of [(sqrt( (x^2+1)/(R^2-x^2) )) ] [sqrt(R^2-x^2)] [alpha1 - alpha0], as delta x goes to zero.

The third significant step:

alpha0 and alpha1 will now be calculated as functions of x.

The steps will be added in a later edit, but here is how we solve this:

We take the eqn of the sphere and the equation of each plane. P0 solves for alpha0 and P1 solves for alpha1. The steps of these two mimic each other.

Combining these two eqns, we get a quadratic in z and a quadratic in y (which mimic each other), where x is treated as a constant within these quadratics. We solve these quadratics.


alpha0 = a bifurcation of two parts: { arctan (solution to z quadratic / solution to y quadratic), iff solution to z quadratic > 0 and solution to y quadratic > 0; 0, otherwise }

If everything was done properly, we'll be able to show that the solutions to y and z quadratic can be complex for values of A, B, C, and x where the P0 does not intersect the sphere at that x, but the solution will otherwise have one positive value and one negative value. Only the positive is useful within the +,+,+ octant. [x will be squared within each of these quadratic solution pairs, meaning one solution corresponds to negative x values and one to positive x values.]

The h value and whether we look at the plane "above" or the plane "below" P0 plays a role here as well (ie, for some h, A, B, and C, we might have to consider "above" plane or else "below" plane, labeled P1 in each case but differentiated by +- value of x0,y0,z0 calculated and discussed earlier).

alpha1 is similar (calculated using P1's eqns that has D') but instead of "0, otherwise" we get "pi/2, otherwise"

Now, let's just list the y solutions for alpha1 (with details left for later edit). The y solutions for alpha0 can be found by replacing D with 0. Also, the z solutions are identical to the corresponding y solns but have B and C reversed.

y solns used to calculate alpha1 = [-b+-sqrt(b^2-4ac)] / 2a

a = 1 + (B^2/C^2) b = (2Bx/C) (Ax/C-D) c = x^2 - R^2 + (Ax/C-D)^2


Well, lots of details to be cleaned up (and maybe changed entirely) and to be filled in, but supposedly we found the integral expression that would yield the surface area desired. This integral expression depends on constants A, B, C, h, and R and also on the integration variable x. Computer algorithm could be used to approximate this surface area by relying on the Riemann sum described and using a small step delta x. Hopefully such a computer program will also be provided in a future edit.


The portion of a sphere between two parallel planes is a spherical "zone". Restricting this portion to that inside one octant means that both planes are in the same hemisphere and the area you want is $1/4$ of the total area of the zone.

Is there a reason to use the techniques of multi-variable calculus to handle this problem? Mathworld's article on spherical zones gives the answer to your question ($2\pi rh$) and includes a derivation. Even though they call their variable of integration $z$, it seems to me like they're using the area of a surface of revolution technique that is taught in single-variable calculus.

  • $\begingroup$ You seem to be assuming a relationship between the orientation of the octant and the orientation of the planes. See my figure. $\endgroup$ Commented May 24, 2011 at 23:11

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