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Work and Geometric Applications · Tutorial 1337 of 1495

Spherical Caps From Washer Slices

Learn to translate a spherical cap’s geometry into slice radii, choose its height bounds, and calculate its volume by integration.

Advanced 9 min read

What You'll Learn

  • Identify a spherical cap and describe its height using the sphere’s centre and cutting plane.
  • Derive the radius and area of a horizontal slice from the sphere equation.
  • Set up volume integrals with bounds that match the chosen vertical coordinate.
  • Derive and apply the volume formula for a spherical cap.
  • Recognize and correct common errors involving slice radii and integration bounds.

From a Sphere to a Cap

A spherical cap is the portion of a sphere on one side of a plane that cuts through it. Its volume depends on both the sphere’s radius and the cap’s height. To calculate that volume, we can slice the cap horizontally: at each height, the slice is a circular disk. The sphere equation tells us the radius of that disk, and integrating its area over the cap’s height gives the volume.

The important modelling choices are geometric. We must identify where the cap begins and ends, express each slice’s radius using the sphere equation, and use integration bounds that describe exactly the cap rather than a different part of the sphere.

Definition: A spherical cap of height \(h\) in a sphere of radius \(R\) is the part of the sphere between its top point and a plane perpendicular to a diameter, where the plane is a distance \(h\) below that top point. For a nonempty cap, \(R>0\) and \(0<h\leq 2R\).

Locate the Cap and Its Slices

Place the sphere’s centre at the origin and its vertical diameter on the \(z\)-axis. The sphere then has equation

$$ x^2+y^2+z^2=R^2. $$

The top of the sphere is at \(z=R\). If the cutting plane is at \(z=R-h\), then the cap runs from \(z=R-h\) to \(z=R\). At a fixed height \(z\), the cross-section is a circle centred on the \(z\)-axis. Its radius, denoted \(r(z)\), satisfies \(x^2+y^2=r(z)^2\). Substituting this into the sphere equation gives

$$ r(z)^2+z^2=R^2, \qquad r(z)^2=R^2-z^2. $$

Each horizontal cross-section is a disk, so its area is \(\pi r(z)^2\). In washer terminology, it is a washer with outer radius \(r(z)\) and inner radius zero. The slice area is therefore

$$ A(z)=\pi\bigl(R^2-z^2\bigr), \qquad R-h\leq z\leq R. $$

The bounds matter: the lower bound is the cutting plane, not necessarily the centre of the sphere. The slice principle for volume says to integrate the cross-sectional area over the interval of heights occupied by the solid. Thus the cap’s volume is

$$ V=\int_{R-h}^{R}\pi\bigl(R^2-z^2\bigr)\,dz. $$

Deriving the Spherical-Cap Formula

We can evaluate the integral directly. Since \(R\) is constant, an antiderivative of \(\pi(R^2-z^2)\) is \(\pi(R^2z-z^3/3)\). Applying the bounds gives

$$ \begin{aligned} V &=\pi\left[R^2z-\frac{z^3}{3}\right]_{R-h}^{R}\\ &=\pi\left(R^3-\frac{R^3}{3} -\left(R^2(R-h)-\frac{(R-h)^3}{3}\right)\right). \end{aligned} $$

To simplify, expand the cubic term:

$$ (R-h)^3=R^3-3R^2h+3Rh^2-h^3. $$

Substituting and collecting terms yields the cap formula.

Theorem: A spherical cap of height \(h\) in a sphere of radius \(R\), with \(R>0\) and \(0<h\leq 2R\), has volume $$ V=\pi h^2\left(R-\frac{h}{3}\right). $$

For completeness, the simplification in the integral is

$$ \begin{aligned} V &=\pi\left(\frac{2R^3}{3}-R^2(R-h)+\frac{(R-h)^3}{3}\right)\\ &=\pi\left(\frac{2R^3}{3}-R^3+R^2h +\frac{R^3-3R^2h+3Rh^2-h^3}{3}\right)\\ &=\pi\left(Rh^2-\frac{h^3}{3}\right)\\ &=\pi h^2\left(R-\frac{h}{3}\right). \end{aligned} $$

This derivation explains why the height bounds and the coordinate choice must agree: the sphere equation supplies the slice radius as a function of \(z\), and the integral adds exactly those disk areas from the cutting plane to the sphere’s top. The formula also has the correct units. Since \(h^2(R-h/3)\) has units of length cubed, \(V\) is a volume.

Worked Calculations

Worked Example: A Cap of Height 2 cm

A sphere has radius \(5\) cm. Find the volume of its top cap with height \(2\) cm, using disk slices.

Put the centre at the origin. The top is at \(z=5\) cm, so the cutting plane is \(2\) cm below the top, at \(z=5-2=3\) cm. The cap’s bounds are \(3\leq z\leq 5\). From the sphere equation, the slice area is \(\pi(25-z^2)\) square centimetres. Therefore

$$ \begin{aligned} V &=\int_3^5\pi(25-z^2)\,dz\\ &=\pi\left[25z-\frac{z^3}{3}\right]_3^5\\ &=\pi\left(\left(125-\frac{125}{3}\right) -\left(75-\frac{27}{3}\right)\right)\\ &=\pi\left(\frac{250}{3}-66\right)\\ &=\frac{52\pi}{3}\text{ cm}^3. \end{aligned} $$

Checking with the cap formula, \(h=2\) cm and \(R=5\) cm give \(V=\pi(2)^2(5-2/3)=4\pi(13/3)=52\pi/3\text{ cm}^3\), as required. Differentiating the antiderivative \(25z-z^3/3\) gives \(25-z^2\), confirming the integrand.

Worked Example: A Cap Cut Above the Centre

A sphere has radius \(6\) m and centre at the origin. A horizontal plane at \(z=2\) m cuts off its top cap. Find the cap’s volume.

The top is at \(z=6\) m, so the cap height is \(h=6-2=4\) m. Its bounds are \(2\leq z\leq6\). The slice area is \(\pi(36-z^2)\) square metres. Integrating gives

$$ \begin{aligned} V &=\int_2^6\pi(36-z^2)\,dz\\ &=\pi\left[36z-\frac{z^3}{3}\right]_2^6\\ &=\pi\left((216-72)-\left(72-\frac{8}{3}\right)\right)\\ &=\pi\left(144-\frac{208}{3}\right)\\ &=\frac{224\pi}{3}\text{ m}^3. \end{aligned} $$

Using the formula instead gives \(V=\pi(4)^2(6-4/3)=16\pi(14/3)=224\pi/3\text{ m}^3\). The two calculations agree. The base radius is also consistent with the sphere equation: at \(z=2\), its square is \(36-2^2=32\text{ m}^2\), so the base radius is \(4\sqrt{2}\) m.

Using a Sphere Whose Centre Is Not at the Origin

The origin is a convenient coordinate choice, not a restriction on the geometry. If the centre is at height \(z=c\), the sphere equation becomes

$$ x^2+y^2+(z-c)^2=R^2. $$

For a top cap cut by the plane \(z=p\), the top is at \(c+R\), the cap height is \(h=c+R-p\), and the integration bounds are \(p\leq z\leq c+R\). At height \(z\), the slice radius satisfies \(r(z)^2=R^2-(z-c)^2\). These statements follow by solving the sphere equation for \(x^2+y^2\); translating the centre changes the coordinate expression but not the cap’s geometry.

Worked Example: A Cap in a Vertically Shifted Sphere

A sphere of radius \(5\) units has centre at \(z=3\). Its top cap is cut by the plane \(z=6\). Find the cap’s volume.

The top is at \(z=3+5=8\), so the cap height is \(8-6=2\) units and the bounds are \(6\leq z\leq8\). The sphere equation is \(x^2+y^2+(z-3)^2=25\), so a slice at height \(z\) has area \(\pi(25-(z-3)^2)\). Hence

$$ \begin{aligned} V &=\int_6^8\pi\bigl(25-(z-3)^2\bigr)\,dz\\ &=\pi\left[25(z-3)-\frac{(z-3)^3}{3}\right]_6^8\\ &=\pi\left(\left(125-\frac{125}{3}\right) -\left(75-\frac{27}{3}\right)\right)\\ &=\pi\left(\frac{250}{3}-66\right)\\ &=\frac{52\pi}{3}\text{ units}^3. \end{aligned} $$

As a check, the height and radius are \(h=2\) and \(R=5\), so the cap formula gives \(\pi(2)^2(5-2/3)=52\pi/3\text{ units}^3\). The antiderivative is also correct: differentiating \(25(z-3)-(z-3)^3/3\) gives \(25-(z-3)^2\), the slice-area factor.

A Useful Change of Coordinate

Sometimes it is clearer to measure depth \(y\) downward from the top of the sphere rather than height \(z\) upward from its centre. In that coordinate, the top is \(y=0\), and a cap of height \(h\) runs from \(y=0\) to \(y=h\). Since the vertical distance from the centre is \(R-y\), the sphere equation gives

$$ r(y)^2=R^2-(R-y)^2=2Ry-y^2. $$

The same volume is then

$$ \begin{aligned} V &=\int_0^h\pi(2Ry-y^2)\,dy\\ &=\pi\left[Ry^2-\frac{y^3}{3}\right]_0^h\\ &=\pi h^2\left(R-\frac{h}{3}\right). \end{aligned} $$

This coordinate choice often makes the bounds especially easy to read: the cap begins at depth zero and ends at depth \(h\). The slice formula changes accordingly. It must use \(2Ry-y^2\), not \(R^2-y^2\), because \(y\) measures depth from the top, not signed height from the centre.

Common Error: Mixing Up Coordinates and Bounds

A frequent mistake is to use the centre-based radius formula \(r^2=R^2-z^2\) with bounds measured from the top. For example, for a radius-\(5\) sphere and a cap of height \(2\), using \(z=0\) to \(2\) with that formula would give

$$ \pi\int_0^2(25-z^2)\,dz =\pi\left[25z-\frac{z^3}{3}\right]_0^2 =\frac{142\pi}{3}. $$

This is not the cap volume: those bounds use the wrong coordinate interval for the centre-based formula. The cap begins at \(z=3\), so its correct centre-based integral is from \(3\) to \(5\), giving \(52\pi/3\). Alternatively, if \(y\) is depth from the top, use the depth-based area \(\pi(2Ry-y^2)\) from \(y=0\) to \(y=2\). Matching the radius formula to its coordinate and bounds prevents this error.

Why the Cap Formula Is Useful

The integral gives a geometric model that can be adapted to different sphere locations and cutting planes. The closed formula is convenient once the cap’s radius and height are known, but deriving it from slices makes its assumptions visible. In particular, a cap is not generally a hemisphere: only when \(h=R\) does the cutting plane pass through the centre. Substituting \(h=R\) gives \(V=2\pi R^3/3\), half of the sphere volume \(4\pi R^3/3\).

The limiting cases also provide useful checks. If \(h=0\), the formula gives zero volume, as expected for a cap with no height. If \(h=2R\), the cutting plane is at the bottom of the sphere and the cap is the whole sphere; the formula gives \(\pi(2R)^2(R-2R/3)=4\pi R^3/3\). These checks support both the formula and the interpretation of the allowed height range.

Check Your Understanding

Use the sphere equation and the cap geometry to answer these questions.

  1. A sphere of radius \(7\) cm has a top cap of height \(3\) cm. What are the centre-based integration bounds?
  2. For a sphere centred at the origin, what is the squared radius of a horizontal slice at height \(z\)?
  3. Write the washer-area integral for a top cap of height \(h\) in a sphere of radius \(R\).
  4. A sphere of radius \(4\) units is cut at \(z=1\), with its centre at the origin. What is the height of the top cap?
  5. Why is \(R^2-y^2\) not the correct squared slice radius when \(y\) measures depth down from the sphere’s top?

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