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Surface Area of Revolution · Tutorial 438 of 998

Generating A Sphere From A Semicircle

Learn to express a sphere’s surface area as a surface-of-revolution integral and evaluate it using either a graph or a parametrized semicircle.

Intermediate 9 min read

What You'll Learn

  • Identify the radius of a semicircle and the axis about which it is revolved.
  • Set up the surface-area integral using distance to the axis and arc length.
  • Differentiate the semicircle’s graph and simplify the resulting integrand.
  • Handle the graph’s vertical tangencies at the endpoints.
  • Use a parametrization to calculate the area of a sphere or hemisphere.

From a Semicircle to a Sphere

A sphere can be generated by revolving a semicircle about its diameter. The semicircle supplies the profile; revolving each point on its curved arc around the diameter traces a circular band. Adding the areas of those bands gives the sphere’s surface area.

The calculation illustrates an important feature of surface area of revolution: the integrand combines two different lengths. One is the distance from the curve to the axis, which determines the band’s circumference. The other is a small length measured along the curve, not merely a horizontal or vertical change.

We will use a semicircle of radius \(R>0\), centered at the origin, with its diameter on the \(x\)-axis. Its upper arc is revolved about that axis. The radius \(R\) is the distance from the center to the curved arc; it is not the diameter \(2R\).

Definition: For a smooth profile curve revolved about an axis, the surface-area element is the circumference of the circular band, \(2\pi r\), multiplied by the small arc length \(ds\). Here \(r\) is the perpendicular distance from the curve to the axis. Thus, when the curve is written as \(y=f(x)\) and revolved about the \(x\)-axis, \(r=|f(x)|\) and \(ds=\sqrt{1+(f'(x))^2}\,dx\).

For a nonnegative graph \(y=f(x)\) revolved about the \(x\)-axis over \(a\le x\le b\), this gives the familiar formula

$$ S=2\pi\int_a^b f(x)\sqrt{1+(f'(x))^2}\,dx. $$

The factor \(2\pi f(x)\) is the circumference of the band, and \(\sqrt{1+(f'(x))^2}\,dx\) is the corresponding arc length. This formula applies to a smooth graph on the interval; if the derivative is unbounded at an endpoint, the integral is interpreted as an improper integral when appropriate.

Setting Up the Semicircle Integral

The upper semicircle of radius \(R\) has equation

$$ x^2+y^2=R^2,\qquad y\ge 0. $$

Solving for \(y\) gives the graph

$$ y=\sqrt{R^2-x^2},\qquad -R\le x\le R. $$

The curve meets the \(x\)-axis at \(x=-R\) and \(x=R\), so its distance from the axis is zero at both endpoints. For \(-R<x<R\), differentiate the graph:

$$ \frac{dy}{dx} =\frac{1}{2}(R^2-x^2)^{-1/2}(-2x) =-\frac{x}{\sqrt{R^2-x^2}}. $$

Consequently, the arc-length factor is

$$ \sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\frac{x^2}{R^2-x^2}} = \sqrt{\frac{R^2-x^2+x^2}{R^2-x^2}} = \frac{R}{\sqrt{R^2-x^2}}. $$

Because \(R>0\), the last square root simplifies to \(R/\sqrt{R^2-x^2}\). Multiplying the arc-length factor by the radius \(y\) cancels the square-root terms:

$$ y\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{R^2-x^2}\, \frac{R}{\sqrt{R^2-x^2}} =R,\qquad -R<x<R. $$

The graph formula therefore reduces to the integral of a constant over an interval of length \(2R\).

Key takeaway: For the upper semicircle \(y=\sqrt{R^2-x^2}\) revolved about its diameter, the profile radius times the arc-length factor is exactly \(R\) at every interior point. The surface-area integral is therefore \(2\pi R\int_{-R}^{R}1\,dx\).

Why the Integral Gives the Area

The semicircle is steep near its endpoints, so \(dy/dx\) is unbounded as \(x\) approaches \(-R\) or \(R\). This does not prevent the area calculation: the graph formula is applied on the open interval and interpreted by limits at the endpoints. The product in the integrand has the finite constant value \(R\) throughout that open interval, so its improper integral exists.

There is also a parametrization that makes both the geometry and the endpoint behavior especially clear. Let \(\theta\) run from \(0\) to \(\pi\). The semicircle can be written as

$$ x=R\cos\theta,\qquad y=R\sin\theta. $$

Its arc-length element follows from the parametric arc-length formula:

$$ ds=\sqrt{\left(\frac{dx}{d\theta}\right)^2+ \left(\frac{dy}{d\theta}\right)^2}\,d\theta =\sqrt{(-R\sin\theta)^2+(R\cos\theta)^2}\,d\theta =R\,d\theta. $$

At a point on this arc, the circular band created by the rotation has radius \(y=R\sin\theta\), so its circumference is \(2\pi R\sin\theta\). Multiplying by \(ds\) and integrating from one endpoint of the arc to the other gives

$$ S=\int_0^\pi 2\pi(R\sin\theta)(R\,d\theta) =2\pi R^2\int_0^\pi\sin\theta\,d\theta. $$

Since an antiderivative of \(\sin\theta\) is \(-\cos\theta\), the integral is \([-\cos\theta]_0^\pi=1-(-1)=2\). Thus the semicircle generates a surface of area \(4\pi R^2\). The parametrized calculation agrees with the graph calculation while avoiding an infinite derivative at the endpoints.

Theorem: Revolving a semicircle of radius \(R\) about its diameter produces a sphere whose surface area is $$ S=4\pi R^2. $$

Worked Calculations

Worked Example: A Sphere of Radius \(5\)

Find the surface area generated by revolving the upper semicircle of radius \(5\) about its diameter.

The profile is \(y=\sqrt{25-x^2}\), with \(-5\le x\le 5\). Differentiating on the interior gives

$$ \frac{dy}{dx}=-\frac{x}{\sqrt{25-x^2}}. $$

Hence the arc-length factor is

$$ \sqrt{1+\left(\frac{dy}{dx}\right)^2} =\sqrt{1+\frac{x^2}{25-x^2}} =\sqrt{\frac{25}{25-x^2}} =\frac{5}{\sqrt{25-x^2}}. $$

The surface-area integral is

$$ \begin{aligned} S &=2\pi\int_{-5}^{5}\sqrt{25-x^2}\, \frac{5}{\sqrt{25-x^2}}\,dx\\ &=2\pi\int_{-5}^{5}5\,dx\\ &=10\pi[x]_{-5}^{5}\\ &=10\pi(5-(-5))\\ &=100\pi. \end{aligned} $$

Check: the sphere formula gives \(4\pi(5)^2=4\pi(25)=100\pi\), agreeing with the integral. The radius is \(5\), not the diameter \(10\).

Worked Example: A Hemisphere from Half of the Semicircle

Use the part of a radius-\(2\) upper semicircle with \(0\le x\le 2\). When this quarter-circle arc is revolved about the \(x\)-axis, it generates the curved surface of a hemisphere. Find its area.

Use \(x=2\cos\theta\), \(y=2\sin\theta\). The range \(0\le x\le 2\) corresponds to \(0\le\theta\le\pi/2\), and the arc-length element is \(ds=2\,d\theta\). The distance to the axis is \(y=2\sin\theta\). Therefore

$$ \begin{aligned} S_{\text{hemisphere}} &=2\pi\int_0^{\pi/2}(2\sin\theta)(2\,d\theta)\\ &=8\pi\int_0^{\pi/2}\sin\theta\,d\theta\\ &=8\pi[-\cos\theta]_0^{\pi/2}\\ &=8\pi(0-(-1))\\ &=8\pi. \end{aligned} $$

Check: the full sphere of radius \(2\) has area \(4\pi(2)^2=16\pi\), and half of that is \(8\pi\). This calculation concerns the curved hemisphere only, not an added flat circular base.

Worked Example: A Sphere of Radius \(3\) in Parametric Form

Set up and evaluate the surface-area integral for the radius-\(3\) semicircle parametrized by \(x=3\cos\theta\), \(y=3\sin\theta\), \(0\le\theta\le\pi\).

Differentiate the coordinates:

$$ \frac{dx}{d\theta}=-3\sin\theta,\qquad \frac{dy}{d\theta}=3\cos\theta. $$

Thus

$$ ds=\sqrt{9\sin^2\theta+9\cos^2\theta}\,d\theta =\sqrt{9}\,d\theta =3\,d\theta. $$

The radius of each band is \(3\sin\theta\). Integrating its circumference times the arc-length element gives

$$ \begin{aligned} S &=2\pi\int_0^\pi(3\sin\theta)(3\,d\theta)\\ &=18\pi\int_0^\pi\sin\theta\,d\theta\\ &=18\pi[-\cos\theta]_0^\pi\\ &=18\pi(1-(-1))\\ &=36\pi. \end{aligned} $$

Check: \(4\pi R^2=4\pi(3)^2=4\pi(9)=36\pi\), as required.

A Common Mistake: Using \(dx\) Instead of Arc Length

A frequent error is to multiply each band’s circumference by \(dx\), as though the curve’s small length were horizontal. For a curved profile, \(dx\) is only the horizontal component of the arc length. In general, the arc-length element is \(ds=\sqrt{1+(dy/dx)^2}\,dx\), which is longer than \(dx\) wherever the curve is not horizontal.

For example, if \(R=3\), omitting the arc-length factor would give the incorrect calculation

$$ 2\pi\int_{-3}^{3}\sqrt{9-x^2}\,dx =2\pi\left(\frac{9\pi}{2}\right) =9\pi^2. $$

The integral inside the parentheses is the area of a semicircle, but it is not the length of the semicircular arc. The correction is to include the arc-length factor. In this case the product simplifies to \(3\), giving \(2\pi\int_{-3}^{3}3\,dx=36\pi\), consistent with the sphere formula.

A separate check is to distinguish radius from diameter. If a problem states that the diameter is \(6\), then \(R=3\), and the area is \(4\pi(3)^2=36\pi\), not \(4\pi(6)^2\).

What the Calculation Shows

The sphere’s area formula is not an unrelated geometric fact inserted into the calculation. It follows directly from the surface-area integral. The semicircle’s changing height determines the circumference of each circular band, while its changing direction determines the arc-length factor. For this particular profile, those two factors simplify so that their product is constant.

The same result can be read geometrically from the parametrization: each small arc length \(R\,d\theta\) produces a band with circumference \(2\pi R\sin\theta\). The bands near the poles have small circumference, while those near the equator have larger circumference. Integrating these different band areas over the entire arc yields \(4\pi R^2\).

When setting up a surface-area integral, identify the axis first, then determine the perpendicular radius and the arc-length element for the chosen curve description. For the semicircle, checking both factors before simplifying makes the cancellation—and the resulting sphere area—transparent.

Check Your Understanding

Use the graph or parametrized method as appropriate. Keep the radius distinct from the diameter, and include the arc-length factor.

  1. Write the upper semicircle of radius \(R\) as a graph and find its derivative for \(-R<x<R\).
  2. For that graph, simplify \(y\sqrt{1+(dy/dx)^2}\). What value does it have throughout the interior of the arc?
  3. Set up the surface-area integral for a sphere generated by a semicircle of radius \(4\), and evaluate it.
  4. What surface is generated by revolving the part of a radius-\(R\) upper semicircle with \(0\le x\le R\)? What is its curved area?
  5. Explain why \(2\pi\int y\,dx\) is not, in general, the surface area generated by revolving a graph about the \(x\)-axis.

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