Surface Area of Revolution

When a curve is rotated about an axis, it sweeps out a surface of revolution. Calculating the area of this surface extends the arc length concept and is an important topic in AQA Further Mathematics.

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1 Surface Area about the x-axis

Derivation

Consider a small element of the curve y = f(x) with arc length ds. When this element is rotated about the x-axis, it sweeps out a thin band (frustum of a cone) with:

• Circumference = 2pi y (the distance from the curve to the x-axis)
• Width = ds

The surface area of this band is approximately:

dS = 2pi y * ds

Since ds = sqrt(1 + (dy/dx)^2) dx, the total surface area is:

S = 2pi * integral from a to b of y * sqrt(1 + (dy/dx)^2) dx

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2 Worked Example 1 — Cone

Find the surface area generated by rotating y = 2x (for 0 <= x <= 3) about the x-axis.

Solution:

dy/dx = 2, so sqrt(1 + 4) = sqrt(5).

S = 2pi * integral from 0 to 3 of 2x * sqrt(5) dx

= 2pi sqrt(5) * integral from 0 to 3 of 2x dx

= 2pi sqrt(5) * [x^2] from 0 to 3

= 2pi sqrt(5) * 9

= 18pi sqrt(5)

This is the lateral surface area of a cone with slant height 3 sqrt(5) and radius 6. Using the cone formula S = pi r l = pi(6)(3 sqrt(5)) = 18pi sqrt(5). Confirmed!

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3 Worked Example 2 — Sphere

Derive the surface area of a sphere of radius r.

Solution:

The sphere is generated by rotating y = sqrt(r^2 - x^2) about the x-axis from x = -r to x = r.

dy/dx = -x / sqrt(r^2 - x^2)

(dy/dx)^2 = x^2 / (r^2 - x^2)

1 + (dy/dx)^2 = (r^2 - x^2 + x^2) / (r^2 - x^2) = r^2 / (r^2 - x^2)

sqrt(1 + (dy/dx)^2) = r / sqrt(r^2 - x^2)

S = 2pi * integral from -r to r of sqrt(r^2 - x^2) * (r / sqrt(r^2 - x^2)) dx

= 2pi * integral from -r to r of r dx

= 2pi r * [x] from -r to r

= 2pi r * 2r

= 4pi r^2

This is the well-known surface area of a sphere.

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4 Worked Example 3 — Catenary

Find the surface area when y = cosh x (for 0 <= x <= a) is rotated about the x-axis.

Solution:

From the arc length lesson, we know sqrt(1 + sinh^2 x) = cosh x.

S = 2pi * integral from 0 to a of cosh x * cosh x dx = 2pi * integral from 0 to a of cosh^2 x dx

Using cosh^2 x = (1 + cosh 2x)/2:

= 2pi * integral from 0 to a of (1 + cosh 2x)/2 dx

= pi * [x + sinh(2x)/2] from 0 to a

= pi(a + sinh(2a)/2)

Using sinh(2a) = 2 sinh a cosh a:

= pi(a + sinh a cosh a)

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5 Surface Area in Parametric Form

If x = x(t), y = y(t), the surface area when rotated about the x-axis is:

S = 2pi * integral from t1 to t2 of y * sqrt((dx/dt)^2 + (dy/dt)^2) dt

Worked Example 4 — Sphere via Parametric Form

Use x = r cos t, y = r sin t for 0 <= t <= pi.

dx/dt = -r sin t, dy/dt = r cos t

sqrt((dx/dt)^2 + (dy/dt)^2) = r

S = 2pi * integral from 0 to pi of r sin t * r dt = 2pi r^2 * integral from 0 to pi of sin t dt