6.6 – Applications of Integration

You probably have learned how to find the area enclosed between the function f(x) and the axes, or between 2 functions. You have also learned the volume of revolution for a function f(x) with the x or y-axis as the axis of rotation. In this section, you’ll be learning 2 new applications, which are the arc length and the surface area of revolution.

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ARC LENGTH

Consider 2 points, P and Q, on a curve. P is the point (x, y) and Q is the point (x + δx, y + δy). Let s be the length of the arc from a point on the y-axis, and δs the length of the arc PQ. Since δs is very small, we can approximate the arc PQ to a straight line. Hence, using Pythagoras’ theorem, we have
(δs)² = (δy)² + (δx)²

Dividing by (δx)², we obtain

As δx → 0, this gives

and after square-rooting both sides, we end up with

The parametric form of s can be obtained by dividing the equation (δs)² = (δy)² + (δx)² with (δt)². While the polar form is probably not in your syllabus, so don’t worry too much. To find the arc length of a particular function, just differentiate it with respect to x, then substitute it in the formula above.

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SURFACE AREA OF REVOLUTION

Let A be the area of the surface formed by rotating the curve y = f(x), between the lines x = a and x = b, about the x-axis. Let the curved surface area of a blue ring shown be δA. Treating the strip as being bounded by 2 cylinders, we have
2πy δs ≤ δA ≤ 2π(y + δy) δs

As δx → 0, δs → 0, so we have

which gives us the formula

Again, differentiate the function, and substitute it into the formula to find the surface area of revolution.

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Note that square root functions are hard to integrate. Therefore, you need to practise more on doing such functions if you want to score well in this section. A typical Further Mathematics paper probably has most questions on this chapter, so it is good that you master it. That’s all for this chapter. ☺