Lengths and Surface Areas Revision Notes for AQA A-Level Further Maths

Lengths and Surface Areas

Introduction to arc lengths and surface areas

Integration can be used to calculate the exact length of a curve between two points and the surface area created when a curve is rotated about an axis. These applications combine differentiation and integration techniques with geometric principles.

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Understanding arc lengths and surface areas requires strong integration skills. Make sure you're comfortable with substitution and trigonometric/hyperbolic identities before tackling these problems.

Arc length of a curve (Cartesian form)

Deriving the arc length formula

When you want to find the length of a curve $y = f(x)$ between two points, you can approximate the curve using small straight-line segments. Consider two nearby points $A$ and $B$ on a curve, separated by small distances $\delta x$ and $\delta y$ in the horizontal and vertical directions respectively.

Using Pythagoras' theorem, the approximate length of the curve segment, $\delta s$ , is:

$(\delta s)^2 = (\delta x)^2 + (\delta y)^2$

Dividing both sides by $(\delta x)^2$ gives:

$\frac{(\delta s)^2}{(\delta x)^2} = 1 + \frac{(\delta y)^2}{(\delta x)^2}$

Taking the square root:

$\frac{\delta s}{\delta x} = \sqrt{1 + \left(\frac{\delta y}{\delta x}\right)^2}$

As $\delta x \to 0$ , we have $\frac{\delta s}{\delta x} \to \frac{ds}{dx}$ and $\frac{\delta y}{\delta x} \to \frac{dy}{dx}$ , so:

$\frac{ds}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}$

Integrating both sides with respect to $x$ gives the total arc length.

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Key formula: Arc length (Cartesian)

The length of the arc of the curve $y = f(x)$ from $x_1$ to $x_2$ is given by:

$s = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

This formula is fundamental for calculating exact curve lengths and must be memorised for the exam.

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Worked Example 1: Arc length of $y = 2x\sqrt{x}$

Question: Calculate the length of the curve $y = 2x\sqrt{x}$ between the points $(0, 0)$ and $(4, 16)$ .

Solution:

First, rewrite the function and find its derivative: $y = 2x^{\frac{3}{2}}$ $\frac{dy}{dx} = 3\sqrt{x}$

Apply the arc length formula from $x = 0$ to $x = 4$ :

$s = \int_0^4 \sqrt{1 + (3\sqrt{x})^2} \, dx$

$= \int_0^4 \sqrt{1 + 9x} \, dx$

To integrate, let $u = 1 + 9x$ , so $\frac{du}{dx} = 9$ , giving $dx = \frac{1}{9}du$ .

When $x = 0$ , $u = 1$ ; when $x = 4$ , $u = 37$ .

$s = \frac{1}{9} \int_1^{37} \sqrt{u} \, du = \frac{1}{9} \int_1^{37} u^{\frac{1}{2}} \, du$

$= \frac{1}{9} \left[\frac{2}{3}u^{\frac{3}{2}}\right]_1^{37} = \frac{2}{27}\left[u^{\frac{3}{2}}\right]_1^{37}$

$= \frac{2}{27}\left(37^{\frac{3}{2}} - 1^{\frac{3}{2}}\right) = \frac{2}{27}(37\sqrt{37} - 1)$

$= \frac{2}{27}(56\sqrt{7} - 1)$

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Exam tip: Always check that you've correctly applied the limits after substitution. A common error is forgetting to change the limits when making a substitution.

Arc length of a curve (Parametric form)

Adapting for parametric equations

When a curve is defined by parametric equations $x = f(t)$ and $y = g(t)$ , you need to modify the arc length formula. Instead of dividing by $(\delta x)^2$ , divide by $(\delta t)^2$ .

Starting from: $(\delta s)^2 = (\delta x)^2 + (\delta y)^2$

Divide by $(\delta t)^2$ : $\left(\frac{\delta s}{\delta t}\right)^2 = \left(\frac{\delta x}{\delta t}\right)^2 + \left(\frac{\delta y}{\delta t}\right)^2$

As $\delta t \to 0$ : $\frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

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Key formula: Arc length (Parametric)

The length of the section of the curve $x = f(t)$ , $y = g(t)$ from $t_1$ to $t_2$ is given by:

$s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

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Worked Example 2: Parametric arc length

Question: Calculate the length of the arc of the curve $x = t^2 - 1$ , $y = \frac{2}{3}t^3 + 1$ from $t = 0$ to $t = 2$ .

Solution:

Find the derivatives with respect to $t$ : $\frac{dx}{dt} = 2t$ $\frac{dy}{dt} = 2t^2$

Apply the parametric arc length formula:

$s = \int_0^2 \sqrt{(2t)^2 + (2t^2)^2} \, dt$

$= \int_0^2 \sqrt{4t^2 + 4t^4} \, dt$

Factor out $4t^2$ : $= \int_0^2 \sqrt{4t^2(1 + t^2)} \, dt = \int_0^2 2t\sqrt{1 + t^2} \, dt$

Let $u = 1 + t^2$ , so $\frac{du}{dt} = 2t$ , giving $dt = \frac{du}{2t}$ .

$s = \int 2t \cdot \sqrt{u} \cdot \frac{du}{2t} = \int \sqrt{u} \, du = \int u^{\frac{1}{2}} \, du$

$= \left[\frac{2}{3}u^{\frac{3}{2}}\right]_0^2$

When $t = 0$ , $u = 1$ ; when $t = 2$ , $u = 5$ .

$s = \frac{2}{3}\left[5^{\frac{3}{2}} - 1^{\frac{3}{2}}\right] = \frac{2}{3}(5\sqrt{5} - 1)$

Surface area of revolution (Cartesian form)

Understanding surface area of revolution

When a curve $y = f(x)$ is rotated through $2\pi$ radians (360°) about the $x$ -axis, it creates a three-dimensional solid. You can calculate the surface area of this solid using integration.

Consider a small section of the curve that, when rotated, creates a frustum (truncated cone). The frustum has:

Lower radius: $y$
Upper radius: $y + \delta y$
Slant height: $\delta s$

Deriving the surface area formula

The curved surface area of a frustum is given by: $S = \pi(r_1 + r_2)l$

where $r_1$ is the base radius, $r_2$ is the top radius, and $l$ is the slant height.

For our small element: $\delta S = \pi(y + (y + \delta y))\delta s = \pi(2y + \delta y)\delta s$

Dividing by $\delta x$ : $\frac{\delta S}{\delta x} = \pi(2y + \delta y)\frac{\delta s}{\delta x}$

As $\delta x \to 0$ , $\delta y \to 0$ and: $\frac{dS}{dx} = 2\pi y \frac{ds}{dx}$

Since $\frac{ds}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}$ : $\frac{dS}{dx} = 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2}$

Integrating gives the total surface area.

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Key formula: Surface area of revolution (Cartesian)

The surface area of revolution when the curve $y = f(x)$ from $x_1$ to $x_2$ is rotated through $2\pi$ radians around the $x$ -axis is given by:

$S = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Note the key difference from arc length: multiply by $2\pi y$ instead of just integrating the arc element.

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Worked Example 3: Surface area of $y = x^3$

Question: The section of the curve with equation $y = x^3$ between $x = 0$ and $x = \frac{1}{2}$ is rotated through $2\pi$ radians around the $x$ -axis. Calculate the exact value of the curved surface area.

Solution:

Find the derivative: $y = x^3 \implies \frac{dy}{dx} = 3x^2$

Apply the surface area formula:

$S = 2\pi \int_0^{\frac{1}{2}} x^3 \sqrt{1 + (3x^2)^2} \, dx$

$= 2\pi \int_0^{\frac{1}{2}} x^3 \sqrt{1 + 9x^4} \, dx$

Let $u = 1 + 9x^4$ , so $\frac{du}{dx} = 36x^3$ , giving $x^3 \, dx = \frac{1}{36}du$ .

When $x = 0$ , $u = 1$ ; when $x = \frac{1}{2}$ , $u = 1 + 9 \cdot \frac{1}{16} = 1 + \frac{9}{16} = \frac{25}{16}$ .

$S = 2\pi \int_1^{\frac{25}{16}} \sqrt{u} \cdot \frac{1}{36} \, du = \frac{2\pi}{36} \int_1^{\frac{25}{16}} u^{\frac{1}{2}} \, du$

$= \frac{\pi}{18} \left[\frac{2}{3}u^{\frac{3}{2}}\right]_1^{\frac{25}{16}} = \frac{\pi}{27}\left[u^{\frac{3}{2}}\right]_1^{\frac{25}{16}}$

$= \frac{\pi}{27}\left[\left(\frac{25}{16}\right)^{\frac{3}{2}} - 1\right]$

$= \frac{\pi}{27}\left[\frac{125}{64} - 1\right] = \frac{\pi}{27} \cdot \frac{61}{64} = \frac{61\pi}{1728}$

Surface area of revolution (Parametric form)

Adapting for parametric equations

When the curve is defined by parametric equations $x = f(t)$ , $y = g(t)$ , the surface area formula becomes:

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Key formula: Surface area of revolution (Parametric)

The surface area of revolution when the curve $x = f(t)$ , $y = g(t)$ from $t_1$ to $t_2$ is rotated through $2\pi$ radians around the $x$ -axis is given by:

$S = 2\pi \int_{t_1}^{t_2} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

Note that $y$ is replaced by $g(t)$ when evaluating the integral.

Integration techniques for lengths and surfaces

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Strategy for solving problems

When calculating arc lengths and surface areas of revolution, you often need to use advanced integration techniques:

Choose the correct formula based on whether the curve is in Cartesian or parametric form.
Use suitable substitutions, ensuring you correctly apply the new limits after substitution.
Apply trigonometric or hyperbolic identities to simplify integrals.

Important hyperbolic identities

The following hyperbolic identities are particularly useful for integration:

$\cosh^2 x = \frac{1}{2}(\cosh 2x + 1)$

$\sinh^2 x = \frac{1}{2}(\cosh 2x - 1)$

Also remember: $\cosh^2 x - \sinh^2 x = 1$

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Worked Example 4: Using hyperbolic identities

Question: The curve $C$ has equation $y = 2x^2$ . Calculate the length of the curve between $(0, 0)$ and $\left(\frac{1}{4}, \frac{1}{8}\right)$ .

Solution:

Find the derivative: $\frac{dy}{dx} = 4x$

Apply the arc length formula:

$s = \int_0^{\frac{1}{4}} \sqrt{1 + (4x)^2} \, dx = \int_0^{\frac{1}{4}} \sqrt{1 + 16x^2} \, dx$

Use the substitution $x = \frac{1}{4}\sinh u$ , so $\frac{dx}{du} = \frac{1}{4}\cosh u$ .

When $x = 0$ , $\sinh u = 0$ , so $u = 0$ .

When $x = \frac{1}{4}$ , $\sinh u = 1$ , so $u = \ln(1 + \sqrt{2})$ .

Substitute: $s = \int_0^{\ln(1+\sqrt{2})} \sqrt{1 + 16 \cdot \frac{1}{16}\sinh^2 u} \cdot \frac{1}{4}\cosh u \, du$

$= \int_0^{\ln(1+\sqrt{2})} \sqrt{1 + \sinh^2 u} \cdot \frac{1}{4}\cosh u \, du$

Using the identity $\cosh^2 x - \sinh^2 x = 1$ , we have $1 + \sinh^2 u = \cosh^2 u$ :

$= \int_0^{\ln(1+\sqrt{2})} \sqrt{\cosh^2 u} \cdot \frac{1}{4}\cosh u \, du = \int_0^{\ln(1+\sqrt{2})} \cosh u \cdot \frac{1}{4}\cosh u \, du$

$= \frac{1}{4}\int_0^{\ln(1+\sqrt{2})} \cosh^2 u \, du$

Using $\cosh^2 u = \frac{1}{2}(\cosh 2u + 1)$ :

$= \frac{1}{4}\int_0^{\ln(1+\sqrt{2})} \frac{1}{2}(\cosh 2u + 1) \, du$

$= \frac{1}{8}\int_0^{\ln(1+\sqrt{2})} (\cosh 2u + 1) \, du$

$= \frac{1}{8}\left[\frac{1}{2}\sinh 2u + u\right]_0^{\ln(1+\sqrt{2})}$

$= \frac{1}{8}\left(\frac{1}{2}(2\sqrt{2}) + \ln(1+\sqrt{2}) - 0\right)$

$= \frac{1}{8}\left(\sqrt{2} + \ln(1+\sqrt{2})\right)$

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Exam tip: When using hyperbolic or trigonometric substitutions, always write down the new limits clearly before substituting. This helps avoid errors and shows clear working to the examiner.

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Common exam pitfalls

Forgetting to square the derivative in the arc length formula.
Confusing arc length and surface area formulas – remember that surface area includes the $2\pi y$ factor.
Not changing limits when making a substitution.
Sign errors when simplifying expressions involving squares and square roots.
Forgetting the factor when substituting (e.g., forgetting the $\frac{1}{36}$ in Example 3).

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Key Points to Remember:

Arc length (Cartesian): $s = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$ – measures the exact length of a curve.
Arc length (Parametric): $s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$ – use when the curve is defined parametrically.
Surface area (Cartesian): $S = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$ – calculates the surface area when rotating about the $x$ -axis.
Surface area (Parametric): $S = 2\pi \int_{t_1}^{t_2} y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$ – parametric version of surface area formula.
Always use substitution or identities to simplify complex integrals involving square roots – look for opportunities to use hyperbolic or trigonometric identities when you see expressions like $1 + x^2$ under a square root.

Lengths and Surface Areas (AQA A-Level Further Maths): Revision Notes