# Arc Length and Surface Area - Problem Solving

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#### Contents

## Arc Length

Main article: Arc Length

If $f'$ is continuous on $[a,b],$ then the **arc length** of the curve $f = f(x)$ between $x = a$ and $b$, is given by

$L=\int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx.$

## Surface Area

If $f'$ is continuous on $[a,b]$, then the surface area of the solid of revolution rotated around the x-axis is

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

## Problem Solving

Find the arc length of the function $y = x^2$ from $x = 4$ to $x = 12$.

Solution:Differentiate with respect to $x$ gives $\frac{dy}{dx} = 2x$, then apply the arc length formula:

$\int_{4}^{12} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = \int_{4}^{12} \sqrt{1+2x} \, dx$

Let $z = 1 + 2x$. When $x = 4, z = 9$, when $x = 12, z = 25$, $dx = \frac12 dz$

$\int_9^{25} \sqrt{z} \, dz = \left . \frac23 z^{\frac32} \right]_9^{25} = \frac{196}3 \ \square$

$(r,\theta)$ as

A hypothetical horn is constructed, which is described by the parameters$H=\left\{(x,y,z)|x=\frac{\cos(\theta)}{r},y=\frac{\sin(\theta)}{r},z=-r,r\in (1,\infty),\theta \in (0,2\pi)\right\}$

If it is required to paint the interior of the horn. If 100 square units of the interior surface can be covered using one cubic unit of paint, of the available options what is the minimum number of cubic units of paint that would ensure that the entire interior of the horn is covered in paint.

###### This problem is not original.

**Cite as:**Arc Length and Surface Area - Problem Solving.

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