Arc Length and Surface Area - Problem Solving

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

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

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

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

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

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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 $x = 12, z = 25$, $dx = \frac12 dz$

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