Question

Find the exact value of the arc length of each function f(x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

$f\left(x\right)=\sqrt{9-x^2}$,$\left[a,b\right]=\left[-3,3\right]$

Short answer

The arc length is$3\pi$.

Step-by-step solution

Step 1. Given information. 

Consider the given function $f\left(x\right)=\sqrt{9-x^2}$,$\left[a,b\right]=\left[-3,3\right]$.

Step 2. Use arc length formula.

The formula for a function to find the arc length from $x=a$to $x=b$is given by role="math" localid="1649165845561" ${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Step 3. Find the arc length. 

$\begin{array}{rcl}\mathrm{Arc}\mathrm{length}& =& {\int }_{-3}^3\sqrt{1+\frac{\mathrm{d}}{\mathrm{dx}}{\left(\sqrt{9-{\mathrm{x}}^2}\right)}^2}\mathrm{dx}\\ & =& {\int }_{-3}^3\sqrt{1+{\left(\frac{1}{2\sqrt{9-{\mathrm{x}}^2}}\left(-2\mathrm{x}\right)\right)}^2}\mathrm{dx}\\ & =& {\int }_{-3}^3\sqrt{1+{\left(-\frac{2\mathrm{x}}{2\sqrt{9-{\mathrm{x}}^2}}\right)}^2}\mathrm{dx}\\ & =& {\int }_{-3}^3\sqrt{1+{\left(-\frac{\mathrm{x}}{\sqrt{9-{\mathrm{x}}^2}}\right)}^2}\mathrm{dx}\\ & =& {\int }_{-3}^3\sqrt{\frac{9-{\mathrm{x}}^2+{\mathrm{x}}^2}{9-{\mathrm{x}}^2}}\mathrm{dx}\\ & =& {\int }_{-3}^3\sqrt{\frac{9}{9-{\mathrm{x}}^2}}\mathrm{dx}\\ & =& {\int }_{-3}^3\frac{3}{\sqrt{9-{\mathrm{x}}^2}}\mathrm{dx}\\ & =& 3{\int }_{-\pi /2}^{\pi /2}1du\\ & =& 3{\left[u\right]}_{-\pi /2}^{\pi /2}\\ & =& 3\left[\frac{\pi }{2}-\left(-\frac{\pi }{2}\right)\right]\\ & =& 3\left[\pi \right]\\ & =& 3\pi \end{array}$