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Answers without the blur. Sign up and see all textbooks for free! Q. 37

Expert-verified Found in: Page 499 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # 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.$f\left(x\right)=\sqrt{9-{x}^{2}}$, $\left[a,b\right]=\left[-3,3\right]$

The arc length is $3\pi$.

See the 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.

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