91Ó°ÊÓ

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

$f\left(x\right)={∫}_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Substitute the known values in the formula.

role="math" localid="1648884641454" $$\begin{gathered} {∫}_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx={∫}_{-3}^3\sqrt{1+{\left(\frac{d}{dx}\sqrt{(9-x^2)}\right)}^2}dx \\ ={∫}_{-3}^3\sqrt{1+{\left(\frac{1}{2}\left({(9-x^2)}^{\frac{-1}{2}}\right)\left(-2x\right)\right)}^2}dx \\ ={∫}_{-3}^3\sqrt{1+{\left(\frac{-x}{\sqrt{\left(9-x^2\right)}}\right)}^2}dx \\ ={∫}_{-3}^3\sqrt{1+\frac{x^2}{\left(9-x^2\right)}}dx \\ ={∫}_{-3}^3\sqrt{\frac{9-x^2+x^2}{\left(9-x^2\right)}}dx \\ ={∫}_{-3}^3\sqrt{\frac{9}{\left(9-x^2\right)}}dx \\ =3{∫}_{-3}^3\sqrt{\frac{1}{\left(9-x^2\right)}}dx \end{gathered}$$

We know that,$∫\frac{1}{\sqrt{r^2-x^2}}dx={\sin }^{-1}\frac{x}{r}+C$.

So, $∫\frac{1}{\sqrt{9-x^2}}dx={\sin }^{-1}\frac{x}{3}+C$.

To solve it we will have to split the integral and consider limits at the endpoints $x=Â\pm 3$.

role="math" localid="1648885792539" $$\begin{gathered} 3{∫}_{-3}^3\frac{1}{\sqrt{9-x^2}}dx=3\left(\underset{A→-3}{\lim }{∫}_A^0\frac{1}{\sqrt{9-x^2}}dx+\underset{B→3}{\lim }{∫}_0^B\frac{1}{\sqrt{9-x^2}}dx\right) \\ =3\left(\underset{A→-3}{\lim }{\left[{\sin }^{-1}\frac{x}{3}\right]}_A^0+\underset{B→3}{\lim }{\left[{\sin }^{-1}\frac{x}{3}\right]}_0^B\right) \\ =3\left(\underset{A→-3}{\lim }\left({\sin }^{-1}0-{\sin }^{-1}\frac{A}{3}\right)+\underset{B→3}{\lim }\left({\sin }^{-1}\frac{B}{3}-{\sin }^{-1}0\right)\right) \end{gathered}$$

Substitute the known values in the obtained expression.

$$\begin{gathered} 3{∫}_{-3}^3\frac{1}{\sqrt{9-x^2}}dx=3\left(0-\left(-\frac{\mathrm{π}}{2}\right)+\left(\frac{\mathrm{π}}{2}-0\right)\right) \\ =3\left(0+\frac{\mathrm{π}}{2}+\frac{\mathrm{π}}{2}\right) \\ =3\left(\frac{2\mathrm{π}}{2}\right) \\ =3\mathrm{π} \end{gathered}$$

Therefore, the length of the graph of$f\left(x\right)=\sqrt{9-x^2}$from$x=-3$to$x=3$is$3\mathrm{Ï€}$units.