Chapter 8 - Section 8.1 - Arc Length - 8.1 Exercises - Page 565: 2

$L = \pi $

$$\eqalign{ & {\text{Let }}y = \sqrt {4 - {x^2}} ,{\text{ }}0 \leqslant x \leqslant 2 \cr & {\text{Differentiating}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\sqrt {4 - {x^2}} } \right] \cr & \frac{{dy}}{{dx}} = \frac{{ - 2x}}{{2\sqrt {4 - {x^2}} }} \cr & \frac{{dy}}{{dx}} = - \frac{x}{{\sqrt {4 - {x^2}} }} \cr & \cr & {\text{Use the arc lenght formula }} \cr & L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr & {\text{Substituting the limits and the derivative of the function}} \cr & L = \int_0^2 {\sqrt {1 + {{\left( { - \frac{x}{{\sqrt {4 - {x^2}} }}} \right)}^2}} dx} \cr & L = \int_0^2 {\sqrt {1 + \frac{{{x^2}}}{{4 - {x^2}}}} dx} \cr & L = \int_0^2 {\sqrt {\frac{{4 - {x^2} + {x^2}}}{{4 - {x^2}}}} dx} \cr & L = \int_0^2 {\frac{2}{{\sqrt {4 - {x^2}} }}dx} \cr & {\text{Integrating using }}\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = \arcsin \left( {\frac{x}{a}} \right) + C} \cr & L = 2\left[ {\arcsin \left( {\frac{x}{2}} \right)} \right]_0^2 \cr & L = 2\left[ {\arcsin \left( {\frac{2}{2}} \right) - \arcsin \left( {\frac{0}{2}} \right)} \right] \cr & L = 2\left( {\frac{\pi }{2} - 0} \right) \cr & L = \pi \cr & \cr & {\text{Checking by the circumference formula}} \cr & {\text{From the graph we have that the radius is }}r = 2 \cr & {\text{The circumference of a circle of radius }}r{\text{ is}} \cr & C = 2\pi r \cr & C = 2\pi \left( 2 \right) \cr & C = 4\pi \cr & {\text{We can see that the arc of the sector is }}\frac{1}{4}{\text{ of a circle}}{\text{, then}} \cr & \frac{1}{4}C = \frac{1}{4}\left( {4\pi } \right) \cr & \frac{1}{4}C = \pi \cr} $$