Page:Calculus Made Easy.pdf/240

This page has been proofread, but needs to be validated.

220

Calculus Made Easy

of all the little triangles making up the required area.

The area of such a small triangle is approximately ${\dfrac {AB}{2}}\times r$ or ${\dfrac {r\,d\theta }{2}}\times r$ ; hence the portion of the area included between the curve and two positions of r corresponding to the angles $\theta _{1}$ and $\theta _{2}$ is given by

${\tfrac {1}{2}}\int _{\theta =\theta _{1}}^{\theta =\theta _{2}}r^{2}\,d\theta$ .

Examples.

(1) Find the area of the sector of $1$ radian in a circumference of radius $a$ inches.

The polar equation of the circumference is evidently $r=a$ . The area is

${\tfrac {1}{2}}\int _{\theta =\theta _{1}}^{\theta =\theta _{2}}a^{2}\,d\theta ={\frac {a^{2}}{2}}\int _{\theta =0}^{\theta =1}d\theta ={\frac {a^{2}}{2}}$ .

(2) Find the area of the first quadrant of the curve (known as “Pascal’s Snail”), the polar equation of which is $r=a(1+\cos \theta )$ .

${\begin{aligned}{\text{Area}}&={\tfrac {1}{2}}\int _{\theta =0}^{\theta ={\frac {\pi }{2}}}a^{2}(1+\cos \theta )^{2}\,d\theta \\&={\frac {a^{2}}{2}}\int _{\theta =0}^{\theta ={\frac {\pi }{2}}}(1+2\cos \theta +\cos ^{2}\theta )\,d\theta \\&={\frac {a^{2}}{2}}\left[\theta +2\sin \theta +{\frac {\theta }{2}}+{\frac {\sin 2\theta }{4}}\right]_{0}^{\frac {\pi }{2}}\\&={\frac {a^{2}(3\pi +8)}{8}}.\end{aligned}}$

This article is issued from Wikisource. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.