Page:Calculus Made Easy.pdf/240

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

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

 

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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 $r=a(1+\cos \theta )$.

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