Integral Calculus Solutions
Topics || Problems
Find the area enclosed by r = 2a Cos2 θ
Solution:
Differential Area
\(dA = \frac{1}{2}{r^2}d\theta \)
\(dA = \frac{1}{2}{(2aCo{s^2}\theta )^2}d\theta \)
Limits:
A figure is formed inbetween 0 to 2π
Thus:
\(A = \frac{1}{2}\int\limits_0^{2\pi } {4{a^2}Co{s^4}\theta d\theta } \)
\({\cos ^4}\theta = {\cos ^2}\theta {\cos ^2}\theta \)
\({\cos ^4}\theta = \left( {1 - {{\sin }^2}\theta } \right){\cos ^2}\theta \)
\({\cos ^4}\theta = {\cos ^2}\theta - {\sin ^2}\theta {\cos ^2}\theta \)
\({\cos ^4}\theta = \frac{1}{2}\left( {\cos 2\theta + 1} \right) - \frac{{{{\sin }^2}2\theta }}{4}\)
\({\cos ^4}\theta = \frac{1}{2}\left( {\cos 2\theta + 1} \right) - \frac{{\frac{{1 - \cos 4\theta }}{2}}}{4}\)
\({\cos ^4}\theta = \frac{1}{2}\left( {\cos 2\theta + 1} \right) - \frac{{1 - \cos 4\theta }}{8}\)
\(A = 2{a^2}\int\limits_0^{2\pi } {\left( {\frac{1}{2}\left( {\cos 2\theta + 1} \right) - \frac{{1 - \cos 4\theta }}{8}} \right)} d\theta \)
\(A = 2{a^2}\left[ {\frac{1}{2}\left( {\frac{{\sin 2\theta }}{2} + \theta } \right) - \frac{1}{8}\left( {\theta - \frac{{\sin 4\theta }}{4}} \right)} \right]_0^{2\pi }\)
\(A = 2{a^2}\left[ {\frac{1}{2}\left( {0 + 2\pi } \right) - \frac{1}{8}\left( {2\pi - 0} \right) - 0} \right]\)
\(A = 2{a^2}\left[ {\pi - \frac{\pi }{4}} \right]\)
\(A = 2{a^2}\left[ {\frac{{3\pi }}{4}} \right]\)
\(A = \frac{{3\pi {a^2}}}{2}\) square unitsAnswer