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    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