Question

Find the area enclosed by the astroid \(x = aco{s^3}\theta ,y = aco{s^3}\theta \)

Short answer

The area enclosed of the given curve is given as: \(A = \frac{{ - 3{a^2}\pi }}{8}\)

Step-by-step solution

Draw the graph of the given expression

The parametric equation of curve is:

\(\begin{array}{l}x = a{\cos ^3}\theta \\y = a{\sin ^3}\theta \end{array}\)

The graph of the ellipse when is shown below:

Find the area

The area enclosed by the curve\(x = f(\theta ),y = g(\theta )\)as \(\theta \)increases from\(\alpha \)to\(\beta \)is:

\(A = \int\limits_\alpha ^\beta {g(\theta )f'(\theta )} d\theta \)

Here\(f(\theta ) = a{\cos ^3}\theta ,g(\theta ) = a{\sin ^3}\theta ,\alpha = 0,\beta = 2\pi \)

Find\(f'(\theta )\)

\(\begin{array}{l}f(\theta ) = a{\cos ^3}\theta \\f'(\theta ) = - 3a{\cos ^2}\theta \sin \theta \end{array}\)

Now we need to find the interval of t.

The area of the curve is:

\(\begin{array}{l}A = \int\limits_\alpha ^\beta {g(\theta )f'(\theta )} d\theta \\A = \int\limits_\alpha ^\beta {a{{\sin }^3}\theta ( - 3a{{\cos }^2}\theta \sin \theta )} d\theta \\A = - 3{a^2}\int\limits_0^{2\pi } {{{\cos }^2}\theta ({{\sin }^4}\theta )} d\theta \end{array}\)

Therecurrence relation is, \({\int {si{n^m}xcos} ^n}xdx = \frac{{si{n^{m - 1}}xco{s^{n + 1}}x}}{{m + n}} + \frac{{m - 1}}{{m + n}}\int {si{n^{m - 2}}xco{s^n}xdx} \)

Let\(m = 4,n = 2\)

\(\begin{array}{l}\int\limits_0^{2\pi } {{{\cos }^2}\theta ({{\sin }^4}\theta )} d\theta = \frac{{{{\sin }^{4 - 1}}x{{\cos }^{2 + 1}}x}}{{4 + 2}} + \frac{{4 - 1}}{{4 + 2}}\int {{{\sin }^{4 - 2}}x{{\cos }^2}xdx} \\\int\limits_0^{2\pi } {{{\cos }^2}\theta ({{\sin }^4}\theta )} d\theta = \frac{{{{\sin }^3}x{{\cos }^3}x}}{6} + \frac{3}{6}\int {{{\sin }^2}x{{\cos }^2}xdx} \\\int\limits_0^{2\pi } {{{\cos }^2}\theta ({{\sin }^4}\theta )} d\theta = \frac{{{{\sin }^3}x{{\cos }^3}x}}{6} + \frac{1}{2}\left( { - \frac{{\sin x{{\cos }^3}x}}{4} + \frac{1}{4}\int {{{\cos }^3}x} dx} \right)\\\int\limits_0^{2\pi } {{{\cos }^2}\theta ({{\sin }^4}\theta )} d\theta = \frac{{{{\sin }^3}x{{\cos }^3}x}}{6} + \frac{1}{2}\left( { - \frac{{\sin x{{\cos }^3}x}}{4} + \frac{1}{4}\int {\frac{{1 + \cos 2x}}{2}} dx} \right)\\\int\limits_0^{2\pi } {{{\cos }^2}\theta ({{\sin }^4}\theta )} d\theta = \frac{{{{\sin }^3}x{{\cos }^3}x}}{6} - \frac{{\sin x{{\cos }^3}x}}{8} + \frac{1}{{16}}x + \frac{1}{{32}}\sin 2x\end{array}\)

Now, the area can be given as:

\(\begin{array}{l}A = - 3{a^2}\int\limits_0^{2\pi } {{{\cos }^2}\theta ({{\sin }^4}\theta )} d\theta \\A = - 3{a^2}\left( {\frac{{{{\sin }^3}x{{\cos }^3}x}}{6} - \frac{{\sin x{{\cos }^3}x}}{8} + \frac{1}{{16}}x + \frac{1}{{32}}\sin 2x} \right)_0^{2\pi }\\A = - 3{a^2}\frac{{2\pi }}{{16}}\\A = \frac{{ - 3{a^2}\pi }}{8}\end{array}\)

Hence, the area enclosed of the given curve is given as: \(A = \frac{{ - 3{a^2}\pi }}{8}\).