Question

Find the area enclosed by the inner loop of the curve \(r = 1 - 3\sin \theta \).

Short answer

The area enclosed by the inner loop of the curve \(r = 1 - 3\sin \theta \) is\(\frac{{11\pi }}{4} - \frac{{11{{\sin }^{ - 1}}(1/3)}}{2} - 3\sqrt 2 \) square units.

Step-by-step solution

Definition of the parametric equation

A parametric equation in mathematics specifies a set of numbers as functions of one or more independent variables known as parameters.

Step 2: Find two consecutive values of \(\theta \)and sketch the graph

The given curve is \(r = 1 - 3\sin \theta \).

Here, we need to find two consecutive values of \(\theta \) in the interval \((0,2\pi )\) for which\(r = 0\).

So, put \(r = 0\) in \(r = 1 - 3\sin \theta \) and solve.

\(\begin{aligned}{c}1 - 3\sin \theta = 0\\sin\theta = \frac{1}{3}\\\theta = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right),\pi - {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\end{aligned}\)

So, the \(r = 1 - 3\sin \theta \) has two solutions in \((0,2\pi )\), which are \(\theta = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\) and\(\theta = \pi - {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\).

Therefore, the required graph is plotted.

 Step 3: Find the area

Find the area of the obtained loop by using\(Area = 10\int_a^b {\frac{{{r^2}}}{2}{\rm{ }}} d\theta \)for \(a = {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\) and\(b = \pi - {\sin ^{ - 1}}\left( {\frac{1}{3}} \right)\).

Then, the area inside the loop is calculated as:

\(\begin{aligned}{c}Area = \frac{1}{2}\int_{{{\sin }^{ - 1}}(1/3)}^{\pi - {{\sin }^{ - 1}}(1/3)} {{{(1 - 3\sin \theta )}^2}} d\theta \\ = \frac{1}{2}\int_{{{\sin }^{ - 1}}(1/3)}^{\pi - {{\sin }^{ - 1}}(1/3)} {{{(1 - 3\sin \theta )}^2}} d\theta \\ = \frac{1}{2}\int_{{{\sin }^{ - 1}}(1/3)}^{\pi - {{\sin }^{ - 1}}(1/3)} {\left( {1 + 9{{\sin }^2}\theta - 6\sin \theta } \right)d\theta } \end{aligned}\)

Use \({\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}\)

\(Area = \frac{1}{2}\int_{{{\sin }^{ - 1}}(1/3)}^{\pi - {{\sin }^{ - 1}}(1/3)} {\left( {1 + \frac{{9 - \cos 2\theta }}{2} - \frac{{9\cos 2\theta }}{2} - 6\sin \theta \,d\theta } \right)} \)

Therefore, the required value is: \(\frac{1}{2}\int_{{{\sin }^{ - 1}}(1/3)}^{\pi - {{\sin }^{ - 1}}(1/3)} 1 + \frac{{9 - \cos 2\theta }}{2} - \frac{{9\cos 2\theta }}{2} - 6\sin \theta \,d\theta \).

Step 4: Evaluate the integral

Now, simplify further,

\(\begin{aligned}{c}Area = \frac{1}{2}\left( {\frac{{11\theta }}{2} - \frac{{9\sin 2\theta }}{4} + 6\cos \theta } \right)_{{{\sin }^{ - 1}}(1/3)}^{\pi - {{\sin }^{ - 1}}(1/3)}\\ = \left( {\frac{{11\theta }}{4} - \frac{{9\sin \theta \cos \theta }}{4} + 3\cos \theta } \right)_{{{\sin }^{ - 1}}(1/3)}^{\pi - {{\sin }^{ - 1}}(1/3)}\end{aligned}\)

Use \(\sin (\pi - \theta ) = \sin \theta ,\;\cos (\pi - \theta ) = - \cos \theta \).

\(Area = \frac{{11\pi }}{4} - \frac{{11{{\sin }^{ - 1}}(1/3)}}{2} + \frac{{9\sin \left( {{{\sin }^{ - 1}}(1/3)} \right)\cos \left( {{{\sin }^{ - 1}}(1/3)} \right)}}{2} - 6\cos \left( {{{\sin }^{ - 1}}(1/3)} \right)\)

Remember that: \(\cos \left( {{{\sin }^{ - 1}}a} \right) = \sqrt {1 - {a^2}} \).

\(Area = \frac{{11\pi }}{4} - \frac{{11{{\sin }^{ - 1}}(1/3)}}{2} - 3\sqrt 2 \).

Therefore, the area enclosed by the inner loop of the curve is\(\frac{{11\pi }}{4} - \frac{{11{{\sin }^{ - 1}}(1/3)}}{2} - 3\sqrt 2 \) square units.