Question

Find the area of one petal of $r = 6\sin 2\theta $?

Answer 1

Hint: Given the value of polar coordinates. We have to find the area of one petal where there are four petals traced by the curve, one petal in each quadrant. Find the limit of integration, Then find the area at the angular position $\theta $ by substituting the values into the formula of area of polar coordinates.

Formula used:
The area in the polar coordinates is given as:
$A = \int_a^b {\dfrac{1}{2}{r^2}d\theta } $
The general solution of the equation, $\sin x = 0$ is given as:
$x = n\pi $

Complete step by step answer:
Let us assume that $f\left( \theta \right) = r = 6\sin 2\theta $

First, we will find the value of $\theta $ by substituting $f\left( \theta \right) = 0$

$ \Rightarrow 6\sin 2\theta = 0$

$ \Rightarrow \sin 2\theta = 0$

Now, we will find the general solution of the equation for $\theta $ by substituting $n = 0,1,2, \ldots $

$ \Rightarrow 2\theta = 0,\pi ,2\pi , \ldots $

Now, divide both sides by $2$.

 $ \Rightarrow \theta = 0,\dfrac{\pi }{2},\dfrac{{2\pi }}{2}, \ldots $

$ \Rightarrow \theta = 0,\dfrac{\pi }{2},\pi , \ldots $

Now we will determine the limits for the single petal which is equal to one quadrant.

$ \Rightarrow a = 0;b = \dfrac{\pi }{2}$

Now, substitute the limits and value of $r$ into the formula of area.

$ \Rightarrow A = \int_0^{\dfrac{\pi }{2}} {\dfrac{1}{2}{{\left[ {f\left( \theta \right)} \right]}^2}d\theta } $

Now, substitute the value of $f\left( \theta \right)$ into the RHS of the expression.

$ \Rightarrow A = \int_0^{\dfrac{\pi }{2}} {\dfrac{1}{2}{{\left( {6\sin 2\theta } \right)}^2}d\theta } $

On simplifying the expression, we get:

$ \Rightarrow A = \int_0^{\dfrac{\pi }{2}} {\dfrac{1}{2} \times {6^2}{{\left( {\sin 2\theta } \right)}^2}d\theta } $

$ \Rightarrow A = \int_0^{\dfrac{\pi }{2}} {3 \times 6{{\left( {\sin 2\theta } \right)}^2}d\theta } $

$ \Rightarrow A = 18\int_0^{\dfrac{\pi }{2}} {{{\sin }^2}2\theta d\theta } $

Now, apply the trigonometric identity, ${\sin ^2}2x = \dfrac{1}{2}\left( {1 - \cos 4x} \right)$to the integral.

$ \Rightarrow A = 18\int_0^{\dfrac{\pi }{2}} {\dfrac{1}{2}\left( {1 - \cos 4\theta } \right)d\theta } $

Now move the constant term out of the integral and cancel out the common term.

$ \Rightarrow A = 18 \times \dfrac{1}{2}\int_0^{\dfrac{\pi }{2}} {\left( {1 - \cos 4\theta } \right)d\theta } $

$ \Rightarrow A = 9\int_0^{\dfrac{\pi }{2}} {\left( {1 - \cos 4\theta } \right)d\theta } $

Now, integrate the expression.

\[ \Rightarrow A = 9\int_0^{\dfrac{\pi }{2}} {1 \cdot d\theta - \cos 4\theta d\theta } \]

\[ \Rightarrow A = 9\left[ {\theta - \dfrac{1}{4}\sin 4\theta } \right]_0^{\dfrac{\pi }{2}}\]

Now, substitute the values of limits into the expression and subtract the lower limit expression from upper limit expression.

\[ \Rightarrow A = 9\left[ {\left( {\dfrac{\pi }{2} - \dfrac{1}{4}\sin 4 \times \dfrac{\pi }{2}} \right) - \left( {0 - \dfrac{1}{4}\sin 4 \times 0} \right)} \right]\]

Simplify the expression, we get:

\[ \Rightarrow A = 9\left[ {\left( {\dfrac{\pi }{2} - \dfrac{1}{4}\sin 2\pi } \right) - \left( { - \dfrac{1}{4}\sin 0} \right)} \right]\]

Substitute \[\sin 2\pi = 0\] and \[\sin 0 = 0\]into the expression.

\[ \Rightarrow A = 9\left[ {\left( {\dfrac{\pi }{2} - \dfrac{1}{4} \times 0} \right) - \left( { - \dfrac{1}{4} \times 0} \right)} \right]\]

\[ \Rightarrow A = 9\left[ {\dfrac{\pi }{2} - \left( 0 \right)} \right]\]

\[ \Rightarrow A = \dfrac{{9\pi }}{2}\]

Hence the area of one petal of the curve is \[\dfrac{{9\pi }}{2}\]

Note: When the trigonometric function is given, apply the integration with the limits equal to the boundary of one petal and substitute the values into the formula.