Q. 68 - Calculus

Question

Use a double integral with polar coordinates to prove that the area enclosed by one petal of the polar rose$r=\cos 3\theta$ is the same as the area enclosed by one petal of the polar rose $r=\sin 3\theta .$

Step-by-Step Solution

Answer

Thus, area of one petal of polar rose$r=\cos 3\theta$ is equal to the area of one petal of polar rose

$r=\sin 3\theta \text{. }$

Step 1: Given information

The objective of this problem is to show that the area of one petal of polar rose $r=\cos 3\theta$ is equal to the area of one petal of polar rose $r=\sin 3\theta .$

Plot the polar rose $r=\cos 3\theta$

Plot of $r=\sin 3\theta$

Step 2: Calculation

Find the tangent at pole of polar rose$r=\cos 3\theta$

Put $r=0$

$\cos 3\theta =0\Rightarrow 3\theta =(2n+1)\frac{\pi }{2}\theta =(2n+1)\frac{\pi }{6}$

where $n=0,1,2,3,4$ and $5$.

Take$n=0$and 5 for one loop. Then tangents at pole are $\theta =\frac{\pi }{6}and\theta =\frac{11\pi }{6}.$

Petal 1 is symmetrical about the initial line ( x - axis).

Area of the region bounded by the one petal of the curve can be expressed as

$A=2{\int }_0^{\pi i6}{\int }_0^{r-\mathrm{cm}3\theta }rdrd\theta$

Integrate with respect to r first.

$A=2{\int }_0^{\pi /6}{\left[\frac{r^2}{2}\right]}_0^{\sin \theta }d\theta$

Put the limits

$$\begin{gathered} A=2{\int }_0^{\pi /6}\left[\frac{(\cos 3\theta {)}^2-0}{2}\right]d\theta \\ A={\int }_0^{\pi /6}{\cos }^{2}3\theta d\theta \end{gathered}$$

$A=\frac{1}{2}{\int }_0^{\pi /6}(1+\cos 6\theta )d\theta \left[{\cos }^{2}x=\frac{1}{2}(1+\cos 2x)\right]$

Integrate with respect to $\theta .$

$A=\frac{1}{2}{\left[\theta +\frac{1}{6}\sin 6\theta \right]}_0^{\approx .6}$

Put the limits

$$\begin{gathered} A=\frac{1}{2}\left[\left(\frac{\pi }{6}+\frac{1}{6}\sin \pi \right)-0\right] \\ A=\frac{\pi }{12} \end{gathered}$$

Step 3: Calculation

Plot the polar race $r=\sin 3\theta$

Plot of $r=\sin 3\theta$

Step 4:Calculation

Find the tangent at pole of polar rose $r=\sin 3\theta$

Put r=0

$$\begin{gathered} \sin 3\theta =0 \\ \sin 3\theta =0 \end{gathered}$$

This implies $3\theta =n\pi$

That is $\theta =\frac{n\pi }{3}$ where n=0,1,2,3,4 and 5 .

Take n=0 and 1 for one loop. Then tangents at pole are \theta=0 and $\theta =\frac{\pi }{3}.$

Area of the region bounded by the one loop of the curve can be expressed as $A={\int }_0^{\pi /3}{\int }_0^{r-sin}3\theta rird\theta$

Integrate with respect to r first.

$A={\int }_0^{\pi /3}{\left[\frac{r^2}{2}\right]}_0^{\sin 3\theta }d\theta$

Put the limits

$$\begin{gathered} A={\int }_0^{\pi /3}\left[\frac{(\sin 3\theta {)}^2-0}{2}\right]d\theta \\ A=\frac{1}{2}{\int }_0^{\pi /3}{\sin }^{2}3\theta d\theta \\ A=\frac{1}{4}{\int }_0^{\pi /3}(1-\cos 6\theta )d\theta \left[2{\sin }^{2}3\theta =1-\cos 6\theta \right] \end{gathered}$$

Integrate with respect to \theta.

$A=\frac{1}{4}{\left[\theta -\frac{1}{6}\sin 6\theta \right]}_0^{\pi /3}$

Put the limits

$$\begin{gathered} A=\frac{1}{4}\left[\left(\frac{\pi }{3}-\frac{1}{6}\sin 2\pi \right)-0\right] \\ A=\frac{\pi }{12} \end{gathered}$$

Thus, area of one petal of polar rose $r=\cos 3\theta$is equal to the area of one petal of polar rose $r=\sin 3\theta .$