Q. 69 - Calculus

Question

Use a double integral with polar coordinates to prove that the combined area enclosed by all of the petals of the polar rose $r=\cos 2n\theta$ is the same for every positive integer $n$.

Step-by-Step Solution

Answer

Area enclosed by eight petals 

$$\begin{gathered} =\frac{\pi }{16}\times 8 \\ =\frac{\pi }{2} \end{gathered}$$

The combined area enclosed by all the petals of the polar rose $r=\cos 2n\theta$ is the same for every positive integer n.

Step 1: Given information

polar rose $r=\cos 2n\theta$

Step 2: Calculation

Plot the polar rose $r=\cos 2n\theta$ for $n=1$

Plot of $r=\cos 2n\theta$

Find the tangent at pole of polar rose $r=\cos 2\theta$

Put $r=0$

$$\begin{gathered} \cos 2\theta =0\Rightarrow 2\theta =(2n+1)\frac{\pi }{2} \\ \theta =(2n+1)\frac{\pi }{4}wheren=0,1,2,3,4,5,6, and 7. \end{gathered}$$

Take $n=0$ and $7$ for one loop. Then tangents at pole are $\theta =\frac{\pi }{4}$ and $\theta =\frac{15\pi }{4}or-\frac{\pi }{4}.$

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 $A=2{\int }_0^{\pi /4}{\int }_0^{r-\cos 2\theta }rdrd\theta$

Integrate with respect to r first.

$A=2{\int }^{\pi /4}{\left[\frac{r^2}{2}\right]}^{\cos 2\theta }rdrd\theta$

Plot of $r=\cos 2\theta$

Find the tangent at pole of polar rose $r=\cos 2\theta$

Put $r=0$

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

$\theta =(2n+1)\frac{\pi }{4}wheren=0,1,2,3,4,5,6,and7$

Take n=0 and 7 for one loop. Then tangents at pole are $\theta =\frac{\pi }{4}$ and $\theta =\frac{15\pi }{4}$ or $-\frac{\pi }{4}$.

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

$A=2{\int }_0^{\pi /4}{\int }_0^{r-\cos 2\theta }rdrd\theta$

Integrate with respect to r first.

$A=2{\int }^{\pi /4}{\left[\frac{r^2}{2}\right]}^{\cos 2\theta }d\theta$

Plot the polar rose $r=\cos 4\theta$

Plot of $r=\cos 4\theta$

Find the tangent at pole of polar rose $r=\cos 4\theta$

Put $r=0$

$\cos 4\theta =0\Rightarrow 4\theta =(2n+1)\frac{\pi }{2}$

$\theta =(2n+1)\frac{\pi }{8}$ where $n=0,1,2,\dots$

Take n=0 for one loop. Then tangents at pole are $\theta =\frac{\pi }{8}$ and $\theta =-\frac{\pi }{8}$.

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

$A=2{\int }_0^{\pi /8}{\int }_0^{r-\cos \theta \theta }rdrd\theta$

Integrate with respect to r first.

$A=2{\int }_0^{\pi /8}{\left[\frac{r^2}{2}\right]}_0^{\cos 4e}d\theta$

Put the limits

$$\begin{gathered} A=2{\int }_0^{\pi /8}\left[\frac{(\cos 4\theta {)}^2-0}{2}\right]d\theta \\ A={\int }_0^{\pi /8}{\cos }^{2}4\theta d\theta \\ A=\frac{1}{2}{\int }_0^{\pi /8}(1+\cos 8\theta )d\theta \left[{\cos }^{2}x=\frac{1}{2}(1+\cos 2x)\right] \end{gathered}$$

Integrate with respect to $\theta$.

$A=\frac{1}{2}{\left[\theta +\frac{1}{8}\sin 8\theta \right]}_0^{x/8}$

Put the limits

$$\begin{gathered} A=\frac{1}{2}\left[\left(\frac{\pi }{8}+\frac{1}{8}\sin \pi \right)-0\right] \\ A=\frac{\pi }{16} \end{gathered}$$

Therefore, area enclosed by eight petals

$$\begin{gathered} =\frac{\pi }{16}\times 8 \\ =\frac{\pi }{2} \end{gathered}$$

Thus, the combined area enclosed by all the petals of the polar rose $r=\cos 2n\theta$ is the same for every positive integer n.