Question

Use a double integral with polar coordinates to prove that the combined area enclosed by all of the petals of the polar rose $r=\sin (2n+1)\theta$is the same for every positive integer $n$.

Short answer

Area of five petals

$$\begin{gathered} =\frac{\pi }{20}\times 5 \\ =\frac{\pi }{4} \end{gathered}$$

The combined area enclosed by all the petals of the polar rose $r=\sin (2n+1)\theta$is the same for every positive integer $n$.

Step-by-step solution

Given information

The polar rose $r=\sin (2n+1)\theta$

Calculation

The objective of this problem is to show that the combined area enclosed by all the petals of the polar rose $r=\sin (2n+1)\theta$is the same for every positive integer $n$.

Plot the polar rose $r=\sin (2n+1)\theta$for $n=1$. For $n=1,r=\sin (2n+1)\theta$shows $r=\sin 3\theta$

Plot of $r=\sin 3\theta$

calculation

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

Put $r=0$

$\sin 3\theta =0$

This implies $3\theta =n\pi$

That is $\theta =\frac{n\pi }{3}$where $n=0,1,2,\dots$

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

$A={\int }_0^{\pi /3}{\int }_0^{\sin 3\theta }rdrd\theta$

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 $\theta$.

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

Put the limits

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

$A=\frac{\pi }{12}$

Therefore, area of three petals

$=\frac{\pi }{12}\times 3$

$=\frac{\pi }{4}$

Calculation

Plot the polar rose $r=\sin (2n+1)\theta$for $n=2$. For $n=2,r=\sin (2n+1)\theta$shows $r=\sin 5\theta$

Plot of $r=\sin 5\theta$

Calculation

To find the tangent at pole of polar rose $r=\sin 5\theta$

Put $r=0$

$\sin 5\theta =0$

This implies $5\theta =n\pi$

That is $\theta =\frac{n\pi }{5}$where $n=0,1,2,\dots$

Take $n=0$and 1 for one loop. Then tangents at pole are $\theta =0$and $\theta =\frac{\pi }{5}$.

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

$A={\int }_0^{n/5}{\int }_0^{\sin \theta \theta }rdrd\theta$

Integrate with respect to $P$first.

$A={\int }_0^{*/5}{\left[\frac{r^2}{2}\right]}_0^{\text{sinse}}d\theta$

Put the limits

$$\begin{gathered} A={\int }_0^{\pi /3}\left[\frac{(\sin 5\theta {)}^2-0}{2}\right]d\theta \\ A=\frac{1}{2}{\int }_0^{n/5}{\sin }^{2}5\theta d\theta \\ A=\frac{1}{4}{\int }_0^{n/5}(1-\cos 10\theta )d\theta \end{gathered}$$

Integrate with respect to $\theta$.

$A=\frac{1}{4}{\left[\left(\theta -\frac{1}{10}\sin 10\theta \right)-0\right]}_0^{x/5}$

Put the limits

$A=\frac{1}{4}\left[\left(\frac{\pi }{5}-\frac{1}{10}\sin \left(2\pi \right)\right)-0\right]$

$A=\frac{\pi }{20}$

Therefore, area of five petals

$$\begin{gathered} =\frac{\pi }{20}\times 5 \\ =\frac{\pi }{4} \end{gathered}$$

Thus, the combined area enclosed by all the petals of the polar rose $r=\sin (2n+1)\theta$is the same for every positive integer $n$.