Q. 9 - Calculus

Question

Consider the three-petaled polar rose defined by $r=\cos 3\theta$. Explain why the iterated integral ${\int }_0^{2\pi }{\int }_0^{\cos 3\theta }rdrd\theta$ calculates twice the area bounded by the petals of this rose.

Step-by-Step Solution

Answer

Integrating the iterated integral, we've verified that the value of the integral is twice the area bounded by petals of the polar rose.

Step 1 : Given Information

Given :

The Integral is${\int }_0^{2\pi }{\int }_0^{\cos 3\theta }rdrd\theta$ and  $r=\cos 3\theta$.

Step 2 : Plotting the polar rose

Plotting the polar rose,$r=\cos 3\theta$ :

Step 3 : Area bounded by the petals

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 $A=2{\int }_0^{\pi /6}{\int }_0^{r-\cos 3\theta }rdrd\theta$

Integrate with respect to $r$ first.

$A=2{\int }_0^{\pi /6}{\left[\frac{r^2}{2}\right]}_0^{\cos 3\theta }d\theta$

Put the limits

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

$A={\int }_0^{\pi /6}{\cos }^{2}3\theta d\theta$

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

Integrate with respect to $\theta$.

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

Put the limits

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

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

Area bounded by three petals

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

Step 4 : Finding the value of the Integral

$I={\int }_0^{2e}{\int }_0^{r-\cos 3\theta }rdrd\theta$

Integrate with respect to $r$ first.

$I={\int }_0^{2\pi }{\left[\frac{r^2}{2}\right]}_0^{ses3e}d\theta$

Put the limits

$I={\int }_0^{2\pi }\left[\frac{(\cos 3\theta {)}^2-0}{2}\right]d\theta$

$I=\frac{1}{2}{\int }_0^{2\pi }{\cos }^{2}3\theta d\theta$

$\left.I=\frac{1}{4}{\int }_0^{2\pi }(1+\cos 6\theta )d\theta {\cos }^{2}x=\frac{1}{2}(1+\cos 2x)\right]$

Integrate with respect to $\theta$.

$I=\frac{1}{4}{\left[\theta +\frac{1}{6}\sin 6\theta \right]}_0^{2\pi }$

Put the limits

$I=\frac{1}{4}\left[\left(2\pi +\frac{1}{6}\sin 12\pi \right)-0\right]$

$I=\frac{\pi }{2}$

Therefore,

$I=2 \mathrm{A}$

Thus, the value of integral is twice the area bounded by petals of polar rose.