Provide an appropriate response.
-True or false? If $\mathrm { n } 0$ is an even integer, then the area of the region enclosed by $\mathrm { r } = \sin \mathrm { n } \theta$ is twice the area of the region enclosed by $r = \sin [ ( n + 1 ) \theta ]$.

Question

Provide an appropriate response.
-True or false? If $\mathrm { n } > 0$ is an even integer, then the area of the region enclosed by $\mathrm { r } = \sin \mathrm { n } \theta$ is twice the area of the region enclosed by $r = \sin [ ( n + 1 ) \theta ]$.

Quizplus · Accepted Answer

This is true. The area enclosed by a polar curve $r = f(\theta)$ can be calculated using the integral $\int_{\alpha}^{\beta} \frac{1}{2} f^2(\theta) d\theta$, where $\alpha$ and $\beta$ are the angles at which the curve intersects itself. For the given curves, we can see that $\sin n\theta = \sin [(n+2)\theta]$ and $\sin[(n+1)\theta] = \sin[(n+3)\theta]$ for any even integer $n$. Therefore, the curves intersect at angles $\alpha = 0, \beta = \frac{\pi}{2n}$ and $\alpha' = \frac{\pi}{2n}, \beta' = \pi$. We can then calculate the areas enclosed by the curves:

\begin{align*}
A &= \int_{0}^{\frac{\pi}{2n}} \frac{1}{2} (\sin n\theta)^2 d\theta + \int_{\frac{\pi}{2n}}^{\pi} \frac{1}{2} (\sin [(n+1)\theta])^2 d\theta \
&= \frac{1}{4n} + \frac{1}{4(n+1)} \
&= \frac{n+n+1}{4n(n+1)} \
&= \frac{1}{2} \cdot \frac{n+1}{2n} \cdot \frac{n}{n+1} \
&= \frac{1}{2} \cdot \frac{n}{2n} \
&= \frac{1}{4} \int_{0}^{\pi} \frac{1}{2} (\sin[(n+1)\theta])^2 d\theta \
&= \frac{1}{4} A'
\end{align*}

Therefore, the area enclosed by $r = \sin n\theta$ is half the area enclosed by $r = \sin[(n+1)\theta]$.

Provide an Appropriate Response n>0\mathrm { n } > 0n>0 Is an Even Integer, Then the Area of the Region

Provide an Appropriate Response $\mathrm { n } > 0$ Is an Even Integer, Then the Area of the Region