91Ó°ÊÓ

A rose is a mathematical curve whose equation is given in the form \(r = a \sin(k\theta))\) or \(r = a \cos(k\theta)\). In our case, we have a rose curve given by the equation \(r = 3 \sin(2\theta)\). The number of leaves (petals) of the rose depends on the integer value of \(k\). If \(k\) is odd, the rose will have \(k\) leaves; if \(k\) is even, the rose will have \(2k\) leaves. In our case, \(k = 2\), so the rose has \(2 * 2 = 4\) leaves.

To sketch the polar curve \(r = 3 \sin(2\theta)\), we can start by noting the key properties. Firstly, our rose curve is symmetric about the origin as it is a sine function rather than a cosine function. Secondly, since it is periodic with period \(\pi\) (as \(k = 2\) here), we can plot its values for \(0 \leq \theta \leq \pi\). Lastly, we know that it has 4 leaves. Now we can plot these leaves using points formed by evaluating the function at different \(\theta\) values within its domain.

The rose curve \(r = 3 \sin(2\theta)\) is a closed curve, and its leaves form the regions we want to calculate the area of. We know that there will be 4 symmetric regions, so we can focus on finding the area of one region and then multiply it by 4. Since the curve is symmetric about the origin, we can choose to work with the region where \(0 \leq \theta \leq \frac{\pi}{4}\).

Now, to find the area of the region within one of the leaves in the bounds \(0 \leq \theta \leq \frac{\pi}{4}\), we'll use the polar area formula, which is given by: Area \(= \frac{1}{2}\int_{\alpha}^{\beta}r(\theta)^2 d\theta\). In our case, \(\alpha = 0\) and \(\beta = \frac{\pi}{4}\). So we have: Area = \(\frac{1}{2}\int_{0}^{\frac{\pi}{4}}(3\sin(2\theta))^2 d\theta\).

First, simplify the integral: Area = \(\frac{1}{2}\int_{0}^{\frac{\pi}{4}}(9\sin^2(2\theta)) d\theta\) To compute this integral, we will use the double angle formula for sine squared function \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\): Area = \(\frac{1}{2}\int_{0}^{\frac{\pi}{4}}(9\cdot\frac{1 - \cos(4\theta)}{2}) d\theta\) Next, we evaluate the integral by solving using u-substitution for \(\cos(4\theta))\): Area = \(\frac{9}{4}\left[\theta - \frac{1}{4}\sin(4\theta)\right]_0^{\frac{\pi}{4}}\) Plug in the limits of integration to find the area of one region: Area = \(\frac{9}{4}\left(\frac{\pi}{4} - \frac{1}{4}\sin(\pi)\right) - \frac{9}{4}(0 - \frac{1}{4}\sin(0)) = \frac{9\pi}{16}\) Now, since there are 4 leaves in total, multiply this value by 4 to find the total area of the regions within all the leaves: Total Area = \(4 \cdot \frac{9\pi}{16} = \frac{9\pi}{4}\).