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Chapter 9: Problem 6

Find the area of the region. One petal of \(r=6 \sin 2 \theta\)

Short Answer

Expert verified
The area of one petal of \(r=6 \sin 2 \theta\) is \(\frac{9\pi}{4}\).

Step by step solution

01

Identify the range of \(\theta\) for one petal

The function \( \sin 2 \theta \) records one full cycle of its value when \( \theta \) changes from 0 to \( \frac{\pi}{2} \). Therefore, one petal of the 4-leaved rose corresponds to this range.
02

Subtract the lower limit of \(\theta\) from the upper limit

The difference is \( \frac{\pi}{2} - 0 = \frac{\pi}{2} \). This is the range over which we will integrate in order to find the area of one petal.
03

Use the formula for the area of a polar curve to integrate over this range

The formula is \(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} (6 \sin 2 \theta)^{2} d\theta\).
04

Simplify the integrand before integrating

Note that \( (6 \sin 2 \theta)^{2} = 36 \sin^{2} 2 \theta = 18(1 - \cos 4 \theta) \). The integral becomes \(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 18 (1 - \cos 4 \theta) d\theta\).
05

Perform the integration

First, integrate \(18\) to get \(18\theta\). Then, integrate \(\cos 4 \theta\) with respect to \(\theta\), which gives \(\frac{1}{4}\sin 4\theta\). Subtract this from \(18\theta\) to get the antiderivative function, \(18\theta - \frac{1}{4}\sin 4\theta\). Evaluate this at \(\frac{\pi}{2}\) and \(0\) to get \(\frac{9\pi}{2} - 0\). Multiply by \(\frac{1}{2}\) to get the area, \(\frac{9\pi}{4}\).
06

Write the final answer

The area of one petal of \(r=6 \sin 2 \theta\) is \(\frac{9\pi}{4}\).

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