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Chapter 3: Problem 3

Find the area of the region bounded by the curve \(\mathrm{r}=\mathrm{a} \cos 4 \varphi\).

Short Answer

Expert verified
Answer: The area of the region bounded by the curve \(r=a\cos(4\varphi)\) is \(\frac{a^2\pi}{4}\).

Step by step solution

01

Graph the curve r = a cos(4桅)

To visualize the shape of the region, we'll graph the curve \(r = a \cos(4\varphi)\). Since the graph is symmetric with respect to the x-axis, we only need to consider the interval \(0 \leq \varphi \leq \pi\).
02

Determine limits of integration for 桅

To find the limits of integration for \(\varphi\), we need to find the points where the curve intersects the origin (\(r=0\)). Solve the following equation: $$a\cos(4\varphi) = 0$$ $$\cos(4\varphi) = 0$$ $$\varphi =\frac{(2n+1)\pi}{8} \quad \text{for}\quad n = 0, 1, 2, ..., 7$$ Since we only consider the interval \(0\leq\varphi\leq\pi\), we only have four possible values of \(\varphi\): $$\varphi = \frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8}$$
03

Calculate area using the polar coordinate system's area formula

Using the polar coordinate area formula: $$\text{Area} = \frac{1}{2} \int_\alpha^\beta r^2 d\varphi,$$ we can calculate the area of the regions bounded by \(r=a\cos(4\varphi)\): Since the graph is symmetric about the x-axis, the area of the entire region will be the sum of the areas of the individual regions. We can find the area of the first region, then multiply it by 8 to account for all the regions: $$\text{Total Area} = 8 \cdot \frac{1}{2}\int_{0}^{\frac{\pi}{8}} (a\cos(4\varphi))^2 d\varphi$$ $$= 4a^2 \int_{0}^{\frac{\pi}{8}} \cos^2(4\varphi) d\varphi.$$
04

Integrate the function

To integrate the function, we can use the double-angle formula for \(\cos^2(\theta) = \dfrac{1+\cos(2\theta)}{2}\): $$\text{Total Area} = 4a^2 \int_{0}^{\frac{\pi}{8}} \frac{1 + \cos(8\varphi)}{2} d\varphi.$$ Integrate the function: $$= 4a^2 \left[\frac{1}{2}\left(\frac{\varphi}{2}\right) +\frac{1}{16}\sin(8\varphi)\right]_{0}^{\frac{\pi}{8}}.$$ Finally, substituting the limits of integration, we get: $$\text{Total Area} = 4a^2 \left[\frac{1}{2}\left(\frac{\pi}{16}\right)+\frac{1}{16}\sin\left(\frac{4\pi}{2}\right)\right] - 4a^2 \left[0\right]$$ $$= \frac{a^2 \pi}{4}.$$ The area of the region bounded by the curve \(r=a\cos(4\varphi)\) is \(\frac{a^2\pi}{4}\).

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Most popular questions from this chapter

Construct the graph of the following functions: (i) \(y=1+x^{2}-0.5 x^{4}\) (ii) \(\mathrm{y}=(\mathrm{x}+1)(\mathrm{x}-2)^{2}\).

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(i) Find the area of the region enclosed by the parabola \(y=2 x-x^{2}\) and the \(x\)-axis. (ii) Find the value of \(\mathrm{m}\) so that the line \(\mathrm{y}=\mathrm{mx}\) divides the region in part (i) into two regions of equal area.
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