Question

For the polar functions of r = V3 Cos 20, calculate the radial lines, graph and calculate the area of the region inside the curve

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Gerhard

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To calculate the radial lines for the polar function r = \sqrt{3} \cos(2\theta) , we need to set r to zero to determine where the function crosses the origin.

1. Set r = 0 :

\sqrt{3} \cos(2\theta) = 0

2. Solve for \theta :

\cos(2\theta) = 0

2\theta = \frac{\pi}{2}, \ \frac{3\pi}{2}

\theta = \frac{\pi}{4}, \ \frac{3\pi}{4}

So, the radial lines are at \frac{\pi}{4} and \frac{3\pi}{4} .

Now, to plot the graph of the polar function r = \sqrt{3} \cos(2\theta) , we have to sketch the curve based on the equation.

Next, to calculate the area of the region inside the curve, we'll integrate the polar function from \theta = \frac{\pi}{4} to \theta = \frac{3\pi}{4} and then double the result, because the region is symmetric about the x-axis.

Now, calculate the area:

A = 2\int_{\pi/4}^{3\pi/4} \frac{1}{2}(\sqrt{3}\cos(2\theta))^2 d\theta

= 2\int_{\pi/4}^{3\pi/4} \frac{3}{2} \cos^2(2\theta) d\theta

= 2\int_{\pi/4}^{3\pi/4} \frac{3}{4}(1 + \cos(4\theta)) d\theta

= 2\left[\frac{3}{4}(\theta + \frac{1}{4}\sin(4\theta))\right]_{\pi/4}^{3\pi/4}

= 2\left[\frac{3}{4}\left(\frac{3\pi}{4} - \frac{\pi}{4} + \frac{1}{4}\sin(3\pi) - \frac{1}{4}\sin(\pi)\right)\right]

= 2\left[\frac{3}{4}\left(\frac{\pi}{2} + \frac{1}{4}(0) - \frac{1}{4}(0)\right)\right]

= 2\left[\frac{3}{4}\cdot\frac{\pi}{2}\right]

= \frac{3\pi}{2}

Therefore, the area of the region inside the curve is \frac{3\pi}{2} .

\boxed{A = \frac{3\pi}{2}}

1. Set

2. Solve for

So, the radial lines are at

Now, to plot the graph of the polar function

Next, to calculate the area of the region inside the curve, we'll integrate the polar function from

Now, calculate the area:

Therefore, the area of the region inside the curve is

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