Question

Find the area of the following region. The region outside the circle r=3 and inside the circle r=-6cos theta

Answer 1

The region outside the circle r = 3 and inside the circle r = - 6cos θ is given below.

$\rm Area = \dfrac{12\pi - 9\sqrt3 }{2}$

What is an area bounded by the curve?

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The region outside the circle r = 3 and inside the circle r = - 6cos θ

Then the value of the θ will be

$\begin{aligned} -6\cos \theta &= 3\\\\\cos \theta &= -\dfrac{1}{2}\\\\\theta &= -\dfrac{2\pi}{3}, \dfrac{2\pi}{3} \end{aligned}$

Then the area will be given as

$\rm Area = \dfrac{1}{2} \int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}} [-6 \cos \theta)^2 - 3^2 ] d \theta \\\\\\Area = \dfrac{1}{2} \int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}} [18 (2\cos ^2 \theta ) - 9]d \theta \\\\\\Area = \dfrac{9}{2} \int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}} [2(1 + \cos 2\theta ) - 1] d \theta \\\\\\Area = \dfrac{9}{2} \int_{-\frac{2\pi}{3}}^{\frac{2\pi}{3}} [2 \cos 2\theta ) + 1] d \theta$

$\rm Area = \dfrac{9}{2} [ \sin 2\theta + \theta ] _{-\frac{2\pi}{3}}^{\frac{2\pi}{3}} \\\\\\ Area = \dfrac{9}{2} [ \sin 2( \frac{2\pi}{3}) +\frac{2\pi}{3} - \sin 2 (\frac{2\pi}{3}) - \frac{2\pi}{3}]$

On simplifying, we have

$\rm Area = \dfrac{12\pi - 9\sqrt3 }{2}$

More about the area bounded by the curve link is given below.

https://brainly.com/question/24563834

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