Question

Find by double integration, the area lying inside the cardioid r = a(1+cos$\theta$ ) and outside the circle r = a.

Answer 1

Given:

The cardioid r = a(1+cos ) and  the circle r = a.

To find:

Find by double integration, the area lying inside the cardioid r = a(1+cos θ) and outside the circle r = a.

Solution:

From given, we have,

The equation of cardioid r = a(1+cos θ) and the equation of circle r = a.

So, we have,

a(1 + cos θ) = a

1 + cos θ = 1

cos θ = 0

θ = cos^{-1} 0

θ = ± 90° = ± π/2

The area lying inside the cardioid r = a(1+cos θ) and outside the circle r = a is given as follows:

$\int _{\theta =\frac{-\pi }{2}}^{\frac{\pi }{2}}\int _{r=a}^{a\left(1+cos\theta \right)}\:\:r\:dr \:d\theta$

Now consider,

$\int _{r=a}^{a\left(1+cos\theta \right)}\:\:r\:dr$

$=\left[\dfrac{r^{1+1}}{1+1}\right]^{a\left(1+\cos \left(\theta \right)\right)}$

$=\left[\dfrac{r^2}{2}\right]^{a\left(1+\cos \left(\theta \right)\right)}$

$=\dfrac{a^2\left(1+\cos \left(\theta \right)\right)^2}{2}-\dfrac{a^2}{2}$

Now consider,

$\int _{\theta =\frac{-\pi }{2}}^{\frac{\pi }{2}}\:\dfrac{a^2\left(1+\cos \left(\theta \right)\right)^2}{2}-\dfrac{a^2}{2}$

$=\int _{\frac{-\pi }{2}}^{\frac{\pi }{2}}\dfrac{a^2\left(1+\cos \left(\theta \right)\right)^2}{2}d\theta -\int _{\frac{-\pi }{2}}^{\frac{\pi }{2}}\dfrac{a^2}{2}d\theta$

$=\dfrac{\left(3\pi +8\right)a^2}{4}-\dfrac{\pi a^2}{2}$

Hence the area lying inside the cardioid r = a(1+cos θ) and outside the circle r = a.

Answer 2

Hope it is helpful

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