How do I find the area inside a limacon?

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How do I find the area inside a limacon?

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Answer 1 · 2018-03-30T17:32:26

The area enclosed by the limaçon $r = b + a cos θ$ $r = b + a cos theta$ is $π (b^{2} + \frac{1}{2} a^{2})$ $pi(b^2+1/2 a^2)$

Explanation:

Consider a limaçon with polar equation:

$r = b + a cos θ$ $r = b + a cos theta$

Since the question is asked in a simple form, I will make a simplifying assumption that the limaçon does not self cross, so $| a | \leq | b |$ $abs(a) <= abs(b)$ .

Dissecting the limaçon into infinitesimal segments about the origin note that each segment has area $\frac{1}{2} r^{2} d θ$ $1/2 r^2 d theta$

So the total area of the limaçon is:

$\int_{0}^{2 π} \frac{1}{2} r^{2} d θ = \int_{0}^{2 π} \frac{1}{2} {(b + a cos θ)}^{2} d θ$ $int_0^(2pi) 1/2 r^2 d theta = int_0^(2pi) 1/2 (b+acos theta)^2 d theta$

$\int_{0}^{2 π} \frac{1}{2} r^{2} d θ = \int_{0}^{2 π} \frac{1}{2} (b^{2} + 2 a b cos θ + a^{2} {cos}^{2} θ) d θ$ $color(white)(int_0^(2pi) 1/2 r^2 d theta) = int_0^(2pi) 1/2 (b^2+2ab cos theta+a^2cos^2 theta) d theta$

$\int_{0}^{2 π} \frac{1}{2} r^{2} d θ = \int_{0}^{2 π} (\frac{1}{2} b^{2} + a b cos θ + \frac{1}{4} a^{2} (1 + cos 2 θ)) d θ$ $color(white)(int_0^(2pi) 1/2 r^2 d theta) = int_0^(2pi) ( 1/2b^2+ab cos theta+1/4a^2(1+cos 2 theta)) d theta$

$\int_{0}^{2 π} \frac{1}{2} r^{2} d θ = {[\frac{1}{2} b^{2} θ + a b sin θ + \frac{1}{4} a^{2} (θ + \frac{1}{2} sin 2 θ)]}_{0}^{2}$ $color(white)(int_0^(2pi) 1/2 r^2 d theta) = [ 1/2b^2 theta+ab sin theta+1/4a^2(theta+1/2sin 2 theta)]_0^(2pi)$

$\int_{0}^{2 π} \frac{1}{2} r^{2} d θ = π (b^{2} + \frac{1}{2} a^{2})$ $color(white)(int_0^(2pi) 1/2 r^2 d theta) = pi(b^2+1/2 a^2)$

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How do I find the area inside a limacon?

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