How to find the area of the loop of the curve $x (x^{2} + y^{2}) = a (x^{2} - y^{2})$ $x(x^2+y^2)=a(x^2-y^2)$ ?

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How to find the area of the loop of the curve $x (x^{2} + y^{2}) = a (x^{2} - y^{2})$ $x(x^2+y^2)=a(x^2-y^2)$ ?

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Answer 1 · 2017-05-11T22:35:49

$\frac{a^{2} (4 - π)}{2}$ $(a^2(4-pi))/2$

Explanation:

Converting to polar using $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ :

$r cos θ (r^{2} {cos}^{2} θ + r^{2} {sin}^{2} θ) = a (r^{2} {cos}^{2} θ - r^{2} {sin}^{2} θ)$ $rcostheta(r^2cos^2theta+r^2sin^2theta)=a(r^2cos^2theta-r^2sin^2theta)$

$r^{3} cos θ ({cos}^{2} θ + {sin}^{2} θ) = a r^{2} ({cos}^{2} θ - {sin}^{2} θ)$ $r^3costheta(cos^2theta+sin^2theta)=ar^2(cos^2theta-sin^2theta)$

Using ${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$ :

$r^{3} cos θ = a r^{2} ({cos}^{2} θ - (1 - {cos}^{2} θ))$ $r^3costheta=ar^2(cos^2theta-(1-cos^2theta))$

$r^{3} cos θ - a r^{2} (2 {cos}^{2} θ - 1) = 0$ $r^3costheta-ar^2(2cos^2theta-1)=0$

$r^{2} (r cos θ - a (2 {cos}^{2} θ - 1)) = 0$ $r^2(rcostheta-a(2cos^2theta-1))=0$

Note that $r^{2} = 0$ $r^2=0$ is just the point $(0, 0)$ $(0,0)$ , so we're left with:

$r cos θ = a (2 {cos}^{2} θ - 1)$ $rcostheta=a(2cos^2theta-1)$

$r = a (2 cos θ - sec θ)$ $r=a(2costheta-sectheta)$

The loop will begin and end when $r = 0$ $r=0$ , which is when

$2 cos θ = sec θ$ $2costheta=sectheta$

$2 {cos}^{2} θ = 1$ $2cos^2theta=1$

$cos θ = \pm \frac{1}{\sqrt{2}}$ $costheta=pm1/sqrt2$

$θ = \pm \frac{π}{4}$ $theta=pmpi/4$

The area of a polar curve $r$ $r$ from $θ = α$ $theta=alpha$ to $θ = β$ $theta=beta$ is given by $\frac{1}{2} \int_{α}^{β} r^{2} d θ$ $1/2int_alpha^betar^2d theta$ , so here the integral for the area is:

$\frac{1}{2} \int_{- \frac{π}{4}}^{\frac{π}{4}} {[a (2 cos θ - sec θ)]}^{2} d θ$ $1/2int_(-pi/4)^(pi/4)[a(2costheta-sectheta)]^2d theta$

Since the loop is symmetric across the $x$ $x$ axis:

$= \int_{0}^{\frac{π}{4}} a^{2} {(2 cos θ - sec θ)}^{2} d θ$ $=int_0^(pi/4)a^2(2costheta-sectheta)^2d theta$

$= a^{2} \int_{0}^{\frac{π}{4}} (4 {cos}^{2} θ - 4 cos θ sec θ + {sec}^{2} θ) d θ$ $=a^2int_0^(pi/4)(4cos^2theta-4costhetasectheta+sec^2theta)d theta$

Using ${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2theta=1/2(1+cos2theta)$ :

$= a^{2} \int_{0}^{\frac{π}{4}} (2 (1 + cos 2 θ) - 4 + {sec}^{2} θ) d θ$ $=a^2int_0^(pi/4)(2(1+cos2theta)-4+sec^2theta)d theta$

$= a^{2} \int_{0}^{\frac{π}{4}} (2 cos 2 θ + {sec}^{2} θ - 2) d θ$ $=a^2int_0^(pi/4)(2cos2theta+sec^2theta-2)d theta$

Integrating term by term:

$= a^{2} (sin 2 θ + tan θ - 2 θ) {∣ ∣}_{0}^{\frac{π}{4}}$ $=a^2(sin2theta+tantheta-2theta)|_0^(pi/4)$

$= a^{2} (sin (\frac{π}{2}) + tan (\frac{π}{4}) - \frac{π}{2}) - a^{2} (sin 0 + tan 0 - 0)$ $=a^2(sin(pi/2)+tan(pi/4)-pi/2)-a^2(sin0+tan0-0)$

$= a^{2} (1 + 1 - \frac{π}{2})$ $=a^2(1+1-pi/2)$

$= \frac{a^{2} (4 - π)}{2}$ $=(a^2(4-pi))/2$

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