Question

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

Answer 1

`(a^2(4-pi))/2`

Explanation:

Converting to polar using `x=rcostheta` and `y=rsintheta`:

`rcostheta(r^2cos^2theta+r^2sin^2theta)=a(r^2cos^2theta-r^2sin^2theta)`

`r^3costheta(cos^2theta+sin^2theta)=ar^2(cos^2theta-sin^2theta)`

Using `cos^2theta+sin^2theta=1`:

`r^3costheta=ar^2(cos^2theta-(1-cos^2theta))`

`r^3costheta-ar^2(2cos^2theta-1)=0`

`r^2(rcostheta-a(2cos^2theta-1))=0`

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

`rcostheta=a(2cos^2theta-1)`

`r=a(2costheta-sectheta)`

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

`2costheta=sectheta`

`2cos^2theta=1`

`costheta=pm1/sqrt2`

`theta=pmpi/4`

The area of a polar curve `r` from `theta=alpha` to `theta=beta` is given by `1/2int_alpha^betar^2d theta`, so here the integral for the area is:

`1/2int_(-pi/4)^(pi/4)[a(2costheta-sectheta)]^2d theta`

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

`=int_0^(pi/4)a^2(2costheta-sectheta)^2d theta`

`=a^2int_0^(pi/4)(4cos^2theta-4costhetasectheta+sec^2theta)d theta`

Using `cos^2theta=1/2(1+cos2theta)`:

`=a^2int_0^(pi/4)(2(1+cos2theta)-4+sec^2theta)d theta`

`=a^2int_0^(pi/4)(2cos2theta+sec^2theta-2)d theta`

Integrating term by term:

`=a^2(sin2theta+tantheta-2theta)|_0^(pi/4)`

`=a^2(sin(pi/2)+tan(pi/4)-pi/2)-a^2(sin0+tan0-0)`

`=a^2(1+1-pi/2)`

`=(a^2(4-pi))/2`