Question

How do you find the area inner loop of `r=4-6sintheta`?

Answer 1

`13pi-26sin^-1(2/3)-12sqrt5`

Explanation:

First we have to find the values of `theta` that constitute one loop. A loop will begin and end at the pole (origin), when `r=0`.

`4-6sintheta=0`

`sintheta=2/3`

Then

`theta=sin^-1(2/3)`

This is the angle in Quadrant `"I"`. The corresponding angle in Quadrant `"II"` where `sintheta=2//3` is given by `pi-sin^-1(2//3)`.

So, we want the area from `alpha=sin^-1(2//3)` to `beta=pi-sin^-1(2//3)`.

The area of a polar curve `r` from `theta_1` to `theta_2` is given by `1/2int_(theta_1)^(theta_2)r^2d theta`. The area of the curve is then:

`A=1/2int_alpha^beta(4-6sintheta)^2d theta`

Also note that `(4-6sintheta)^2=(2(2-3sintheta))^2=4(2-3sintheta)^2`.

`=2int_alpha^beta(2-3sintheta)^2d theta`

`=2int_alpha^beta(4-12sintheta+9sin^2theta)d theta`

Use the simplification `sin^2theta=1/2(1-cos2theta)`.

`=2int_alpha^beta(4-12sintheta+9/2-9/2cos2theta)d theta`

`=int_alpha^beta(13-24sintheta-9cos2theta)d theta`

Integrate term-by-term. Depending on how comfortable you are with integration, you may want to use the substitution `u=2theta` for the last term.

`=[13theta+24costheta-9/2sin2theta]_alpha^beta`

Using `sin2theta=2sinthetacostheta`:

`=[13theta+24costheta-9sinthetacostheta]_alpha^beta`

Recall that `sinalpha=sinbeta=2//3`. Thus, `cosalpha=sqrt(1-sin^2alpha)=sqrt5//3`. However, since `beta` is in Quadrant `"II"`, `cosbeta=-sqrt5//3`.

`=13beta+24cosbeta-9sinbetacosbeta-13alpha-24cosalpha+9sinalphacosalpha`

`=13(beta-alpha)+24(-sqrt5/3)-9(2/3)(-sqrt5/3)-24(5/sqrt3)+9(2/3)(sqrt5/3)`

`=13(pi-sin^-1(2/3)-sin^-1(2/3))-16sqrt5+4sqrt5`

`=13pi-26sin^-1(2/3)-12sqrt5`