Question

Find the area of a loop of the curve `r=a sin3theta`?

Answer 1

`(pia^2)/12`

Explanation:

Where `a=1`, the curve looks like:

![desmos.com]()

Increasing or decreasing the value of `a` will only change the radius of the curve.

To find when the curve begins and ends, set `r=0`, since this is where the curve is at the origin.

If `asin3theta=0`, then `sin3theta=0`. Since `sintheta=0` at `theta=0,pi,2pi...` we see that for `sin3theta`, it will be `0` at `0,pi//3,2pi//3...`

So, the curve in the first quadrant varies from `theta=0` to `theta=pi//3`.

The expression for the area of any polar equation `r` from `theta=alpha` to `theta=beta` is given by `1/2int_alpha^betar^2d theta`.

For one loop of the given equation, the corresponding integral is then `1/2int_0^(pi//3)(asin3theta)^2d theta`.

Working this integral:

`1/2int_0^(pi//3)(asin3theta)^2d theta=1/2int_0^(pi//3)a^2(sin^2 3theta)d theta`

Use the identity `cos2alpha=1-2sin^2alpha` to rewrite for `sin^2alpha`, showing that `sin^2alpha=1/2(1-cos2alpha)`.

We can use this to say that `sin^2 3theta=1/2(1-cos6theta)`. Then the integral reduces:

`=1/2int_0^(pi//3)a^2(1/2(1-cos6theta))d theta=a^2/4int_0^(pi//3)(1-cos6theta)d theta`

Integrating term by term:

`=a^2/4(theta-1/6sin6theta)|_0^(pi//3)`

`=a^2/4[pi/3-1/6sin2pi-(0-1/6sin0)]`

`=a^2/4(pi/3)`

`=(pia^2)/12`