Question

Find the area of a single loop in curve `r=\sin(6\theta)`?

I am told the formula is `A=1/2\int_a^br^2d\theta`, so
`A=1/2\int_a^b(\sin(6\theta))^2d\theta`

But what are the bound values, `a` and `b`?

Answer 1

The area of 1 loop of the given polar curve is `pi/24` square units.

Explanation:

Start by drawing the polar curve. It helps to picture it.

As you can see, each loop starts and ends when `r = 0`. Thus our bounds of integration will be consecutive values of `theta` where `r = 0`.

`sin(6theta) = 0`

`6theta = 0 or 6theta = pi`

`theta = 0 or theta = pi/6`

Thus we will be finding the value of `1/2 int_0^(pi/6) sin^2(6x) dx` to find the area.

`A = 1/2int_0^(pi/6) sin^2(6x)dx`

Recall that `cos(2x) = 1 - 2sin^2(x)`, thus `cos(12x) = 1 - 2sin^2(6x)`, and it follows that `sin^2(6x) = (cos(12x) - 1)/(-2) = 1/2 - cos(12x)/2`

`\color(maroon)(A=1/2\int_0^(\pi/6)(1/2-\cos(12x)/2)dx)`
`\color(maroon)(A=1/2{:[1/2x-1/2(1/12\sin(12x))]|:}_0^(\pi/6))`
`A = 1/2{:[1/2x - 1/24sin(12x)]|:}_0^(pi/6)`
`A = 1/2(pi/12)`
`A = pi/24`

Hopefully this helps!