Question

What is the area enclosed by `r=-thetasin(-16theta^2+(7pi)/12)` between `theta in [0,(pi)/4]`?

Answer 1

`A_"Total" = 0.0597217`

Explanation:

To determine an area in polar coordinates, it first helps to look at the graph that is depicted by the given parameters:

`r=-thetasin(-16theta^2+(7pi)/12)` with `theta in [0, pi/4]`

![Desmos.com and MS Paint]()

Not every bit of the graph in within `[0,pi/4]` is an enclosed area or loop. Once we find the `r` and `theta` values corresponding to the three loops, we can use them as limits of integration. Each area formed by a loop can be determined by the formula:

`A=int_a^b1/2f[theta]^2 d theta`

Three loops, means three integrals. The roots of radius, `r`, will give us the limits of integration.

`-thetasin(-16theta^2+(7pi)/12) = 0`

The trivial solution is when `theta=0`. Because the sine function is zero at `pi n`, we can say `sin(-16 theta^2 +(7pi)/12)=0` whenever

`-16 theta^2 + (7pi)/12 = pi n`

`-16 theta^2 = pi n - (7pi)/12`

`theta^2 = -(pi n)/16 + (7 pi)/192`

`theta = +-sqrt(-(pi n)/16 + (7 pi)/192)`

We know `theta` is restricted to the interval `[0, pi//4]`. Because our interval involves only positive values, then we can write:

`0 <= sqrt(-(pi n)/16 + (7 pi)/192) <= pi/4`

`0 <= -(pi n)/16 + (7 pi)/192 <= pi^2/16`

`0 <= -pi n + (7 pi)/12 <= pi^2`

`-(7 pi)/12 <= -pi n <= pi^2 - (7 pi)/12`

`7/12 >= n >= -pi + 7/12`

`-2.55826 <= n <= 0.583333`

The integers, `n`, that lie on this interval are `-2, -1, 0`. The respective angles, `theta`, with those values of `n` are `theta=0` and:

`theta ~~0.3384, 0.5576, 0.7122`

Now we have our limits of integration for three integrals.

`A_1=int_0^0.3384 1/2[-thetasin(-16theta^2+(7pi)/12)]^2 d theta`

`A_2=int_0.3384^0.5576 1/2[-thetasin(-16theta^2+(7pi)/12)]^2 d theta`

`A_3=int_0.5576^0.7122 1/2[-thetasin(-16theta^2+(7pi)/12)]^2 d theta`

Using a computer algebra system, we get

`A_1 = 0.0057972`

`A_2 = 0.0225589`

`A_3 = 0.0313656`

`A_"Total" = 0.0597217`