Hey, they look like my teeth

We'll have to find the lowest points of the "molar" plot, and we can do so by differentiation:

$f'(x) = -\cos(x)*(\cos(1/\sin x))/(\sin x)^2 = 0$

There are many values that give f'(x) = 0, but the 2 corresponding coordinates to the figure are (0.2138, -1) and (2.9278, -1).

If we set a new function g(x) = \sin(\csc(x))+1, the "molar" will rise above the x-axis, and we can simply calculate the area by using definite integral from 0.2138 to 2.9278:

So the area is roughly 4.74.

f'(x) = -cos(x)*(cos(1/sin x))/(sin x)^2 = 0

There are many values that give f'(x) = 0, but the 2 corresponding coordinates to the figure are (0.2138, -1) and (2.9278, -1).

If we set a new function g(x) = sin(csc(x))+1, the "molar" will rise above the x-axis, and we can simply calculate the area by using definite integral from 0.2138 to 2.9278:

Lowest points we can calculate easier by solving equation $\sin(\frac1{\sin(x)})=-1$ , and get simple equation $\sin(x)=\frac2{3π}$ . Integral I have solved numerically