First we need to solve for the intersections between the graphs.
17sintheta=9-sintheta
17sintheta+sintheta=9
18sintheta=9
sintheta=1/2
theta=sin^-1(1/2)
theta=pi/6
Now, the thing with trig functions is that they are periodic, so we need to find the next point that has sintheta=1/2 from theta=pi/6. In this case, sin((5pi)/6)=1/2.
So now we have the interval in which we have the area above y_1=9-sintheta and the area below y_2=17sintheta (Yellow area, Figure 1).
![Desmos]()
Figure 1
Now, we can integrate the two areas.
A_1=int_(pi/6)^(5/6pi)9-sintheta \quad d theta
A_2=int_(pi/6)^(5/6pi)17sintheta \quad d theta
A=A_2-A_1
Now, just a side note that the blue shaded area is just A_1+A_2
Let's integrate A_1 and ignore the constant
int \quad 9-sintheta \quad d theta=int \quad 9 \quad d theta - int \quad sintheta \quad d theta=x+costheta
So
int_(pi/6)^(5/6pi)9-sintheta \quad d theta=x+costheta \quad |_(pi/6)^(5/6pi)
=cos(5/6pi)-cos(pi/6)+5/6pi-pi/6
=-sqrt(3)-2/3pi
Let's integrate A_2 and ignore the constant
int\quad 17sintheta \quad d theta
=-17costheta
int_(pi/6)^(5/6pi)17sintheta \quad d theta=-17costheta \quad |_(pi/6)^(5/6pi)=17sqrt(3)
So, the area is 18sqrt(3)-2/3pi
