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).
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#
