Let' consider a circumference with centre in #C(0;-1)# and radius 9. Its equation is #x^2+(y+1)^2=81# and its vertical diameter is 8 long in the positive y semiplane (for #y> 0#) just the height of our cap.
The intersect #OI# of it with the x-axys is the height of the rectangle triangle inscripted inside the vertical semicircunference where the projection of the two catheti are just 8 and 10. As a result, according to the second euclid's theorem #IO=root2(80)=4root2(5)#.
Notice that #2*OI# is the side of the prism's base, i.e. #L=8root2(5)# whereas its height is #H=15#, so the prism's volume is
#V_(prism)=L^2*H=(8*root2(5))^2*15=4800 #
Moving to the cap's volume, it can be seen as the volume of the solid generated by the arc of circumference in the first quarter of the cartesian axis, whose explicit equation is
#y=root2(81-x^2)-1# for #0 < x<4root2(5)#
It results from the second theorem of Guldino that
#V_("cap")=2pi*x_("barycentre")*area_("section")#
#V_("cap")= pi*(int_0^(4root2(5))(root2(81-x^2)-1)^2dx)/(int_0^(4root2(5))dx)*int_0^(4root2(5))dx#
that can be simplified as
#V_("cap")=pi*int_0^(4root2(5))(-2root2(81-x^2)+1+81-x^2)dx#
Solved the integral we get
#V_("cap")=pi(-2*(1/2xroot2(81-x^2)+81*arcsin(x/9))+82x-x^3/3)|_0^(4root2(5))#
#V_("cap")=pi(-4root2(5)-162*arcsin4root2(5)/9+82*4root2(5)-(4root2(5))^3/3)#
