Look at the image!A solid region R , whose boundary contains the surface in the figure, is bounded by z=2y^(2),z=6,x=0, and x=6. To find the centroid of R you would compute _(R)\delta (x,y)dV=\int_a^b \int_c^d \int_e^f dzdydx,\int_a^b \int_c^d \int_e^f xdzdydx,\int_a^b \int_c^d \int_e^f ydzdydx and \int_a^b \int_c^d \int_e^f zdzdydx where a= b= d= e= f=\int_a^b \int_c^d \int_e^f dzdydx=\int_a^b \int_c^d \int_e^f xdzdydx=\int_a^b \int_c^d \int_e^f ydzdydx=\int_a^b \int_c^d \int_e^f zdzdydx= and finally the centroid is \bar{x}=()/(bar)(y)=()/(bar)(z)=