Hydrostatic Force of A Cylindrical Tank!!! PLEASE HELP! CALCULUS 2

Question

A cylindrical tank with a diameter of 6m is placed so that the axis of the cylinder is horizontal. Find the fluid force on a circular end of the tank if the tank is half full. The density of water is 1000kg/m^3 and the acceleration of gravity is 9.8m/s^2.

1. Set up the integral to determine the force.
I got the integral of ((1000)(9pi/2)(9.8)(3-y))dy from 0 to 3 but I'm not sure if this is correct. 

I integrated it, but I didn't get the answer that my teacher gave us, which was 176,400 newtons

Can someone help explain to me how to set up the integral!

Thank you so much!

Answer 1

Consider looking at the cylinder from the end.  We see we have a circle.  Draw a circle whose center is at the origin on an xy-coordinate system with the positive y-axis pointing down.  Note the length of the cylinder does not matter.  Writing the equation of a circle we have x² + Y² = 3².  The surface of the water is the x-axis.  Draw in a cross section.  Find the width of a cross section by noting that w = 2x.  Solving for x we have x =  sqrt(9 - y²) so w = 2sqrt(9 - y²).  Area of cross section A = wΔy = 2sqrt(9 - y²)Δy.  Depth to cross section is y.  Pressure at depth y is P = (1000)(9.8)y.  Force is F = PA = 1000(9.8)(y)(2sqrt( 9 - y²)Δy.  Write the integral

 

F = integral from 0 to 3 of 1000(9.8)(2)ysqrt(9 - y²)dy = 176400 N