Fluid Force Problem- Caclulus II

[lovelylila]

Problem:
A porthole on a vertical side of a submarine submerged in sea water has a diameter of 2r feet. Sea water has a weight density of 64 pounds per cubic foot. Find the fluid force on the porthole, assuming that the center of the porthole is d feet below the surface.

Relevant Equations:
fluid pressure= weight * height aka p=wh
fluid force= p*A, A being the area the force is exerted on.

Attempt at Solution:

I know its going to be the integral of p*A bound from zero to some number. I thought that p might equal 64y (the weight times the height), but that seems too simple and I think "d" has to be incorporated into that part of the integral somehow. For A, I used the equation of the circle, x^2 + y^2 = r^2, A=y= square root (r^2-x^2). That's where I stopped, because none of that seems right, I feel like I need to have everything in terms of one or two variables in order to integrate and actually find the fluid force. P

Please help, I am so stuck and frustrated, I drew a diagram but it's not helping much! I desperately need to figure out the correct answer to this problem!

 

[HallsofIvy]

"A porthole on a vertical side". The porthole is vertical and different horizontal lines drawn across the porthole have different depths. Your diagram should be a circle with at least one horizontal line across it representing a section of the porthole at a constant depth. Since the center of the porthole is at d feet, and has radius r feet, the depth of each part of the porthole ranges from d-r feet (top) to d+r feet (bottom). Those should be the limits of integration. The equation of a circle of radius r with center at (0, d) is, of course, x^2+ (y-d)^2= r^2 or x= \sqrt{r^2- (y-d)^2}. A horizontal strip at depth y, thickness dy, will have area \sqrt{r^2- (y-d)^2}dy. The pressure at depth y is 64y so the total force on that strip is 64y\sqrt{r^2- (y-d)^2}dy. That's what you need to integrate.

 

[lovelylila]

thank you VERY much, but how does one go about integrating that? help, I'm very stuck, i tried using u-substitution for the square root, but i feel that somehow there should be an easy way to integrate the square root function, because its a circle...