Calculating the work for pumping water from a hemispherical tank can be a bit tricky, but it looks like you're on the right track with your integral. However, there are a few things you may be doing wrong that are causing you to get stuck.
Firstly, it's important to note that the integral you are evaluating is for the volume of the tank, not the work needed to pump the water out. In order to calculate the work, we need to consider the weight of the water being pumped out of the tank.
To do this, we can use the formula W = mgh, where W is the work, m is the mass of the water being pumped, g is the acceleration due to gravity (32.2 ft/s^2), and h is the height the water is being pumped to. In this case, h would be the radius of the tank, or 5 feet.
Next, we need to calculate the mass of the water being pumped. Since we know the weight of water per cubic foot (62.5 pounds), we can use the formula m = V * p, where m is the mass, V is the volume, and p is the density. In this case, the volume of water being pumped would be the volume of the hemispherical tank, or 2/3 * pi * r^3, where r is the radius of the tank (5 feet). So the mass of the water being pumped would be (2/3 * pi * 5^3) * 62.5 = 1304.16 pounds.
Now, we can plug in these values into the formula W = mgh to get the work needed to pump the water out of the tank. This would be (1304.16 pounds) * (32.2 ft/s^2) * (5 feet) = 210,309.12 foot-pounds.
So it looks like the work needed to pump the water out of the lip of the tank would be 210,309.12 foot-pounds. I hope this helps clarify the process for you. Keep practicing and don't get discouraged, math can be tricky at times but with practice and patience, you'll get it!
