Complete Question
A circular swimming pool has a diameter of 16 m. The circular side of the pool is 4 m high, and the depth of the water is 3.5 m. (The acceleration due to gravity is 9.8 m/s2 and the density of water is 1000 kg/m3.) How much work (in Joules) is required to: 
(a) pump all of the water over the side
(b) pump all of the water out of an outlet 3 m over the side?
Answer:
a
The work required is 
b
The work is   
 
Explanation:
From the question we are told that 
       The diameter is  
 
        The height of the circular side is 
       The depth of water is  
        The acceleration due to gravity is 
         The density of water is   
 
The radius of the swimming pool is 
              
 The work is mathematically given as 
              
Now force is mathematically given as 
             
                
Now the work done to pump all of the water over the side
is 
           
                
                
               ![= 64000 * 9.8 \pi [4x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.](https://tex.z-dn.net/?f=%3D%2064000%20%2A%209.8%20%5Cpi%20%5B4x%20-%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20%5D%5Cleft%20%7B3.5%7D%20%5Catop%20%7B0%7D%20%5Cright.)
               ![= 64000 *9.8 \pi [4(3.5) - \frac{3.5^2}{2} ]](https://tex.z-dn.net/?f=%3D%2064000%20%2A9.8%20%5Cpi%20%5B4%283.5%29%20-%20%5Cfrac%7B3.5%5E2%7D%7B2%7D%20%5D)
               
 Now for 3 meters over the side the new height would be 
              
                        
                       
Now applying the formula for work that is  above 
            
                  
                  ![= 64000 * 9.8 \pi [7x - \frac{x^2}{2} ]\left {3.5} \atop {0} \right.](https://tex.z-dn.net/?f=%3D%2064000%20%2A%209.8%20%5Cpi%20%5B7x%20-%20%5Cfrac%7Bx%5E2%7D%7B2%7D%20%5D%5Cleft%20%7B3.5%7D%20%5Catop%20%7B0%7D%20%5Cright.)
                 ![= 64000 *9.8  \pi[7(3.5) - \frac{(3.5)^2}{2} ]](https://tex.z-dn.net/?f=%3D%2064000%20%2A9.8%20%20%5Cpi%5B7%283.5%29%20-%20%5Cfrac%7B%283.5%29%5E2%7D%7B2%7D%20%5D)
                 