Calculus Early Transcendentals (7th Edition) James Stewart

Page 450, #22 Solved by Andy Xiong

A tank is full of water. Find the work required to pump the water out of the spout.

3m O

1m

 x

y

Let the tank be graphed on a x-y coordinate plane where the center of the tank is at (0,0). The bottom of the tank is at (0, -3), and the top of the tank (not considering the spout) is at (0, 3). The radius is 3m. To obtain certain values for calculating work, we must arbitrarily choose a cross-section where the tank is filled with water, but not to its maximum capacity. The cross-section is a circle. In Figure A, the water is filled close to the top of the tank as we have chosen. Considering the cross section, which is a circle, we must realize that there is a right triangle that can be related to it in terms of its radius.

Figure  A    

y  3m  

Figure  B  r  

Figure C is the cross-section, which is a circle with radius r. Figure B is the right triangle, which relates to the cross-section with radius r. In order to calculate work, we must consider our integration bounds. The bottom of the tank is at (0, -3), and the top of the tank is at (0, 3). Therefore, it is plausible to integrate from [-3, 3] in respect to y. Now that we understand that our integration is in respect to y, we must express the area of the cross-section in terms of y. Recall that

𝑉 =  ∫ 𝐴(𝑦)  𝑑𝑦!! , where V is the volume and A(y) is the area of the cross-section.

𝑉 =  ! 𝐴(𝑦)  𝑑𝑦 = 𝐴(𝑦)∆𝑦!

!

Figure  C    

r  

Note that 𝐴(𝑦)∆𝑦 is just another way of expressing volume. We need this.

We need to find the area of the cross-section, which is A(y).

𝐴 𝑦 =  𝜋𝑟!

In order to find the area, we need solve for r from Figure B with the Pythagorean Theorem.

𝑟!  +  𝑦! = (3)! 𝑟! = 9−  𝑦! 𝑟 =   9− 𝑦!

Incorporate r into A(y). 𝐴 𝑦 =  𝜋( 9− 𝑦!)! 𝐴(𝑦) =  𝜋(9− 𝑦!)

Now that we have A(y), we need to reconfigure A(y) in terms of volume. We start by multiplying both sides of the previous expression by Δ𝑦 in order to do this. (Volume is the product of area and ‘a little thickness’. The area is A(y), and ‘a little thickness’ is Δ𝑦.)

                                                                 𝑉 =   𝐴 𝑦  𝑑𝑦 = 𝐴(𝑦)∆𝑦!

!

                       𝑉 =  𝐴(𝑦)∆𝑦

                                               𝐴 𝑦 Δ𝑦 =  𝜋(9− 𝑦!)Δ𝑦                                            𝑉 =  𝜋(9− 𝑦!)Δ𝑦

**Note that the unit of this volume is cubed meters.

𝑉 =  𝜋 9− 𝑦! Δ𝑦  𝑚! _________________________________________________________________ Recall that

𝑊𝑜𝑟𝑘 = 𝐹𝑜𝑟𝑐𝑒  ×  𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 and that

𝐹𝑜𝑟𝑐𝑒 = 𝑚𝑎𝑠𝑠  ×  𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝐹𝑜𝑟𝑐𝑒 = 𝑚𝑎𝑠𝑠  ×  𝑔𝑟𝑎𝑣𝑖𝑡𝑦

**Note that gravity is contributing to the production of force in this problem. Therefore, its value would be 9.8 N/kg. However, we still need mass. We need to find the mass of water at some y.

𝑚𝑎𝑠𝑠!"#$% = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦  !"#$%×  𝑣𝑜𝑙𝑢𝑚𝑒!"#$%

The density of water is 1000  𝑘𝑔/𝑚!.

𝑚𝑎𝑠𝑠!"#$% = 1000𝑘𝑔𝑚! [𝜋 9− 𝑦! Δ𝑦  𝑚!]

𝑚𝑎𝑠𝑠!"#$% = 1000𝜋 9− 𝑦! Δ𝑦  𝑘𝑔 Now we look at the equation for force.

 

𝐹𝑜𝑟𝑐𝑒 = 𝑚𝑎𝑠𝑠!"#$%×  𝑔𝑟𝑎𝑣𝑖𝑡𝑦

𝐹𝑜𝑟𝑐𝑒 = [1000𝜋 9− 𝑦! Δ𝑦  𝑘𝑔](9.8𝑁𝑘𝑔)

𝐹𝑜𝑟𝑐𝑒 = 9800𝜋 9− 𝑦! Δ𝑦  𝑁

Incorporate force into the equation of work.

𝑊𝑜𝑟𝑘 = 𝐹𝑜𝑟𝑐𝑒  ×  𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑊𝑜𝑟𝑘 = [9800𝜋 9− 𝑦! Δ𝑦  𝑁](𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)

We need to find the distance water travels or is moved when it is pumped out of the tank. See Figure D.

3m O

1m

x

y

Figure  D     y  

4  

The variable y is the distance from the arbitrary surface of the water to the top of the tank (excluding the spout). The value 4 is the total distance of the radius of the tank and the spout. Therefore, when water is pumped out through the spout, the total distance traveled is

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 4 − 𝑦   The unit of this distance is in meters.

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = (4− 𝑦)  𝑚

Calculate the value of work. 𝑊𝑜𝑟𝑘 = 𝐹𝑜𝑟𝑐𝑒  ×  𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑊𝑜𝑟𝑘 = [9800𝜋(9− 𝑦!)Δ𝑦  𝑁][(4 − 𝑦)  𝑚] 𝑊𝑜𝑟𝑘 = 9800𝜋(9 − 𝑦!)(4 − 𝑦)  Δ𝑦 J

**Note that the product of Newtons and meters is Joules. Hence, the unit J (Joules).

This is the value of work done at some y. However, this does not give us the total work. The total work is the energy required to pump out all of the water from point a to point b. And the tank is full with water. All of that water needs to be pumped out, so we need to integrate from [a,b] to get the total work. In our case, we need to integrate from [-3, 3] in respect to y. Again, the y-value of -3 is at the bottom of the tank, and the y-value of 3 is at the top of the tank (excluding the spout).

𝑊𝑜𝑟𝑘!"#$% =   𝑔 𝑦  𝑑𝑦!

!

𝑊𝑜𝑟𝑘!"#$% =   9800𝜋 9− 𝑦! 4− 𝑦  𝑑𝑦!

!!

𝑊𝑜𝑟𝑘!"#!" =  9800𝜋 9− 𝑦! 4− 𝑦  𝑑𝑦!

!!

𝑊𝑜𝑟𝑘!"#$% =  9800𝜋 (36− 9𝑦 − 4𝑦! + 𝑦!)  𝑑𝑦!

!!

𝑊𝑜𝑟𝑘!"#$% =  9800𝜋  [36𝑦 −92𝑦

! −43𝑦

! + 𝑦!]!!!  

𝑊𝑜𝑟𝑘!"#$% = 1,411,200𝜋  𝐽𝑜𝑢𝑙𝑒𝑠