Math  /  Calculus

QuestionA vertical right-circular cylindrical tank measures 20 ft high and 20 ft in diameter. It is full of liquid weighing 50.6lb/ft350.6 \mathrm{lb} / \mathrm{ft}^{3}. How much work does it take to pump the liquid to the level of the top of the tank?
Let y=0y=0 correspond to the bottom of the tank. Set up the integral that gives the work required, in ftlb\mathrm{ft}-\mathrm{lb}, to pump the liquid to the top of the tank. \square w=dyw=\int \square \mathrm{dy} \square (Type exact answers, using π\pi as needed.)

Studdy Solution

STEP 1

What is this asking? We need to figure out the total work needed to lift all the liquid in a cylindrical tank to the top. Watch out! Remember that the weight of a slice of liquid changes depending on how deep it is in the tank.

STEP 2

1. Define the problem variables
2. Calculate the weight of a liquid slice
3. Calculate the work to lift a liquid slice
4. Set up the integral
5. Evaluate the integral

STEP 3

Alright, let's **define** some variables!
Let yy be the depth of a liquid slice from the bottom of the tank.
The tank is **20 ft** high and **20 ft** in diameter, so the radius is **10 ft**.
The liquid's weight density is **50.6 lb/ft³**.

STEP 4

Imagine a super thin horizontal slice of liquid at depth yy.
It's like a super-thin pancake!
Its thickness is dy\mathrm{d}y, and its circular area is π(radius)2=π(10)2=100π\pi \cdot (\text{radius})^2 = \pi \cdot (10)^2 = 100\pi ft2\text{ft}^2.

STEP 5

So, the volume of this tiny liquid pancake is 100πdy100\pi \, \mathrm{d}y ft3\text{ft}^3.
Since the liquid weighs **50.6 lb/ft³**, the weight of this slice is 50.6100πdy=5060πdy50.6 \cdot 100\pi \, \mathrm{d}y = 5060\pi \, \mathrm{d}y lb\text{lb}.

STEP 6

Now, how much work does it take to lift this little pancake to the top?
Work is force (weight) times distance.
The top of the tank is at y=20y = 20, and our slice is at depth yy, so the distance to lift is 20y20 - y ft\text{ft}.

STEP 7

Therefore, the work to lift this slice is (20y)5060πdy=5060π(20y)dy(20 - y) \cdot 5060\pi \, \mathrm{d}y = 5060\pi(20 - y) \, \mathrm{d}y ft-lb.

STEP 8

To find the **total work**, we need to add up the work to lift all the little liquid pancakes from the bottom (y=0y = 0) to the top (y=20y = 20).
This is where integration comes in!

STEP 9

Our integral is: W=0205060π(20y)dyW = \int_0^{20} 5060\pi(20 - y) \, \mathrm{d}y

STEP 10

Let's **evaluate** this integral: W=5060π020(20y)dyW = 5060\pi \int_0^{20} (20 - y) \, \mathrm{d}y W=5060π[20y12y2]020W = 5060\pi \left[ 20y - \frac{1}{2}y^2 \right]_0^{20}W=5060π[(202012202)(2001202)]W = 5060\pi \left[ \left(20 \cdot 20 - \frac{1}{2} \cdot 20^2\right) - \left(20 \cdot 0 - \frac{1}{2} \cdot 0^2\right) \right]W=5060π[(400200)(0)]W = 5060\pi \left[ (400 - 200) - (0) \right]W=5060π200W = 5060\pi \cdot 200W=1012000πW = 1012000\pi

STEP 11

The total work required to pump the liquid to the top of the tank is 1012000π1012000\pi ft-lb.
So, the integral is: w=0205060π(20y)dyw = \int_0^{20} 5060\pi(20-y) \, \mathrm{d}y

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