A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs $62.5 \, \text{lb/ft}^3$. (Assume $a = 3 \, \text{ft}, b = 4 \, \text{ft},$ and $c = 5 \, \text{ft}$).

Neetesh Kumar | October 12, 2024

Calculus Homework Help

This is the solution to Work done to pump the water out of the tank
Assignment Question
Contact me if you need help with Homework, Assignments, Tutoring Sessions, or Exams for STEM subjects.
You can see our Testimonials or Vouches from here of the previous works I have done.

Get Homework Help

---

Step-by-step solution:

The work done to pump the water out of the tank can be computed using the formula:

$W = \int_0^a \text{(Weight density of water)} \times \text{Area of cross section at height } y \times \text{Distance water has to be pumped} \ \, dy$

Given:

• The weight density of water is $62.5 \, \text{lb/ft}^3$.
• Dimensions of the tank: $a = 3 \, \text{ft}, b = 4 \, \text{ft}, c = 5 \, \text{ft}$.

Step 1: Set up the problem

The base area of the tank is $b \times c = 4 \times 5 = 20 \, \text{ft}^2$.

The water at height $y$ needs to be pumped a distance $(3 - y)$.

Step 2: Set up the integral

The work done to pump the water out is:

$W = \int_0^3 62.5 \times 20 \times (3 - y) \, dy$

Step 3: Simplify and solve the integral

First, simplify:

$W = 1250 \int_0^3 (3 - y) \, dy$

Now, compute the integral:

$\int_0^3 (3 - y) \, dy = \left[ 3y - \frac{y^2}{2} \right]_0^3$

Evaluating the limits:

$= \left[ 3(3) - \frac{3^2}{2} \right] - \left[ 3(0) - \frac{0^2}{2} \right]$ $= \left[ 9 - \frac{9}{2} \right]$ $= 9 - 4.5 = 4.5$

Step 4: Compute the work

$W = 1250 \times 4.5 = 5625 \, \text{ft-lb}$

Final Answer:

$\boxed{5625 \, } \ \text{ft-lb}$

Please comment below if you find any error in this solution.
If this solution helps, then please share this with your friends.
Please subscribe to my Youtube channel for video solutions to similar questions.
Keep Smiling :-)