The ends of a "parabolic" water tank are the shape of the region inside the graph of y = x2 for 0 ≤ y ≤ 9 ; the cross sections parallel to the top of the tank (and the ground) are rectangles. At its center the tank is 9 feet deep and 6 feet across. The tank is 6 feet long. Rain has filled the tank and water is removed by pumping it up to a spout that is 4 feet above the top of the tank. Set up a definite integral to find the work W that is done to lower the water to a depth of 5 feet and then find the work.
Added by Vicente R.
Step 1
The volume of the water in the tank when it is full is given by the integral of the area of the cross sections from 0 to 9. The area of each cross section is the length times the width, which is 6x for a cross section at height y (since the width is 2x and the Show more…
Show all steps
Close
Your feedback will help us improve your experience
Tony Hartman and 75 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The ends of a "parabolic" water tank are the shape of the region inside the graph of y = x2 for 0 ≤ y ≤ 9 ; the cross sections parallel to the top of the tank (and the ground) are rectangles. At its center the tank is 9 feet deep and 6 feet across. The tank is 10 feet long. Rain has filled the tank and water is removed by pumping it up to a spout that is 5 feet above the top of the tank. Set up a definite integral to find the work W that is done to lower the water to a depth of 3 feet and then find the work. [Hint: You will need to integrate with respect to y.] W =_______________ = _____(foot-pounds)
Aman G.
The ends of a "parabolic" water tank are the shape of the region inside the graph of y = x2 for 0 ≤ y ≤ 4 ; the cross sections parallel to the top of the tank (and the ground) are rectangles. At its center the tank is 4 feet deep and 4 feet across. The tank is 8 feet long. Rain has filled the tank and water is removed by pumping it up to a spout that is 3 feet above the top of the tank. Set up a definite integral to find the work W that is done to lower the water to a depth of 3 feet and then find the work. [Hint: You will need to integrate with respect to y.]
Tony H.
A rectangular swimming pool $50 \mathrm{ft}$ long, $20 \mathrm{ft}$ wide, and $10 \mathrm{ft}$ deep is filled with water to a depth of $9 \mathrm{ft}$. Use an integral to find the work required to pump all the water out over the top.
Using the Definite Integral
Applications to Physics
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD