• Support PF! Buy your school textbooks, materials and every day products Here!

Area of cross-sections.

  • Thread starter 7yler
  • Start date
  • #1
31
0
The ends of a "parabolic" water tank are the shape of the region inside the graph of y = [tex]x^{2}[/tex] 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. It is 5 feet long. Rain has filled the tank and water is removed by pumping it up to a pipe 2 feet above the top of the tank. Set up an iterated integral to find the work W that is done to lower the water to a depth of 2 feet and then find the work. [Hint: You will need to integrate with respect to y.]

So I came up with [tex]\int_2^4 62.5(6-y)(4-y^{2})dy[/tex] , and I'm not sure why this integral isn't correct. The water is being pumped to 6 feet, so it's 6-y, right?
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
250
hi 7yle! :wink:

your (6 - y) is correct, but I have no idea what your (4 - y2) is supposed to be :confused:

your slice is 5 ft long, dy high, and … wide? :smile:
 

Related Threads on Area of cross-sections.

  • Last Post
Replies
3
Views
676
Replies
5
Views
2K
Replies
3
Views
8K
Replies
1
Views
8K
Replies
1
Views
3K
Replies
1
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
16
Views
6K
Replies
3
Views
2K
Top