Question

A water tank is in the form of a right circular cylinder with height 20 ft and radius 6 ft. If the tank is half full of water, find the work required to pump all of it over the top rim.

For this one I am slightly confused.
I know it should be something like:
int(0,20) 62.4 * 36pi * (10-y) dy
That gives me 0 though.
I've tried integrating from 10 to 20 also.
I can't figure it out, the answer is 1058591.1 ft-lbs

Answer 1

how much does water weigh?

Answer 2

a right circular cylindar is a tin can; the crossection at any given moment is just the area of the circle

Answer 3

62.4 lb/ft is the weight of water
the area of one slice of the cylinder is pi*r^2 which is 36pi

Answer 4

Work = Force times distance
= (2pi r)(h)(weight)

Answer 5

2 pi 6 = 12pi, not 36 pi

Answer 6

why not 36pi? I need pi*r^2 * dy for the volume of a slice of water.
If I did 2 pi 6, wouldn't that give me the circumference?

Answer 7

pfft, yeah youve got the better idea :)

Answer 8

the height is the part I can't figure out. It's not (10-y) because that makes the integral 0.
\[\int\limits_{0}^{20}62.4\times36\pi(height)\]

Answer 9

area * height

pi r^2 dh; 36pi dh

from 10 to 20, since it is half full

Answer 10

whoops forgot dy

Answer 11

It looks like each slice should be raised 20-y where y is the height of the slice
integrate from 0 to 10 (the highest slice)

Answer 12

\[64.4(36pi)\int_{10}^{20}h~dh\]

Answer 13

ah @phi yes! no wonder. thank you for your help too @amistre64

Answer 14

\[62.4(36pi)(\frac{20^2-10^2}{2})\]
\[62.4(18pi)(20^2-10^2)\]
\[62.4(18pi)(300)=1,058,591.06\]

Answer 15

I'll be posting another question rather similar to this in just a second, for clarification.

Answer 16

with a change of variables (and limits) you get amistre's set up