# Thread: Work Using the Integral? Help!

1. ## Work Using the Integral? Help!

Question:
?
A tank in the form of a hemispherical bowl of radius 4 m is full of water. Find the work required to
pump all of the water to a point 2 m above the tank.??

i dont know how to start this, please i need a starting point anyone much appreciated!

2. Originally Posted by zangestu888
Question:
?
A tank in the form of a hemispherical bowl of radius 4 m is full of water. Find the work required to
pump all of the water to a point 2 m above the tank.??

i dont know how to start this, please i need a starting point anyone much appreciated!
$work = \int WALT$

W = weight density of the fluid
A = cross-sectional area as a function of y
L = lift distance of a representative horizontal cross section as a function of y
T = cross sectional thickness ... dy

sketch the graph of the circle $x^2 + y^2 = 4^2$ ... use the lower half to represent the edge of the bowl.

radius of a representative horizontal "slice" of liquid is r = x, so cross-sectional area ... $A = \pi x^2 = \pi(4^2 - y^2)$

lift distance of a representative 'slice" ... $L = 2-y$

weight density of water is $9800 \, N/m^3$

all the "slices" of water lie between $y = -4$ and $y = 0$

work in Joules is given by ...

$Work = \int_{-4}^0 9800 \pi(4^2 - y^2) (2 - y) \, dy$

3. I dont understand how you get the lift distance?, and also the weight density of the water, and the limits of intergration? how? thank you

4. Originally Posted by zangestu888
I dont understand how you get the lift distance?, and also the weight density of the water, and the limits of intergration? how? thank you
a representative horizontal slice of water has position y ... this representative slice is to be lifted to the position y = 2 ... therefore, the distance of lift is 2-y.

water has a mass density of 1000 kg/m^3
the weight of one kg of water is 9.8 N
so ... weight density of water is 9800 N/m^3

as stated in my initial response, all the horizontal slices of water reside between the positions y = -4 and y = 0 ... the integral sums the work required to lift all these slices between y = -4 and y = 0.