1. ## Calculus 2: Excercise about "Work Needed"?

I'm not sure if this should be in this thread or in the Physics one, but it was given to me by my Calc2 professor. Thing is, I'm not sure if I'm in the right direction.

"An aquarium measures 2m long, 1m wide and 1m deep and it is full of water. Calculate the work needed to empty it halfway. (Remember that density of water is given by $1000 \frac{kg}{m^3}$)"

Here's what I have.

Volume ==> $V = 2m^3$

If we want half the volume, then $V(\frac{1}{2}) = 2m^3(\frac{1}{2})$, so the volume I want is $1m^3$

Would that make my integral go from 2 to 1? How would I go about the function I'm supposed to integrate?

Thank you!

2. Originally Posted by Gretchy
I'm not sure if this should be in this thread or in the Physics one, but it was given to me by my Calc2 professor. Thing is, I'm not sure if I'm in the right direction.

"An aquarium measures 2m long, 1m wide and 1m deep and it is full of water. Calculate the work needed to empty it halfway. (Remember that density of water is given by $1000 \frac{kg}{m^3}$)"

Here's what I have.

Volume ==> $V = 2m^3$

If we want half the volume, then $V(\frac{1}{2}) = 2m^3(\frac{1}{2})$, so the volume I want is $1m^3$

Would that make my integral go from 2 to 1? How would I go about the function I'm supposed to integrate?

Thank you!
take a representative horizontal "slice" of water ... length 2m , width 1m , thickness $dy$

volume of the slice is $2 \, dy$

mass of the slice is $2000 \, dy$

weight of the slice is $2000g \, dy$ , where $g = 9.8 \, m/s^2$

work to lift one slice to the top of the tank (force times distance) is $2000g \cdot (1-y) \, dy$

work (in Joules) to lift all the slices in the top half of the tank to the top of the tank ...

$\displaystyle W = \int_{\frac{1}{2}}^1 2000g \cdot (1-y) \, dy$