# Thread: Integral to calculate work

1. ## Integral to calculate work

Problem on review sheet for my son's calculus final:

A right circular canonical tank of height 25 and radius 5 has its vertex at ground level and a vertical axis has oil weighing 50 lbs/cubic foot. Set up an integral that represents the amount of work done in pumping the oil up to the top of the tank. Include an approximation for the work required to move one disk of water to the top, but do not evaluate the integral.

Answer: $\displaystyle \int_{0}^{25}{2 \pi x^2 (25-x)\;dx}$

The way we tried to solve it (imitating an example in the textbook):

Force is weight times volume. Divide the canonical tank into circular slices of height delta y:
$\displaystyle \\\Delta F =(weight)(volume) \\=(50)(\pi x^2 \Delta y)$

A circle with radius 5, centered at (0,0) has the form
$\displaystyle x^2+y^2=25$ which can be solved for
$\displaystyle x^2=25-y^2.$
I don't completely understand how (or why) they picked the center for the example we're working from, so we just chose (0,0).

The distance to the top of the cone is height of the cone minus height of the oil: $\displaystyle 25-y.$

Next, the work incremental is the force incremental times the distance
$\displaystyle \\\Delta W = \Delta F(25-y) \\= (50)(\pi x^2 \Delta y)(25-y) \\=50 \pi (y^3-25y^2-25y+625) \Delta y.$

Integrate from height of 0 to 25, so the work must be
$\displaystyle W=\int_{0}^{25}{50 \pi (y^3-25y^2-25y+625)\;dy}.$

But that doesn't equal the answer given to us.

Thanks.

2. ## Re: Integral to calculate work

work = $\displaystyle \int WALT$

$\displaystyle W$ = weight density

$\displaystyle A$ = cross-sectional area of a representative "slice"

$\displaystyle L$ = lift distance for a representative "slice"

$\displaystyle T$ = "slice" thickness

$\displaystyle dW = 50 \cdot \pi \left(\frac{y}{5}\right)^2 \cdot (25-y) \cdot dy$

integrating $\displaystyle dW$ sums all the "slices" residing from y = 0 to y = 25 ...

try it again.

3. ## Re: Integral to calculate work

Originally Posted by skeeter
work = $\displaystyle \int WALT$

$\displaystyle W$ = weight density

$\displaystyle A$ = cross-sectional area of a representative "slice"

$\displaystyle L$ = lift distance for a representative "slice"

$\displaystyle T$ = "slice" thickness

$\displaystyle dW = 50 \cdot \pi \left(\frac{y}{5}\right)^2 \cdot (25-y) \cdot dy$

integrating $\displaystyle dW$ sums all the "slices" residing from y = 0 to y = 25 ...

try it again.
My textbook has the formula $\displaystyle W=\int_a^b \rho yA(y)dy$.

Upon substitution, I get

$\displaystyle W=\int_0^{25} 50*y*\pi\left(\frac{y}{5}\right)^2dy$

$\displaystyle =\int_0^{25} 2\pi y^3dy$

Where have I gone wrong?

4. ## Re: Integral to calculate work

the lift distance for a representative "slice" is (25-y) , not y

5. ## Re: Integral to calculate work

Originally Posted by skeeter
the lift distance for a representative "slice" is (25-y) , not y
Why is that? Doesn't each representative slice have to be lifted from the ground to a height y?

6. ## Re: Integral to calculate work

the slice starts at height "y" and is pumped up to a height of 25. this means it travels 25-y.

7. ## Re: Integral to calculate work

Originally Posted by alexmahone
Why is that? Doesn't each representative slice have to be lifted from the ground to a height y?
no ... from the y-value where the slice initially resides to the y-value at the top of the cone, y = 25

look at the sketch again ...

8. ## Re: Integral to calculate work

I think I misunderstood the question to mean that the tank was being filled. Actually it is being emptied?

9. ## Re: Integral to calculate work

Originally Posted by alexmahone
I think I misunderstood the question to mean that the tank was being filled. Actually it is being emptied?
...

Set up an integral that represents the amount of work done in pumping the oil up to the top of the tank.

10. ## Re: Integral to calculate work

Originally Posted by skeeter
work = $\displaystyle \int WALT$

$\displaystyle W$ = weight density

$\displaystyle A$ = cross-sectional area of a representative "slice"

$\displaystyle L$ = lift distance for a representative "slice"

$\displaystyle T$ = "slice" thickness

$\displaystyle dW = 50 \cdot \pi \left(\frac{y}{5}\right)^2 \cdot (25-y) \cdot dy$

integrating $\displaystyle dW$ sums all the "slices" residing from y = 0 to y = 25 ...

try it again.
Thanks, skeeter. I guess the only thing we did wrong was the substitution for the height it's being pumped - the one part of the example I didn't understand.

Is it a property of cones that the ratio of the height and radius is constant? So this cone's base's radius is 5 and it's height is 25, or 1:5.

A new question: What would this equation look like if the cone were oriented with the vertex at the top and the base at the bottom. That would require a different amount of work to empty (more oil further from the hole and less oil close to it). Where does that show up in this equation?

Thanks.

11. ## Re: Integral to calculate work

...
A new question: What would this equation look like if the cone were oriented with the vertex at the top and the base at the bottom. That would require a different amount of work to empty (more oil further from the hole and less oil close to it). Where does that show up in this equation?

Thanks.
Instead of the line $\displaystyle y=5x$ representing the surface of the tank in its cross-section, we would have the line:

$\displaystyle y=-5x+25$ thus the radius of each differential circular slice would be:

$\displaystyle x=\frac{25-y}{5}$

and the work would then be:

$\displaystyle W=\int_0^{25}50\cdot\pi\left(\frac{25-y}{5}\right)^2(25-y)\,dy=\int_0^{25}2\cdot\pi(25-y)^3\,dy$

Do you see how inverting the tank replaced y with 25 - y in the squared term? It takes 3 times as much work to empty the tank now.