# Thread: application integral word problem

1. ## application integral word problem

I need some help on this word problem.

A water tank is in the shape of a right circular cone of altitude 10 feet and base radius 5 feet, with it's vertex at the ground. If the tank is full, find the work done in pumping all of the water out the top of the tank.

Note: The weight of water is not given so I assume it is the standard 62.4pi.

Would I need to slice the interval [0,10] or do I need to factor in the base radius?

Is this close $\frac{62.4\pi}{4}\int(y^2)(10-y)dy$

2. Originally Posted by gammaman
I need some help on this word problem.

A water tank is in the shape of a right circular cone of altitude 10 feet and base radius 5 feet, with it's vertex at the ground. If the tank is full, find the work done in pumping all of the water out the top of the tank.

Note: The weight of water is not given so I assume it is the standard 62.4pi.

Would I need to slice the interval [0,10] or do I need to factor in the base radius?

Is this close $\frac{62.4\pi}{4}\int(y^2)(10-y)dy$

sketch the lines $y = 2x$ and $y = -2x$ starting at the origin, up to the points $(5,10)$ and $(-5,10)$

weight of a representative slice is ...

$62.4 \, dV = 62.4 \pi \cdot x^2 \, dy = 62.4 \pi \cdot \frac{y^2}{4} \, dy
$

the slice needs to be lifted a distance $(10 - y)$ ... work in raising the slice is

$dW = 62.4 \pi \cdot \frac{y^2}{4}(10 - y) \, dy$

total work to raise all slices ...

$W = 15.6 \pi \int_0^{10} y^2(10 - y) \, dy$

3. sketch the lines and starting at the origin, up to the points and
why are we doing this?

4. Originally Posted by gammaman
why are we doing this?
it give you a side view of the cone so that you may determine dV and the limits of integration.

5. Where does the base radius come into play? Also I am confused why (y^2) is. Does it just come from the formula for a cone 1/3pi*r^2h. If so why do my notes say to divide by 4?

6. Originally Posted by gammaman
Where does the base radius come into play? Also I am confused why (y^2) is. Does it just come from the formula for a cone 1/3pi*r^2h. If so why do my notes say to divide by 4?
another "why" for sketching a diagram ...

a horizontal slice of the cone's liquid is a cylinder with radius $x$ and thickness $dy$.

since $y = 2x$ , $x = \frac{y}{2}$

$dV = \pi x^2 \, dy = \pi \left(\frac{y}{2}\right)^2 \, dy = \frac{\pi}{4} y^2 \, dy$