# How do I use the integral of work to solve the hemisphere pump problem?

I am in Calculus 2 and ran across this problem.

The tank (hemisphere) is full of water. Using the fact that the weight of water is 62.5lb/ft3, find the work (in ft-lbs) required to pump the water out of the outlet. Make sure your answer is correct to within ten ft-lbs. The radius of the hemisphere is 5.

Ok so I've learned: $$Work = (Force)(Distance)$$

The problem gives me a force: $$62.5 \frac{lb}{ft^3}$$

The problem also says the hemisphere is full, so to remove the water the distance is: $$5 - y$$

So my work to pump out the water becomes: $$Work = (62.5)(Volume)(5-y)$$

Taking a infinitely thin slice around the hemisphere, I can take the area: $$A = \pi r^2$$

I know the function for this hemisphere: $$y^2 + x^2 = 25$$ $$x^2 = 25 - y^2$$ $$x = \sqrt{25 - y^2}$$

I'll use this for the radius of my slice, so I can take the area as: $$A = \pi(25 - y^2)$$

Then to take the volume of this infinitely small slice: $$V = \pi(25 - y^2)dy$$

Substituting back into my work function, I get the amount of work to move a infinitely small slice: $$Work = (62.5)(\pi)(25 - y^2)(5-y)dy$$

I expand terms: $$Work = 62.5\pi (125-25y-5y^2 + y^3)dy$$

Then to get the full amount of work, I add all of these infinitely quantities of work by taking the integral:

$$Work = 62.5\pi\int_{0}^{5} {(125-25y-5y^2 + y^3)} dy$$

Evaluating the integral, I get: $$Work = 62.5\pi\int_{0}^{5} {(125-25y-5y^2 + y^3)} dy$$ $$Work = 62.5\pi (125y - \frac{25}{2}y^2 - \frac{5}{3}y^3 + \frac{1}{4}y^4) \Big|_0^5 = 51132.69293$$

However this is wrong. Where am I going wrong?

## 2 Answers

You have $y=0$ as the bottom, and $y=5$ as the top. But the center is at the top of the circle, so instead for the equation of the circle you want $$x^2+(y-5)^2=25$$

• What is the intuition behind this? The way I've been trying to work the problems has been to put my the top of my shape at 0, and the bottom of my shape at height. In this case 0 to 5. Did I not do this right? Commented Jan 31, 2016 at 2:21

The height in the formula and the region of integration in the geometry must be consistent, which means they must use the same origin of coordinates. One choice is: the bottom of the tank of radius $$R$$ is at $$z= -R$$, so the tank wall is the sphere $$x^2+y^2+z^2=R^2$$ and the integration means to lift water from levels $$-R\le z\le 0$$ to $$z=0$$ to get it (at least) above the rim. The work is the total potential energy difference between the water levels, and the choice of origins made above means that if all water is out, its energy is zero, so that cancels and we need only to calculate the (absolute value of the negative) potential energy of the filled tank to get the work needed. The potential energy $$P$$ is force times altitude $$z$$, the force is mass $$m$$ times acceleration $$g$$ (which is $$g\approx 9.806$$ m/s$$^2\approx 32.172$$ ft/s$$^2$$ on Earth) and mass is density $$\rho$$ times volume $$V$$. $$P= \rho g V z.$$ Split the water into small infinitesimal volumes $$dV$$ and integrate over the tank $$P= \rho g \int\int\int dV z$$ $$= \rho g \int_{z=-R}^0 dz \int_{y=-\sqrt{R^2-z^2}}^{\sqrt{R^2-z^2}}dy \int_{x=-\sqrt{R^2-y^2-z^2}}^{\sqrt{R^2-y^2-z^2}} dx z$$ $$= \rho g \int_{z=-R}^0 dz \int_{y=-\sqrt{R^2-z^2}}^{\sqrt{R^2-z^2}}dy 2\sqrt{R^2-y^2-z^2} z$$ $$= \rho g \int_{z=-R}^0 \pi (R^2-z^2) dz z = -\rho g \pi R^4/4.$$ The absolute value is $$62.5 \frac{lb}{ft^3}\times 32.172 \frac{ft}{s^2}\times 3.14159\times (5 ft)^4/4 = 987024 \frac{lb\cdot ft^2}{s^2}.$$