# Thread: Chk Anwser - How much work is done.....

1. ## Chk Anwser - How much work is done.....

A 10-ft trough filled with water has semicircular cross section of diameter 4 ft. How much work is done in pumping all the water over the edge of the trough? Assume that water weighs $\displaystyle 62.5 lb/ft^3$.

$\displaystyle \int_{0}^{2}10*62.5*2x*dy = \int_{0}^{2}1250*\sqrt{1 -y^2}dy$$\displaystyle =\frac{1250(1-y^2)^(3/2)}{3y}=5356 lb.ft. 2. Originally Posted by Yogi_Bear_79 A 10-ft trough filled with water has semicircular cross section of diameter 4 ft. How much work is done in pumping all the water over the edge of the trough? Assume that water weighs \displaystyle 62.5 lb/ft^3. \displaystyle \int_{0}^{2}10*62.5*2x*dy = \int_{0}^{2}1250*\sqrt{1 -y^2}dy$$\displaystyle =\frac{1250(1-y^2)^(3/2)}{3y}=5356$lb.ft.
The width of a layer $\displaystyle x\ \mbox{ft}$ below the rim of the trough is:

$\displaystyle w(x)=2\sqrt{4-x^2}\ \ \mbox{ft}$

So the volume of a layer of thickness $\displaystyle dx$ $\displaystyle x\ \mbox{ft}$ below the rim is:

$\displaystyle V(x)=w(x)\times 10\ dx=20\times \sqrt{4-x^2}\ dx\ \ \mbox{ft^3}$

and so its mass is:

$\displaystyle M(x)=1250\times \sqrt{4-x^2}\ dx \ \ \mbox{lb}$.

The work needd to lift this mass to the height of the rim is:

$\displaystyle W(x)=1250\times x\sqrt{4-x^2}\ dx\ \ \mbox{lb.ft}$.

So the total work is:

$\displaystyle \int_0^2 W(x) = \int_0^2 1250\times x\sqrt{4-x^2}\ dx\ \ \mbox{lb.ft}$

RonL

3. I know I am missing something, but by Captain Black's anwser I come up with $\displaystyle 0$ $\displaystyle lb.ft.$ work done?

4. Originally Posted by CaptainBlack
$\displaystyle \int_0^2 W(x) = \int_0^2 1250\times x\sqrt{4-x^2}\ dx\ \ \mbox{lb.ft}$
You integrated wrong. Note that the graph of $\displaystyle y = x \sqrt{4-x^2}$ is always positive on [0,2], thus the integral can't be zero.

To continue:

Let z = x^2. Then dz = 2x dx.

$\displaystyle \int_0^2 1250\times x\sqrt{4-x^2}\ dx = \int_0^4 1250 \times \frac{1}{2}\sqrt{4-z} \, dz$

Now let y = 4 - z, so dy = -dz.

$\displaystyle \int_0^2 1250\times x\sqrt{4-x^2}\ dx = - \int_4^0 \frac{1250}{2} \times \sqrt{y} \, dy$ $\displaystyle = \int_0^4 \frac{1250}{2} \times \sqrt{y} \, dy$

$\displaystyle = \frac{1250}{2} \times \frac{2}{3}y^{3/2} | _0^4$ $\displaystyle = \frac{1250}{2} \times \frac{2}{3} (8 - 0)$

$\displaystyle = \frac{1250 \cdot 8}{3}$ ft lbs.

-Dan