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

**Yogi_Bear_79** · August 27th 2006, 09:53 AM

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 $62.5 lb/ft^3$ .

$\int_{0}^{2}10*62.5*2x*dy = \int_{0}^{2}1250*\sqrt{1 -y^2}dy$ $=\frac{1250(1-y^2)^(3/2)}{3y}=5356$ lb.ft.

**CaptainBlack** · August 27th 2006, 10:15 AM

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 $62.5 lb/ft^3$ .

$\int_{0}^{2}10*62.5*2x*dy = \int_{0}^{2}1250*\sqrt{1 -y^2}dy$ $=\frac{1250(1-y^2)^(3/2)}{3y}=5356$ lb.ft.

The width of a layer $x\ \mbox{ft}$ below the rim of the trough is:

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

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

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

and so its mass is:

$ 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:

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

So the total work is:

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

RonL

**Yogi_Bear_79** · September 5th 2006, 02:50 PM

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

**topsquark** · September 5th 2006, 03:35 PM

Originally Posted by CaptainBlack

$ \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 $y = x \sqrt{4-x^2}$ is always positive on [0,2], thus the integral can't be zero.

To continue:

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

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

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

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

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

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

-Dan

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