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

  1. #1
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    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 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.
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  2. #2
    Grand Panjandrum
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    Quote 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:

    <br />
w(x)=2\sqrt{4-x^2}\ \ \mbox{ft}<br />

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

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

    and so its mass is:

    <br />
M(x)=1250\times \sqrt{4-x^2}\ dx \ \ \mbox{lb}<br />
.

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

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

    So the total work is:

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

    RonL
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  3. #3
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    I know I am missing something, but by Captain Black's anwser I come up with 0 lb.ft. work done?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack
    <br />
\int_0^2 W(x) = \int_0^2 1250\times x\sqrt{4-x^2}\ dx\ \ \mbox{lb.ft}<br />
    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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