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Math Help - Hard question

  1. #1
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    Hard question

    A region of the xy plane is defined by the inequalities
    0\leq y\leq \sin x, 0\leq x\leq \pi

    Find:
    (a) the area of the region
    (b) The first moment of area about the x-axis,
    (c) the coordinates of the centroid of this area.

    Find Also:
    (d)The volume obtained when this area is rotated completely about the x-axis,
    (e) The first moment of volume about the y-axis,
    (f) The centroid of this volume.


    My attempt:
    I found (a) correct, its value is 2.
    First moment of area about Ox is Ay\simeq \sum^{\pi}_{x=0} y^2\delta x
    2y=\int^{\pi}_0(\frac{1}{2}-\frac{1}{2}\cos 2x)dx
    2y=[\frac{x}{2}-\frac{1}{4}\sin 2x]^{\pi}_0
    2y=\frac{\pi}{2}
    y=\frac{\pi}{4}
    This gives me the correct answer, but when i proceed to (c) i have problems. The x-coordinate of the centroid would be \frac{\pi}{2} and the y-coordinate would be \frac{\pi}{4} but the answer is \frac{\pi}{8}

    I also found (d) correctly it is \frac{\pi^2}{2}
    Now (e) we have First moment of \delta V about Oy is \simeq(\pi y^2 \delta x)x
    Vx\simeq\sum^{\pi}_{x=0}(\pi y^2 x \delta x)
    \frac{\pi^2}{2}x=\int^{\pi}_0 \pi \sin^3x dx
    \frac{\pi^2}{2}x=\pi \int^{\pi}_0\sin x(1-\cos^2x)dx
    \frac{\pi^2}{2}x=\pi [-\cos x+\frac{1}{3}\cos^3x]^{\pi}_0
    \frac{\pi^2}{2}x=\pi(\frac{4}{3})
    x=\frac{8}{3\pi}
    answer is \frac{\pi^3}{4}
    I can do the last one.

    Thanks!
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  2. #2
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    deleted
    Last edited by jiboom; February 10th 2010 at 06:12 AM. Reason: pressed submit before ready
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  3. #3
    MHF Contributor Calculus26's Avatar
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    For c)

    obtains 2y = pi/2 not pi^2/2 as you have x not x^2

    so Moment is pi/4

    and the the y coordinate is pi/8
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  4. #4
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    the first moment  M_y is
    <br />
M_y= \int^{\pi}_0(\frac{y^2}{2}) dx

     <br />
2M_y=\int^{\pi}_0(\frac{1}{2}-\frac{1}{2}\cos 2x)dx

     <br />
2M_y=\frac{\pi}{2}<br />

    so
    first moment M_y= \frac{\pi}{4}

    Y co-ord of centroid is \frac{M_y}{A}=\frac{\frac{\pi}{4}}{2}=\frac{\pi}{8  }


    for volume:

    first moment of volume

     <br />
M_y=\int^{\pi}_0 \pi y^2 x dx

     <br />
M_y=\int^{\pi}_0 \pi x\sin x^2 dx

    which is

     <br />
\pi( [\frac{x^2}{4}-\frac{x\sin(2x)}{4}-\frac{\cos(2x)}{8}]^{\pi}_0 )<br />

     <br />
=\pi([\frac{\pi^2}{4}-0-\frac{1}{8}] -  [-\frac{1}{8}])<br />

     <br />
=\frac{\pi^3}{4}<br />
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  5. #5
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    I've got another similar question and this time the borders of the area are y=x^2 and y=4. The parts are about the same.
    first i was asked to get the area which i found correctly.
    then i am asked to find the first moment of area about the x-axis. so i did this:
    M_x=\int^2_{-2}\frac{(4-x^2)^2}{2}dx=\frac{1}{2}[16x-\frac{8x}{3}+\frac{x^5}{5}]^2_{-2}=\frac{256}{15}
    the answer is \frac{128}{5} as far as i can see my logic is correct and so is the working.
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  6. #6
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    Quote Originally Posted by arze View Post
    I've got another similar question and this time the borders of the area are y=x^2 and y=4. The parts are about the same.
    first i was asked to get the area which i found correctly.
    then i am asked to find the first moment of area about the x-axis. so i did this:
    M_x=\int^2_{-2}\frac{(4-x^2)^2}{2}dx=\frac{1}{2}[16x-\frac{8x}{3}+\frac{x^5}{5}]^2_{-2}=\frac{256}{15}
    the answer is \frac{128}{5} as far as i can see my logic is correct and so is the working.
    i think you better of integrating between the cure and y -axis here as you have stuff between the curve and x-axis you dont want.
    so we want (as we both sides of the y-axis i think the 2 outside integral is ok,it gives the right answer so its promising.

     <br />
2\int_0^4y(f(y))<br />

     <br />
= 2\int_0^4 y(y^{1/2})<br />

     <br />
=2 [\frac{2x^{5/2}}{5}]_0^4<br />

     <br />
=128/5<br />
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