Results 1 to 6 of 6

Thread: Hard question

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    Hard question

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

    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 $\displaystyle Ay\simeq \sum^{\pi}_{x=0} y^2\delta x$
    $\displaystyle 2y=\int^{\pi}_0(\frac{1}{2}-\frac{1}{2}\cos 2x)dx$
    $\displaystyle 2y=[\frac{x}{2}-\frac{1}{4}\sin 2x]^{\pi}_0$
    $\displaystyle 2y=\frac{\pi}{2}$
    $\displaystyle y=\frac{\pi}{4}$
    This gives me the correct answer, but when i proceed to (c) i have problems. The x-coordinate of the centroid would be $\displaystyle \frac{\pi}{2}$ and the y-coordinate would be $\displaystyle \frac{\pi}{4}$ but the answer is $\displaystyle \frac{\pi}{8}$

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

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2009
    Posts
    83
    deleted
    Last edited by jiboom; Feb 10th 2010 at 05:12 AM. Reason: pressed submit before ready
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2009
    Posts
    83
    the first moment $\displaystyle M_y$ is
    $\displaystyle
    M_y= \int^{\pi}_0(\frac{y^2}{2}) dx $

    $\displaystyle
    2M_y=\int^{\pi}_0(\frac{1}{2}-\frac{1}{2}\cos 2x)dx$

    $\displaystyle
    2M_y=\frac{\pi}{2}
    $

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

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


    for volume:

    first moment of volume

    $\displaystyle
    M_y=\int^{\pi}_0 \pi y^2 x dx $

    $\displaystyle
    M_y=\int^{\pi}_0 \pi x\sin x^2 dx $

    which is

    $\displaystyle
    \pi( [\frac{x^2}{4}-\frac{x\sin(2x)}{4}-\frac{\cos(2x)}{8}]^{\pi}_0 )
    $

    $\displaystyle
    =\pi([\frac{\pi^2}{4}-0-\frac{1}{8}] - $$\displaystyle [-\frac{1}{8}])
    $

    $\displaystyle
    =\frac{\pi^3}{4}
    $
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338
    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:
    $\displaystyle M_x=\int^2_{-2}\frac{(4-x^2)^2}{2}dx=\frac{1}{2}[16x-\frac{8x}{3}+\frac{x^5}{5}]^2_{-2}=\frac{256}{15}$
    the answer is $\displaystyle \frac{128}{5}$ as far as i can see my logic is correct and so is the working.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2009
    Posts
    83
    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:
    $\displaystyle M_x=\int^2_{-2}\frac{(4-x^2)^2}{2}dx=\frac{1}{2}[16x-\frac{8x}{3}+\frac{x^5}{5}]^2_{-2}=\frac{256}{15}$
    the answer is $\displaystyle \frac{128}{5}$ 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.

    $\displaystyle
    2\int_0^4y(f(y))
    $

    $\displaystyle
    = 2\int_0^4 y(y^{1/2})
    $

    $\displaystyle
    =2 [\frac{2x^{5/2}}{5}]_0^4
    $

    $\displaystyle
    =128/5
    $
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. hard question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jan 25th 2010, 04:31 PM
  2. Hard question.
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: Dec 8th 2009, 12:48 PM
  3. Hard cal question Help please
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 10th 2009, 01:48 PM
  4. 2 hard question....help plz!!!!
    Posted in the Math Topics Forum
    Replies: 13
    Last Post: Jul 18th 2006, 08:28 AM
  5. need help hard question
    Posted in the Algebra Forum
    Replies: 5
    Last Post: May 29th 2006, 05:18 AM

Search Tags


/mathhelpforum @mathhelpforum