Results 1 to 3 of 3

Math Help - volume question + Simpson's rule

  1. #1
    Member
    Joined
    May 2008
    Posts
    77

    volume question + Simpson's rule

    Two Questions, one I'm pretty confident I've done and just want confirmation, the other I'm stuck on:

    First Question:
    Use Simpson's Rule, n=4, to approximate:
    \int^5_2 \sqrt(x-1).dx
    and find a bound on the error.
    I won't bore you all with the full result:
    h=3/4
    k=0, x0=2, y0=2, w0=1, yw=2
    k=1, x1=2.75, y1=2.658, w1=4, yw=10.583
    k=2, x2=3.5, y2=3.162, w2=2, yw=6.325
    k=3, x3=4.25, y3=3.605, w3=4, yw=14.422
    k=4, x4=5, y4=4, w5=1, yw=4
    Simpson's approximation = 3/12*37.330 = 9.332 (3dp)
    Hope that's correct!
    Error:
    f''''(x)=\frac{-15}{16}(x-1)^{\frac{-7}{2}}
    x=5, this is -0.007.
    Error = (-0.007*243)/(180*256) = -0.0000386
    also correct?

    Okay onto the volume question I'm stuck on:
    Base of a solid V is the region bounded by y=x^2 and y=2-x^2. Find the volume if V has square cross sections.
    tbh, I'm really stuck on this one and haven't the foggiest!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Dr Zoidburg View Post

    Okay onto the volume question I'm stuck on:
    Base of a solid V is the region bounded by y=x^2 and y=2-x^2. Find the volume if V has square cross sections.
    tbh, I'm really stuck on this one and haven't the foggiest!
    Find the intersection points. They will define the upper and lower bounds.

    Now for the volume.

    Recall that V(x)=\int_a^b A(x)\,dx

    Note the region that we are given.



    We see that the width of the base is the distance between the 2 graphs:

    D=2-x^2-x^2=2-2x^2

    Since A_{\square}=D^2, we see that A(x)=(2-2x^2)^2

    Therefore, V(x)=\int_{-1}^{1}(2-2x^2)^2\,dx

    Does this make sense?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    77
    yes it does!
    Thanks very much for that.
    I'd got as far as finding the intersect points but no further. I missed your next step so when I tried integrating I got zero! D'oh!
    Thanks again. That's another assignment finished, tg.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More Simpson Rule Question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 2nd 2010, 10:52 PM
  2. Simpson Rule Question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 27th 2010, 05:13 AM
  3. Simpson's Rule Question! Very hard!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 12th 2010, 04:09 AM
  4. Simpson's Rule, quick question!
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 26th 2009, 10:31 PM
  5. Replies: 1
    Last Post: April 19th 2009, 12:19 AM

Search Tags


/mathhelpforum @mathhelpforum