Results 1 to 5 of 5

Math Help - Another volume integral

  1. #1
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1

    Another volume integral

    Let R be the region in the first quadrant bounded by the curves y = x^3 and y = 2x-x^2. Calculate the following quantities.

    (a) The area of R
    (I did this already and got 37/12 units squared)

    (b) The volume obtained by rotating R about the x-axis
    (Having trouble with this one. I feel like I know what I'm doing, but can't seem to get a satisfactory answer. I split it into two integrals:
    V = \pi \int_0^1((2x-x^2)^2 -(x^3)^2)~dx + \pi\int_{-2}^0((x^3)^2 -(2x-x^2)^2) ~dx

    But the second integral keeps returning negative volume, and I don't understand why. I am pretty sure I put the correct function on top, and that I took it over the correct interval. :/ I could just switch the sign (and hope that I did everything else correctly, it's an even problem so don't know the answer) but I wouldn't have any idea why I was doing that other than the method I thought should work didn't.

    (c) The volume obtained by rotating R about the y-axis
    (haven't even attempted this yet, can't get past b)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Ok

    Quote Originally Posted by angel.white View Post
    Let R be the region in the first quadrant bounded by the curves y = x^3 and y = 2x-x^2. Calculate the following quantities.

    (a) The area of R
    (I did this already and got 37/12 units squared)

    (b) The volume obtained by rotating R about the x-axis
    (Having trouble with this one. I feel like I know what I'm doing, but can't seem to get a satisfactory answer. I split it into two integrals:
    V = \pi \int_0^1((2x-x^2)^2 -(x^3)^2)~dx + \pi\int_{-2}^0((x^3)^2 -(2x-x^2)^2) ~dx

    But the second integral keeps returning negative volume, and I don't understand why. I am pretty sure I put the correct function on top, and that I took it over the correct interval. :/ I could just switch the sign (and hope that I did everything else correctly, it's an even problem so don't know the answer) but I wouldn't have any idea why I was doing that other than the method I thought should work didn't.

    (c) The volume obtained by rotating R about the y-axis
    (haven't even attempted this yet, can't get past b)
    you should have that V=\pi\int_{-2}^{0}x^6-(2x-x^2)^2dx +\pi\int_0^1{(2x-x^2)^2-x^6dx}...is the volume \frac{531\pi}{35}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by Mathstud28 View Post
    you should have that V=\pi\int_{-2}^{0}x^6-(2x-x^2)^2dx +\pi\int_0^1{(2x-x^2)^2-x^6dx}...is the volume \frac{531\pi}{35}?
    I don't know, it's an even problem so the book doesn't say.

    I have a theory, though: Since when x <1, y<1. This means that even though the function y=x^3 appears to be on top, it is actually the inner function (once they are flipped around the x-axis, it will be on the bottom. This means that I should be integrating outer - inner all the way along, and so I can do it with a single integral \int_{-2}^1 ((2x-x^2)^2-(x^3)^2)~dx

    I am feeling pretty confident in this answer, it makes sense to me, and explains why the integral returned negative volume earlier (because I was taking the inner radius - the outer radius). I think I got confused because I was mixing concepts for areas with concepts for volumes.


    EDIT: For the last one, rotated about the y-axis, I got (63/10)*pi Does anyone have any way of checking whether this is correct?
    Last edited by angel.white; April 13th 2008 at 01:39 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Since they're asking for the first quadrant, we only need to integrate from 0 to 1.

    Here are both methods for both axes.

    rotate about x-axis:

    Shells: 2{\pi}\int_{0}^{1}y(y^{\frac{1}{3}}-(1-\sqrt{1-y}))dy=\frac{41{\pi}}{105}

    Washers: {\pi}\int_{0}^{1}\left[(2x-x^{2})^{2}-x^{6}\right]=\frac{41{\pi}}{105}dx

    Rotate about y-axis:

    Shells: 2{\pi}\int_{0}^{1}x(2x-x^{2}-x^{3})dx=\frac{13\pi}{30}

    Washers: {\pi}\int_{0}^{1}\left[y^{\frac{2}{3}}-(1-\sqrt{1-y})^{2}\right]dy=\frac{13\pi}{30}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by galactus View Post
    Since they're asking for the first quadrant, we only need to integrate from 0 to 1.
    well . Why don't I ever read the instructions?

    Thank you so much, you saved me from getting -2 points (we get negative points if we do the wrong problem)

    Going back over it, I was able to get the same answers as you ^_^ I appreciate it, you've been very helpful today.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 18th 2011, 02:12 AM
  2. set up an integral for the volume cut out.
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 2nd 2010, 06:35 PM
  3. Volume integral to a surface integral ... divergence theorem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: October 7th 2010, 07:11 PM
  4. [SOLVED] integral volume
    Posted in the Calculus Forum
    Replies: 13
    Last Post: September 11th 2010, 11:26 PM
  5. Volume As An Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 18th 2009, 07:54 PM

Search Tags


/mathhelpforum @mathhelpforum