Results 1 to 3 of 3

Math Help - Volumes from definite integrals

  1. #1
    Member
    Joined
    Oct 2007
    Posts
    159

    Volumes from definite integrals

    I have read and reread the section on the subject of volumes in my calculus book more times than I can remember. When I read it, it makes sense and then when I attempt the problems my integrals and answers do not match the text's. If there is someone out there that can explain what I am missing I would greatly appreciate it!!!!!

    For example, there is a cone that is 5 cm tall and has a diameter at the base of 4 cm.
    So, I take the integral from 0 to 5 cm of pi times (2 - x)^2 x being the change in the radius as you approach the top of the cone.
    I get pi times the integral of 4 + y^2 - 4y dy which has an antidervative of
    4y + y^3/3 - 2y^2 and a answer of 35pi/3

    My text says the integral should be from 0 to 5 4pi/25 y^2 dy and the volume is 20pi/3 Can someone please tell me why I am incorrect?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Oct 2008
    From
    Seattle, Washington
    Posts
    63
    Hello Frostking:

    I'm not sure how you arrived at 2 - x for the radius.

    If you integrate along the y-axis, then your radius needs to be in terms of y.


    The vertical cross-section yields a right-triangle with height 5 and base 2. The line passing through the hypotenuse has the following equation.

    y = -(5/2)*x + 5

    Taking the inverse gives us the distance x in terms of the height along the y-axis (i.e., the radius of the cylindrical disks).

    r = -(2/5)*y + 2

    The circular area of each disk is Pi*(-2/5*y + 2)^2

    The value of V below is 20*Pi/3.


    V \;=\; \pi \cdot \int_0^5 \frac{4}{25} \cdot y^2 \; - \; \frac{8}{5} \cdot y \; + \; 4 \;\; dy

    The line passing through the hypotenuse can be shifted downward so that it passes through the origin. The relationship between x and y will not change. In other words, we could have started with the following equation and then taken its inverse to find an expression for r in terms of y.

    y = -(5/2)x

    This is why the integrand in your text contains only the y-squared term.

    Cheers,


    ~ Mark
    Last edited by mmm4444bot; November 23rd 2008 at 01:27 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2007
    Posts
    159

    Volume from definite integral explaination

    Mark, I have used this site for discrete math and calculus and your answer is the BEST I have received and I have had many helpful answers. You have saved me hours of confusion and I appreciate it. I hope that others will also be assisted by it! Frostking
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: December 5th 2011, 06:21 PM
  2. Replies: 4
    Last Post: April 13th 2011, 03:08 AM
  3. Minimizing Volumes and Triple Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 26th 2009, 02:12 PM
  4. Volumes by multiple integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 3rd 2009, 07:23 PM
  5. volumes of solids using integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 2nd 2008, 12:07 PM

Search Tags


/mathhelpforum @mathhelpforum