Results 1 to 5 of 5

Math Help - Finding the volume of a solid...

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    182

    Finding the volume of a solid...

    Find the volume of the solid lying beneath the surface z=xy and above the triangle in the x-y-plane with vertices (1,2),(1,4),(5,2).

    I can picture this in my head, and while I initially thought to thought to make the shape into a cuboid, and find its area, then half it - the related problems are all integration related, so I figured perhaps I ought to use calculus instead, perhaps some double integral...

    Beyond that, though, I'm lost...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by Unenlightened View Post
    Find the volume of the solid lying beneath the surface z=xy and above the triangle in the x-y-plane with vertices (1,2),(1,4),(5,2).

    I can picture this in my head, and while I initially thought to thought to make the shape into a cuboid, and find its area, then half it - the related problems are all integration related, so I figured perhaps I ought to use calculus instead, perhaps some double integral...

    Beyond that, though, I'm lost...
    The tricky part is figuring out the bounds. Thankfully, two of the sides of the triangle are parallel to the axes, so that will make things easier.

    Let x=1...5.

    It's clear that the lower bound of y is 2. To find the upper bound, finding the equation of the line through (1,4) and (5,2) gives y=-\frac{1}{2}x+\frac{9}{2}.

    So you need to integrate

    \int_1^5\int_2^{-\frac{1}{2}x+\frac{9}{2}}xy\,dy\,dx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Here's a hint:

    Find the equation of the line that passes through the points (1,4) and (5,2). The base of the triangle is the line y=2.

    It is y=\frac{-1}{2}x+\frac{9}{2}

    The x limits range form 1 to 5.

    That is your x limits.

    So, we have \int_{1}^{5}\int_{2}^{\frac{-1}{2}x+\frac{9}{2}}xy \;\ dydx
    Last edited by galactus; November 24th 2009 at 12:14 PM. Reason: Oops. redsoxfan beat me. well, we agree:):)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2008
    Posts
    182
    So the inner part gives us
    \frac{xy^{2}}{2} between -\frac{1}{2}x+\frac{9}{2} and 2?

    \frac{x(-\frac{1}{2}x+\frac{9}{2})^{2}}{2}-\frac{x(2)^{2}}{2}

    =\frac{x^{3}}{8}+\frac{x}{4}+\frac{81}{4}

    and then I integrate that with respect to x, giving me

    \frac{x^{4}}{32}+\frac{x^{2}}{8}+\frac{81x}{4} between 1 and 5?

    which works out to be 103.5...


    Does that seem reasonable..?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Assuming you did all your arithmetic right, that's correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the volume of a solid.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 26th 2010, 07:43 PM
  2. Finding the volume of Solid on a graph.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 25th 2010, 05:50 PM
  3. Finding volume of a solid
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 22nd 2009, 06:00 AM
  4. Finding the volume of a solid(3d)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 22nd 2009, 03:14 PM
  5. Replies: 4
    Last Post: August 17th 2006, 08:01 PM

Search Tags


/mathhelpforum @mathhelpforum