Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volume of Sold with a Known Crosssection

  1. Apr 7, 2007 #1
    the base is bounded by x^2 + y^2 = 4
    the solid is formed by isosceles right triangles perpendicular to x axis. the hypotenuse of each triangle is sitting on the base of the figure

    i just need to know what area formula i would start out with, is it just bh/2 ?
    Last edited: Apr 7, 2007
  2. jcsd
  3. Apr 7, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    An "isosceles right triangle perpendicular to the x axis" has hypotenuse y. Since it is isosceles, a2+ a2= 2a2= y2. That is, each leg has length [itex]a= y\sqrt{2}[/itex] and so its area is (1/2)bh= (1/2)(a*a)= (1/2)(2y)= y.
  4. Apr 9, 2007 #3
    hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be
    (1/4)y^2. Right?
  5. Apr 10, 2007 #4
    yes, you are right.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Volume of Sold with a Known Crosssection
  1. Volume of a cube (Replies: 2)

  2. Volumes by slicing (Replies: 4)

  3. Volume Function (Replies: 4)