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Another volume by cross section.

  1. Oct 31, 2009 #1
    The base of a solid is the triangular region bounded by the y-axis and the lines x + 2y = 4, x - 2y =4. Find the volume of the solid given that the cross sections perpendicular to the x-axis are isosceles right triangles with hypotenuse on the xy-plane.

    So what I did is:

    [tex]2\int_{x=0}^2 (4-2y)^2/(\sqrt{2})^2*1/2[/tex]

    I set [tex]A(x) = (4-2y)^2/(\sqrt{2})^2*1/2[/tex]

    and it comes from 1/2 * base * height.

    I assumed that the other two angles when the hypotenuse is 4-2y to be pi/4. I do not know why it is pi/4. I researched this forum and found out that it is supposed to be like that.

    However, the answer is wrong, so what did I do wrong here?
    (and why do you assume that the two angles are pi/4 ?)

    Thanks for reading, and thank all of you who answered my previous question.
  2. jcsd
  3. Oct 31, 2009 #2


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    How in the world did you get that integral? I can see that "4- 2y" is from the line x+ 2y= 4, but why are you not using the other line, x- 2y= 4, which is equivalent to x= 4+2y? And I have no idea how you got that "[itex]\sqrt{2}[/itex]" in there!

    Since you are told that the cross sections perpendicular to the x-axis are isosceles right triangles, you need their "thickness" along the x-axis and so should be integrating with respect to x, not y.

    The top line is given by y2= 2- (1/2)x and the bottom line by y1= -2+ (1/2)x. One the lengths of the legs of the triangles are given by the difference of those two y, y2-y1, values at each x. The area of such a triangle is, of course (1/2)(y2-y1)2. The volume of each "slab" is that are times the thickness, dx. That is what you integrate.

    You were told that the triangles were isosceles right triangles. That means that the two legs are the same length and, therefore, that the two base angles are the same. Since any triangle has angles summing to [itex]\pi[/itex] radians and one angle you know is [math]\pi/2[/itex], that leaves [itex]\pi- \pi/2= \pi/2[/itex] radians for the other two angles. And since they are the same, each is [itex](\pi/2)/2= \pi/4[/itex] radians. Of course, you don't need to know those angles for this problem.
  4. Oct 31, 2009 #3
    Yes. How in the world did I get that equation? That is so wrong.... It is so obvious for the isosceles right triangle to have two angles same....

    However, according to your answer,
    y2-y1 = 4-x &[tex]\int_{x=0}^4(1/2)(4-x)^2[/tex]

    This is wrong. [tex](4-x)^2[/tex] is supposed to be hypotenuse so the

    correct integral is:


    with [tex]A(x) = ((4-x)/\sqrt{2})^2[/tex]

    Thank you.
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