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Math Help - Urgent Help with volumes and integraiton! Please!

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    Urgent Help with volumes and integraiton! Please!

    The base of S is the triangular region with vertices (0,0) (1,0) and (0,1). Cross-sections perpendicular to the y-axis are equilateral triangles.

    WHat's the volume of the solid?

    (I have no idea!!!)
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  2. #2
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    Quote Originally Posted by Andreamet View Post
    The base of S is the triangular region with vertices (0,0) (1,0) and (0,1). Cross-sections perpendicular to the y-axis are equilateral triangles.

    WHat's the volume of the solid?
    I'm confused:
    a) (0, 0), (1, 0) and (0, 1) are the vertices of an isosceles right triangle. To get equilateral triangles as cross-sections the solid has to be distorted heavily.
    b) the height of the solid isn't mentioned so the volume could be zero or unlimited ...

    Please send the exact wording of your problem.
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    Quote Originally Posted by earboth View Post
    I'm confused:
    a) (0, 0), (1, 0) and (0, 1) are the vertices of an isosceles right triangle. To get equilateral triangles as cross-sections the solid has to be distorted heavily.
    b) the height of the solid isn't mentioned so the volume could be zero or unlimited ...

    Please send the exact wording of your problem.
    Thanks a lot for you help...

    Unfortunately that is the exact wording of the problem. The base of the shape is the right triangle described in the problem. The height is described by an equilateral triangle that is perpendicular to the Y axis. I think i can visualize it, but I don't know if I made it a little bit clear.
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  4. #4
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    Quote Originally Posted by Andreamet View Post
    Thanks a lot for you help...

    Unfortunately that is the exact wording of the problem. The base of the shape is the right triangle described in the problem. The height is described by an equilateral triangle that is perpendicular to the Y axis. I think i can visualize it, but I don't know if I made it a little bit clear.
    Does it look like this?
    Attached Thumbnails Attached Thumbnails Urgent Help with volumes and integraiton! Please!-schiefe_pyramid.gif  
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  5. #5
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    Quote Originally Posted by earboth View Post
    Does it look like this?
    yah!!!
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  6. #6
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    Quote Originally Posted by Andreamet View Post
    yah!!!
    Well then it is much easier to calculate the volume if you take the red equilateral triangle as base:

    The height in an equilateral triangle with side 1 is (use Pythagorean theorem):

    h= \frac12 \cdot \sqrt{3}

    The base area then is:

    A= \frac12 \cdot h \cdot 1= \frac12 \cdot \frac12 \cdot \sqrt{3} \cdot 1= \frac14 \cdot \sqrt{3}

    The height H of the solid is the edge(?) form (0, 0) to (0, 1) that means H = 1

    Use the formula of the volume of a pyramide:

    V_{pyr} = \frac13 \cdot A \cdot H . Plug in all values you know:

    V_{solid} = \frac13 \cdot \frac14 \cdot \sqrt{3} \cdot 1 = \frac1{12} \cdot \sqrt{3}


    EDIT: You probably have noticed that all the nice calculations are superfluous

    1. The height in the equilateral triangle has the same length as the height of the solid.
    2. The base area (painted blue) is half of a square with side 1.
    3. Use the volume formula of the pyramide:

    V_{solid} = \frac13 \cdot \frac12 \cdot \frac12 \cdot \sqrt{3} = \frac1{12} \cdot \sqrt{3}
    Last edited by earboth; February 7th 2008 at 11:27 PM.
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  7. #7
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    Quote Originally Posted by Andreamet View Post
    The base of S is the triangular region with vertices (0,0) (1,0) and (0,1). Cross-sections perpendicular to the y-axis are equilateral triangles.

    WHat's the volume of the solid?
    According to your headline you are supposed to use integrals:

    The length of the side of one cross-sectional equilateral triangle is (1-y). Therefore the area of such atringle is:

    A=\frac14 \cdot (1-y)^2 \cdot \sqrt{3}

    Consider the volume of the solid as a lot of thin slices of equilateral triangles. The thickness of one slice is dy. The volume is the sum of all possible slices:

    V=\int_1^0\left(\frac14 \cdot (1-y)^2 \cdot \sqrt{3}\right)dy = \left. \frac1{12} \cdot (1-y)^3 \cdot \sqrt{3}\right|_1^0 = \frac1{12} \cdot \sqrt{3} - 0
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  8. #8
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    cross-sections are squares

    What is the answer to this problem if let's say the cross-sections are squares?
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