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Math Help - Finding Volume of Pyramid (Integration)?

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    Finding Volume of Pyramid (Integration)?

    It's easy to find the volume of a cone, but how would you find the volume of a pyramid using integration? With height h and base length a. And if it's a SQUARE BASED pyramid.

    Also...what's the antiderivative of (e^x)^2?
    Last edited by SportfreundeKeaneKent; July 9th 2008 at 09:23 PM.
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    It's easy to find the volume of a cone, but how would you find the volume of a pyramid using integration? With height h and base length a. And if it's a SQUARE BASED pyramid.

    Also...what's the antiderivative of (e^x)^2?
    1. Read this: Calculus/Volume - Wikibooks, collection of open-content textbooks

    2. Read this thread: http://www.mathhelpforum.com/math-he...functions.html
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    Mr. F, I think you misread

    Quote Originally Posted by SportfreundeKeaneKent View Post
    Also...what's the antiderivative of (e^x)^2?
    \int(e^x)^2\,dx = \int e^{2x}\,dx

    =\frac12\int2e^{2x}\,dx=\frac12e^{2x} + C
    Last edited by Reckoner; July 9th 2008 at 10:43 PM. Reason: Forgot my constant!
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    It's easy to find the volume of a cone, but how would you find the volume of a pyramid using integration? With height h and base length a. And if it's a SQUARE BASED pyramid.

    Also...what's the antiderivative of (e^x)^2?
    Here is one way.

    We divide the pyramid into 4 congruent right triangular pyramids.
    Meaning, cut the square-based pyramid along the two diagonals of the square base, vertically. (If you understand that, you're good ).

    If the square base is a by a, then each diagonal is sqrt(2)* a long.
    Since the diagonals of a square are perpendicular and they bisect each other, then anyone of those 4 congruent right triangular pyramids has a base that is a right triangle whose hypotenuse is a and whose legs are both sqrt(2) a/2 = a/sqrt(2) each.

    We get the volume of one of these right triangular pyramids, multiply that by 4, and we get the volume of the right square-based pyramid.

    In one of the right triangular pyramids:
    Set the dV ---the element of the volume---anywhere from y=0 and y=h.
    >>>dA = area of the dV
    Say dV is a y below the apex, and the hypotenuse of the dA is x.
    So, the area of the dA is (1/2)(x /srt(2))(x /sqrt(2)) = (x^2)/4
    And the dV is [(x^2)/4]dy

    -------------------------------------------------
    We find the relationship of y and x.

    In one vertical face of the right triangular pyramid, there are two right triangles;
    Bigger right triangle has base = a/sqrt(2), and height = h.
    Smaller right triangle has base = x/sqrt(2), and height = y.
    By proportion,
    [ a/sqrt(2)] /h = [x/sqrt(2) /y]
    Cross multiply,
    a*y = h*x
    So, x = ( a/h)y

    ------------------
    Hence, dV = [(x^2)/4]dy = (1/4)[ a/h)^2 *y^2]dy
    dV = (1/4)( a^2 /h^2)[y^2]dy

    And so, for the whole right square-based pyramid,
    V = 4{(1/4)( a^2 /h^2)INT(0 to h)[y^2]dy} ...going down is positive.

    V = ( a^2 h^2)INT(0 to h)[y^2]dy
    V = ( a^2 /h^2)[(y^3)/3](0 to h)
    V = ( a^2 /h^2)[(h^3)/3 -0]
    V = (1/3)( a^2)h

    That is it.
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    I don't really understand the first part ticbol...is there a diagram to explain it?

    What I'm also trying to do is explain why the cross sections parallel to the base will be A(x)= (a^2/h^2)*x^1
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    Yes, here is a diagram I made best I could in Paint. Hope it helps.

    The cross sections perpendicular to the y-axis(parallel to the x-axis) are squares. If s is a side length of this square then:

    \frac{\frac{s}{2}}{\frac{a}{2}}=\frac{h-y}{h}

    or

    s=\frac{a}{h}(h-y)

    \text{area of square}=A(y)=s^{2}=\frac{a^{2}}{h^{2}}(h-y)^{2}

    V=\int_{0}^{h}A(y)dy=\int_{0}^{h}\frac{a^{2}}{h^{2  }}(h-y)^{2}dy=\frac{1}{3}a^{2}h
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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