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Math Help - Cross sections of a region

  1. #1
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    Find the volume of the described solid

    Find the volume of the described solid S

    The base of S is an elliptical region with boundary curve 9x^2 + 4y^2 = 36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

    Some help please
    Last edited by amiv4; October 2nd 2008 at 08:07 PM.
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  2. #2
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    i think all i need is the formula for the isosceles right triangle. So can anyone tell me what taht is. and would the integral go from -6 to 6
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  3. #3
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    Given 9x^2 + 4y^2 = 36, you have

    4y^2 = 36 - 9x^2

    y^2 = \frac{36 - 9x^2}{4} = 9 - \frac{9x^2}{4}

    Now, the length of each triangle base is 2y.

    So the area of each triangle is \frac{1}{2}y\sqrt{2}\cdot y\sqrt{2} = y^2, which is 9 - \frac{9x^2}{4}.

    Can you do the rest?
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  4. #4
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    what would the integral be evaluated from
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  5. #5
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    Quote Originally Posted by amiv4 View Post
    what would the integral be evaluated from
    What are the extreme left and right points on the ellipse?
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