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    volume calculus problem

    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. Find the voume of S.
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    Quote Originally Posted by twilightstr View Post
    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. Find the voume of S.
    1. From your equation you get:

    9x^2 + 4y^2=36~\implies~\dfrac{x^2}{4} + \dfrac{y^2}{9}=1

    That means the ellipse intersect the x-axis at x = -2 or x = 2.

    2. An isosceles right triangle with the base b has an area of

    a=\frac14 b^2

    3. With your question b = 2y~\implies~a_{right\ triangle} = \frac14(2y)^2 = y^2

    4. y^2=9-\frac94x^2

    5. Therefore the volume is:

    V=\int_{-2}^2\left( 9-\frac94x^2 \right) dx = \left[ 9x-\frac34 x^3\right]_{-2}^2 = 24
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