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Math Help - Find the volume of the solid

  1. #1
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    Find the volume of the solid

    Hi,

    I'm having trouble with this question as I'm sure how the do the problem when it involves the isosceles triangle. I be great if someone could possibly draw me a diagram and show me how to form the equation to use and evaluate the definite integral that is obtained from the intercepts that I already know to be the square root of 1/2 and -1/2.

    Thanks,
    Daddy_Long_Legs
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Daddy_Long_Legs View Post
    Hi,

    I'm having trouble with this question as I'm sure how the do the problem when it involves the isosceles triangle. I be great if someone could possibly draw me a diagram and show me how to form the equation to use and evaluate the definite integral that is obtained from the intercepts that I already know to be the square root of 1/2 and -1/2.

    Thanks,
    Daddy_Long_Legs
    The base of each isosceles triangle is x^2-(1-x^2)=2x^2-1.

    Since the height of each isosceles triangle is the same as its base, then for any x\in\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right], the area of each slice is A(x)=\frac{1}{2}\left(2x^2-1\right)^2.

    So to find the volume, evaluate V=\frac{1}{2}\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}(2x^2-1)^2\,dx=\int_{0}^{\frac{1}{\sqrt{2}}}(2x^2-1)^2\,dx

    Can you continue?
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  3. #3
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    Thanks a lot Chris L T521.

    I managed to find A(x) but I found I made a mistake as I cancelled the and the square for some reason so I only had A(x) as (2x^2-1)

    I then found V but I just wasn't sure that it was equal to the other definite integral.

    I found the answer 0.3771 to be the same evaluating either integral so I assume this is correct?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Daddy_Long_Legs View Post
    Thanks a lot Chris L T521.

    I managed to find A(x) but I found I made a mistake as I cancelled the and the square for some reason so I only had A(x) as (2x^2-1)

    I then found V but I just wasn't sure that it was equal to the other definite integral.

    I found the answer 0.3771 to be the same evaluating either integral so I assume this is correct?
    That would be correct. The exact answer would be preferred: \frac{4}{15}\sqrt{2}
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