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Thread: How do you do this inverse trig question?

  1. #1
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    How do you do this inverse trig question?

    Snip20171010_1.png

    I cannot figure out how to do part b

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    Re: How do you do this inverse trig question?

    Remember: Volume = pi. int[x^2 dy] (from y=a to y=b)

    If you rearrange y = 1/sqrt(1-x^2) to make x^2 the subject, you will get: x^2 = 1 - 1/y^2

    And borders:
    x=0 y=1
    x=1/2 y=sqrt(2)

    Hence,
    Volume = pi.int[ 1 - 1/y^2 dy ] (from y=1 to y=sqrt(2))

    I think you should be able to do this question from here.
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    Re: How do you do this inverse trig question?

    Quote Originally Posted by fluffchuck View Post
    Remember: Volume = pi. int[x^2 dy] (from y=a to y=b)

    If you rearrange y = 1/sqrt(1-x^2) to make x^2 the subject, you will get: x^2 = 1 - 1/y^2

    And borders:
    x=0 y=1
    x=1/2 y=sqrt(2) ????

    Hence,
    Volume = pi.int[ 1 - 1/y^2 dy ] (from y=1 to y=sqrt(2))

    I think you should be able to do this question from here.
    Is that correct?
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    Re: How do you do this inverse trig question?

    Quote Originally Posted by fluffchuck View Post
    Remember: Volume = pi. int[x^2 dy] (from y=a to y=b)

    If you rearrange y = 1/sqrt(1-x^2) to make x^2 the subject, you will get: x^2 = 1 - 1/y^2

    And borders:
    x=0 y=1
    x=1/2 y=sqrt(2)

    Hence,
    Volume = pi.int[ 1 - 1/y^2 dy ] (from y=1 to y=sqrt(2))

    I think you should be able to do this question from here.
    When x = 1/2 I get y = 2/sqrt(3)? (Just saw Drongoski's question so I tried doing it myself)
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    Re: How do you do this inverse trig question?

    Quote Originally Posted by Sp3ctre View Post
    When x = 1/2 I get y = 2/sqrt(3)? (Just saw Drongoski's question so I tried doing it myself)
    Yes, sorry, my bad.
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