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Math Help - Volumes of a solid of revolutions

  1. #16
    Member Awsom Guy's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Yes, I'm sure.

    [x^{1/3}]^2=x^{2/3}
    Can you show me how you solved that I am sort of confused with that thanks. DO you add the powers... explain thanks
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  2. #17
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Awsom Guy View Post
    Can you show me how you solved that I am sort of confused with that thanks. DO you add the powers... explain thanks
    (x^m)^n=(x^m)(x^m)...ntimes Therefore, (x^m)^n=x^{nm}. so you multiply powers.
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  3. #18
    Member Awsom Guy's Avatar
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    Oh man... I have forgotton my basics thanks heaps for that explanation I have got the correct answer thank you.
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  4. #19
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    Quote Originally Posted by Awsom Guy View Post
    Can you show me how you solved that I am sort of confused with that thanks. DO you add the powers... explain thanks
    What he's doing is squaring the expression, which is what is required in the volumes of revolution formulas. It's an application of laws of indices if you look at it in that kind of way.

    For instance, your value for y might also be something like 3sin^2 \theta - here, you would get (3sin^2 \theta)^2 which leads to 9sin^4 \theta.

    I think what you might need to familiarise yourself with though is the whole concept of x^a . x^b = x^{ab}. Here, what you've just basically witnessed was the a, which in your case was 1/3, multiplied by 2. And two thirds is...2/3. Which is what you get - and then when you integrate x to that power, you're adding one and getting a coefficient of 3/5.
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  5. #20
    Member Awsom Guy's Avatar
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    yes thank you I understand.
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