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Volume integration Watch

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    Prove that the volume of a solid generated by completely rotating the function y = 6 / (2x + 1)^1/2 around the x axis between x = 0 and x = 4 equals 36 pi ln3 units cubed


    I get so far then I don't know where to go next this what I've got

    y^2 = 36(2x + 1)^1/4

    V = pi [72/5 (2x + 1)^5/4 ] between 4 and 0, 4 being at top of bracket don't know how to write that on here

    So pi {[72/5(9)^5/4]-[72/5]}
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    Not sure how you've squared it. If  y=\frac{6}{(2x+1)^{\frac{1}{2}}} then y^2 =\frac{36}{2x+1}


    as you have to use the multiplying power rule for the denomoinator square. [(2x+1)^{\frac{1}{2}}]^2
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    (Original post by NotNotBatman)
    Not sure how you've squared it. If  y=\frac{6}{(2x+1)^{\frac{1}{2}}} then y^2 =\frac{36}{2x+1}


    as you have to use the multiplying power rule for the denomoinator square. [(2x+1)^{\frac{1}{2}}]^2
    oh thanks that makes more sense
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    (Original post by NotNotBatman)
    Not sure how you've squared it. If  y=\frac{6}{(2x+1)^{\frac{1}{2}}} then y^2 =\frac{36}{2x+1}


    as you have to use the multiplying power rule for the denomoinator square. [(2x+1)^{\frac{1}{2}}]^2
    ive done what youve said and when ive integrated and subbed the x's in I get...

    V= pi {[24(9)^3/4]-[24]}
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    (Original post by Custardcream000)
    ive done what youve said and when ive integrated and subbed the x's in I get...

    V= pi {[24(9)^3/4]-[24]}
    There's a problem with how you've integrated. Remember \int{\frac{f'(x)}{f(x)}} dx = ln|f(x)| +c

    What did you get as your integral?
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    (Original post by NotNotBatman)
    There's a problem with how you've integrated. Remember \int{\frac{f'(x)}{f(x)}} dx = ln|f(x)| +c

    What did you get as your integral?
    oh duh yeah i brought it up thats why
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    (Original post by Custardcream000)
    oh duh yeah i brought it up thats why
    so you get....

    v= pi {[18ln(9)^1/4]-[18 ln(1)]}
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    (Original post by NotNotBatman)
    Not sure how you've squared it. If  y=\frac{6}{(2x+1)^{\frac{1}{2}}} then y^2 =\frac{36}{2x+1}


    as you have to use the multiplying power rule for the denomoinator square. [(2x+1)^{\frac{1}{2}}]^2

    One thing is wrong, the power.
    [(2x+1)^{\frac{1}{2}}]^2

    I think you're doing 2 x 1/2 =1/4, but 2 x 1/2 = 1

    So you've got the integration right except for the power of 1/4
 
 
 
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