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Math Help - Volume of the solid revolving about y-axis

  1. #1
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    Volume of the solid revolving about y-axis

    This involves trig functions so I am not sure if I am on the right track

    y = (tan^2 x)/x 0 <= x <= pi/4

    a = 0; b=pi/4 on the x-axis

    so I think I should use the formula 2pi (integral sign "S") x f(x) dx

    v = 2 pi S a=0; b=pi/4 x (tan^2 x)/x dx

    so I am thinking the x's cancel each other out and I have
    v= 2 pi S tan^2 x dx

    v = 2 pi ((1/3)sec^6)

    Am I on the right track??

    Thanks

    Calculus Beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    This involves trig functions so I am not sure if I am on the right track

    y = (tan^2 x)/x 0 <= x <= pi/4

    a = 0; b=pi/4 on the x-axis

    so I think I should use the formula 2pi (integral sign "S") x f(x) dx

    v = 2 pi S a=0; b=pi/4 x (tan^2 x)/x dx

    so I am thinking the x's cancel each other out and I have
    v= 2 pi S tan^2 x dx

    v = 2 pi ((1/3)sec^6)

    Am I on the right track??

    Thanks

    Calculus Beginner
    You are except for the last part.

     <br />
V = 2 \pi \int_0^{\pi/4}\tan ^2 x\, dx = 2 \pi \int_0^{\pi/4}\left(\sec ^2 x - 1\right)\, dx<br />
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  3. #3
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    next step

    Okay so 2pi S a=0; b= pi/4 sec^2 x - 1 dx

    = 2pi (tan x - x) pi/4 and 0

    2pi (1 - (pi/4) - 0 - 0) = 2pi - (1/2)pi^2

    Is this correct?

    Thanks

    Calculus beginner
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  4. #4
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    Quote Originally Posted by calcbeg View Post
    Okay so 2pi S a=0; b= pi/4 sec^2 x - 1 dx

    = 2pi (tan x - x) pi/4 and 0

    2pi (1 - (pi/4) - 0 - 0) = 2pi - (1/2)pi^2

    Is this correct?

    Thanks

    Calculus beginner
    Looks good to me
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