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Math Help - Surface area of revolution question

  1. #1
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    Surface area of revolution question

    The arc of the parabola y^2 = 4x joining the points (0, 0) and (3, 2\sqrt{3}) is rotated through 2\pi radians about the x-axis. Find the area of the curved surface generated giving the answer in the form \frac{k\pi}{3} where k is an integer.

    My attempt at answering it is written down here:



    I'm not sure if it's right because if it wants an answer in that form, then it would suggest that the integer k isn't divisible by 3 - otherwise it could be simplified. Through that logic, I believe I have made some kind of mistake. Could you please take a look and suggest any area where I have slipped up?

    Thanks for your help and time if you can answer
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  2. #2
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    y = 2\sqrt{x}

    y' = \frac{1}{\sqrt{x}}

    (y')^2 = \frac{1}{x}


    2\pi \int_0^3 2\sqrt{x} \cdot \sqrt{1 + \frac{1}{x}} \, dx

    4\pi \int_0^3 \sqrt{x+1} \, dx

    4\pi \left[\frac{2}{3}(x+1)^{\frac{3}{2}}\right]_0^3

    \frac{8\pi}{3}(8 - 1)

    \frac{56\pi}{3}
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  3. #3
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    Quote Originally Posted by db5vry View Post
    The arc of the parabola y^2 = 4x joining the points (0, 0) and (3, 2\sqrt{3}) is rotated through 2\pi radians about the x-axis. Find the area of the curved surface generated giving the answer in the form \frac{k\pi}{3} where k is an integer.

    My attempt at answering it is written down here:



    I'm not sure if it's right because if it wants an answer in that form, then it would suggest that the integer k isn't divisible by 3 - otherwise it could be simplified. Through that logic, I believe I have made some kind of mistake. Could you please take a look and suggest any area where I have slipped up?

    Thanks for your help and time if you can answer
    y^2=4x \rightarrow x=\frac{y^2}{4}
    Let f(y)=\frac{y^2}{4}.
    Then, the surface area = 2\pi \int_0^{2\sqrt{3}} y \, \sqrt{ 1 + [ \, f'(y) \, ]^2 } \, \, dy.
    I dont know what is the point of that \frac{k\pi}{3} !
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  4. #4
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    Quote Originally Posted by skeeter View Post
    y = 2\sqrt{x}

    y' = \frac{1}{\sqrt{x}}

    (y')^2 = \frac{1}{x}


    2\pi \int_0^3 2\sqrt{x} \cdot \sqrt{1 + \frac{1}{x}} \, dx

    4\pi \int_0^3 \sqrt{x+1} \, dx

    4\pi \left[\frac{2}{3}(x+1)^{\frac{3}{2}}\right]_0^3

    \frac{8\pi}{3}(8 - 1)

    \frac{56\pi}{3}
    Thanks for your solution! I can see what you've done for the most part, only the part which is in the fourth and fifth lines I can't seem to understand. I see you differentiated it after reducing y^2 to y but differentiating it implicitly gives the same thing once simplified - I see what has happened by there.

    I cannot see how you have got it into that particular form. Please could you show me what you have done there more closely?

    I can see you took the 2 out as a constant to get 4\pi but can't see how it simplifies to \sqrt{x+1}.
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  5. #5
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    Quote Originally Posted by db5vry View Post
    Thanks for your solution! I can see what you've done for the most part, only the part which is in the fourth and fifth lines I can't seem to understand. I see you differentiated it after reducing y^2 to y but differentiating it implicitly gives the same thing once simplified - I see what has happened by there.

    I cannot see how you have got it into that particular form. Please could you show me what you have done there more closely?

    I can see you took the 2 out as a constant to get 4\pi but can't see how it simplifies to \sqrt{x+1}.
    note ... \sqrt{a} \cdot \sqrt{b} = \sqrt{ab}

    \sqrt{x} \cdot \sqrt{1 + \frac{1}{x}} = \sqrt{x + 1}<br />
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