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Thread: Curved surface area of a cone

  1. #1
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    Curved surface area of a cone

    Hi,

    I've run into (another) difficulty in the chapter entitled 'further differentiation' in my book. Could anyone give me a pointer as to how to deal with the following problem?

    A right circular cone with base radius r cm and height h cm has volume [tex]\frac{1}{3}\pi r^2h cm^3[/tex] and curved surface area [tex]\pi r (r^2 + h^2)^\frac{1}{2} cm^2[/tex]

    Show that the curved surface area of the cone with volume [tex]\frac{4 \pi} {3} cm^3[/tex] is given by:

    [tex]S^2 = \pi ^2(r^4 + \frac {16}{r^2})[/tex]

    I can see no way in which these equations could be linked and am at a complete loss as to what to do here! It seems completely unlike anything covered in the book up till now so I'm completely lost.

  2. #2
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    Quote Originally Posted by jonnburton View Post
    Hi,

    I've run into (another) difficulty in the chapter entitled 'further differentiation' in my book. Could anyone give me a pointer as to how to deal with the following problem?

    A right circular cone with base radius r cm and height h cm has volume [tex]\frac{1}{3}\pi r^2h cm^3[/tex] and curved surface area [tex]\pi r (r^2 + h^2)^\frac{1}{2} cm^2[/tex]

    Show that the curved surface area of the cone with volume [tex]\frac{4 \pi} {3} cm^3[/tex] is given by:

    [tex]S^2 = \pi ^2(r^4 + \frac {16}{r^2})[/tex]

    I can see no way in which these equations could be linked and am at a complete loss as to what to do here! It seems completely unlike anything covered in the book up till now so I'm completely lost.
    Please tell us which equation you have been taught to use to find surface area through integration.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    Quote Originally Posted by jonnburton View Post
    Hi,

    I've run into (another) difficulty in the chapter entitled 'further differentiation' in my book. Could anyone give me a pointer as to how to deal with the following problem?

    A right circular cone with base radius r cm and height h cm has volume [tex]\frac{1}{3}\pi r^2h cm^3[/tex] and curved surface area [tex]\pi r (r^2 + h^2)^\frac{1}{2} cm^2[/tex]

    Show that the curved surface area of the cone with volume [tex]\frac{4 \pi} {3} cm^3[/tex] is given by:

    [tex]S^2 = \pi ^2(r^4 + \frac {16}{r^2})[/tex]
    This formula has an "r" but no "h". From formula for volume, above, [tex]\frac{1}{3}\pi r^2 h= \frac{4\pi}{3}[/tex]. Solve that for h in terms of r and then replace h, in the surface area formula, by that. However, I think your formula is incorrect. The formula I get is [tex]S^2= \pi^2(r^4+ 16)[/tex], without the [tex]r^2[/tex] under the 16.

    I can see no way in which these equations could be linked and am at a complete loss as to what to do here! It seems completely unlike anything covered in the book up till now so I'm completely lost.
    Last edited by HallsofIvy; 02-01-2013 at 06:53 PM.

  4. #4
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    Thanks for both of your replies and advice. I haven't been taught any equation to find surface area through integration... I will try what HallsofIvy said and see how that works out.

  5. #5
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    I have been working on this and did the following:

    [tex]\frac{1}{3}\pi r^2h = \frac{4 \pi} {3}[/tex]


    Multiply both sides by 3:

    [tex]\pi r^2h = 4 \pi [/tex]


    [tex]h = \frac {4 \pi}{\pi r^2}[/tex]

    [tex]h = \frac {4}{r^2}[/tex]


    Area

    [tex]\pi r (r^2 + h^2)^\frac{1}{2} [/tex]

    Substitute value of h into this:

    [tex]\pi r (r^2 + \frac{16}{r^4})^\frac{1}{2} [/tex]

    Assuming I haven't made any mistakes up to here, this is the point at which I get stuck again. I am tempted to get the square root of the terms in the brackets but I can see this doesn't lead any closer to what the solution should be:

    [tex]\pi r (r + \frac{4}{r^2})^\frac{1}{2}[/tex]

    [tex]\pi r^2 + \frac{8}{r^2}[/tex]

    Can anyone tell me what the next step is?

  6. #6
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    Quote Originally Posted by jonnburton View Post
    ...
    [tex]\pi r (r^2 + \frac{16}{r^4})^\frac{1}{2} [/tex]

    ...
    Can anyone tell me what the next step is?
    This term is correct, that means you have now:

    [tex]S=\pi r (r^2 + \frac{16}{r^4})^\frac{1}{2} [/tex]

    To get rid of the square-root square both sides of the equation:

    [tex]S^2=\pi^2 r^2 (r^2 + \frac{16}{r^4}) [/tex]

    Now drag the factor r˛ into the brackets:

    [tex]S^2=\pi^2 \left(r^4 + \frac{16}{r^2}\right) [/tex]

  7. #7
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    That's great Pappus, I understand that now. Thanks!

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