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  1. #1
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    Question surface area

    Find the surface area of the portion of the sphere of radius 4 that lies inside the cylinder x^2 +y^2 =12 and above the xy-plane.

    Could you please show me the steps of solving this question? Thank you very much.
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  2. #2
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    Quote Originally Posted by kittycat View Post
    Find the surface area of the portion of the sphere of radius 4 that lies inside the cylinder x^2 +y^2 =12 and above the xy-plane.

    Could you please show me the steps of solving this question? Thank you very much.
    \int_0^{2\pi} \int_0^{\pi/4} \int_0^4 \rho^2 \sin \phi d\rho \ d\phi \ d\theta

    Because the cone creates a 45 degree opening angle with the z axis.
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  3. #3
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    Quote Originally Posted by kittycat View Post
    Find the surface area of the portion of the sphere of radius 4 that lies inside the cylinder x^2 +y^2 =12 and above the xy-plane.

    Could you please show me the steps of solving this question? Thank you very much.
    In spherical polars the surface area element of a sphere is r^2 \sin(\phi) ~d\phi d \theta , where \phi is the zenithal angle
    of the patch and \theta the azimuthal angle.

    The part of the surface of the sphere that is inside the cylinder is that part for which \phi \in [\arcsin(\sqrt{12}/4), \pi/2]=[\pi/6, \pi/2]

    So the required area is:

    <br />
A=\int_{\phi=\pi /6}^{\pi}\int_{\theta=0}^{2\pi} 16 \sin(\phi) ~d\theta d \phi<br />

    RonL
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  4. #4
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    Hello, kittycat!

    The problem has be solved "head on".


    Find the surface area of the portion of the sphere of radius 4
    that lies inside the cylinder x^2 +y^2 \:=\:12 and above the xy-plane.

    Turn the diagram 90 "clockwise".
    Code:
                    |
                  * * *
              *     |     *
            *-------*-------*A
           *       B|     * :*
                    |   *   :
          *         | *     : *
      - - * - - - - + - - - + * - -
          *         |O      C *
                    |
           *        |        *
            *-------+-------*
              *     |     *
                  * * *
                    |

    In right triangle ABO\!:\;\;OA = 4,\:OB = \sqrt{12}

    . . Hence: . BA\,=\,OC\,=\,2 .**


    We want the area of the "polar cap" formed by the segment at the far right.


    The formula for the area of a surface of revolution is: . S \;=\;2\pi\int^b_a y\,\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx


    The equation of the circle is: . x^2+y^2\:=\:16\quad\Rightarrow\quad y \:=\:\sqrt{16-x^2}

    Then: . \frac{dy}{dx}\:=\:\frac{-x}{\sqrt{16-x^2}}

    . . . . 1 + \left(\frac{dy}{dx}\right)^2\;=\;1 + \frac{x^2}{16-x^2} \;=\;\frac{16}{16-x^2}

    . . . . \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\frac{4}{\sqrt{16-x^2}}


    Hence, we have: . S \;=\;2\pi\int^4_2\sqrt{16-x^2}\cdot\frac{4}{\sqrt{16-x^2}}\,dx \;=\;8\pi\int^4_2\!\!dx \;=\;8\pi x\bigg]^4_2

    . . . . = \;8\pi(4 - 2) \;=\;\boxed{16\pi}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **
    Note that: . \angle AOC \,=\,60^o


    Fascinating bit of trivia:

    A central cone with vertex angle 120 subtends the area of the sphere.

    With such a cone at the north pole and the south pole,
    . . the "equatorial band" has half the area of the sphere.

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