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Math Help - area

  1. #1
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    area

    Find the area of the cone  z^2 = a^2(x^2 + y^2) between the planes  z = 1 and  z = 2 .

    I think you use cylindrical coordinates.

    So:  z^{2} = a^{2} r^{2}

    And is it:  \int_{0}^{2 \pi} \int_{0}^{\frac{z^{2}}{a^2}} \int_{1}^{2} a^{2}r^{2} \ dz \ dr \ d \theta ?
    Last edited by shilz222; November 6th 2007 at 09:17 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shilz222 View Post
    Find the area of the cone  z^2 = a^2(x^2 + y^2) between the planes  z = 1 and  z = 2 .

    I think you use cylindrical coordinates.

    So:  z^{2} = a^{2} r^{2}

    And is it:  \int_{0}^{2 \pi} \int_{0}^{\frac{z^{2}}{a^2}} \int_{1}^{2} a^{2}r^{2} \ dz \ dr \ d \theta ?
    no. this is not a triple integral question, and you want the surface area.

    use: \mbox{Area} = \iint_D \sqrt{1 + \left( \frac {\partial z}{\partial x} \right)^2 + \left( \frac {\partial z}{\partial y} \right)^2}~dA

    D is the region we're integrating over. use polar coordinates
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  3. #3
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    So you let  z = \sqrt{a^{2}x^{2} + a^{2}y^{2}} and then:

     f_x = \frac{a^{2}x^{2}}{2}\left (a^{2}x^{2} + a^{2}y^{2}\right)^{-\frac{1}{2}}


     f_y = \frac{a^{2}y^{2}}{2}\left(a^{2}x^{2} + a^{2}y^{2}\right )^{-\frac{1}{2}}


    So we have:  \int_{0}^{2 \pi} \int_{1/a}^{2/a} \sqrt{1 + \frac{a^{4}x^{4}}{4}(a^{2}x^{2} + a^{2}y^{2})^{-1}+  \frac{a^{4}y^{4}}{4}(a^{2}x^{2} + a^{2}y^{2})^{-1}}

    Then convert the inside into polar coordinates (which is a bit messy). Is this ok?
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  4. #4
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    Actually i ended up getting  \int_{0}^{2 \pi} \int_{1/a}^{2/a} \sqrt{1+a} \ r \ dr \ d \theta = \int_{0}^{2 \pi} d \theta \int_{1/a}^{2/a} r \sqrt{1+a^{2}} \ dr

    This equals  \frac{3 \pi \sqrt{1+a^{2}}}{a^{2}}


    Is this correct?
    Last edited by shilz222; November 7th 2007 at 02:53 AM.
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