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Thread: area

  1. #1
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    area

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

    I think you use cylindrical coordinates.

    So: $\displaystyle z^{2} = a^{2} r^{2} $

    And is it: $\displaystyle \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; Nov 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 $\displaystyle z^2 = a^2(x^2 + y^2) $ between the planes $\displaystyle z = 1 $ and $\displaystyle z = 2 $.

    I think you use cylindrical coordinates.

    So: $\displaystyle z^{2} = a^{2} r^{2} $

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

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

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


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


    So we have: $\displaystyle \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 $\displaystyle \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 $\displaystyle \frac{3 \pi \sqrt{1+a^{2}}}{a^{2}} $


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