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Math Help - Surface area integral

  1. #1
    Senior Member mollymcf2009's Avatar
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    Surface area integral

    Find surface area of solid formed by rotating the following curve about the y-axis:

    x = \sqrt{a^2 - y^2} on 0 \leq y \leq \frac{a}{2}

    *The "a" in there kinda screwed me up, but this is my work:

    When I solved for the arc length, I ended up getting \sqrt{2}, like this:

    x = \sqrt{a^2 - y^2}

    x' = \frac{2a - 2y}{2\sqrt{a^2-y^2}}

    = \frac{a-y}{a-y} = 1

    ds = \sqrt{1 + (1)^2}

    **Is this right?

    So my integral is:

    \int\limits^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^2 - y^2}) (\sqrt{2})

    Then I pulled my square root out in front:

    2\sqrt{2} \pi \int\limits^{\frac{a}{2}}_{0} \sqrt{a^2 - y^2} dy

    2\sqrt{2} \pi \int\limits^{\frac{\pi}{6}}_{0} asin \theta \cdot acos \theta d\theta

    I have more, but can some one check this up until this point?
    Thanks!! Molly
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  2. #2
    MHF Contributor
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    Quote Originally Posted by mollymcf2009 View Post
    Find surface area of solid formed by rotating the following curve about the y-axis:

    x = \sqrt{a^2 - y^2} on 0 \leq y \leq \frac{a}{2}

    *The "a" in there kinda screwed me up, but this is my work:

    When I solved for the arc length, I ended up getting \sqrt{2}, like this:

    x = \sqrt{a^2 - y^2}

    x' = \frac{2a - 2y}{2\sqrt{a^2-y^2}}

    = \frac{a-y}{a-y} = 1 ****

    ds = \sqrt{1 + (1)^2}

    **Is this right?

    So my integral is:

    \int\limits^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^2 - y^2}) (\sqrt{2})

    Then I pulled my square root out in front:

    2\sqrt{2} \pi \int\limits^{\frac{a}{2}}_{0} \sqrt{a^2 - y^2} dy

    2\sqrt{2} \pi \int\limits^{\frac{\pi}{6}}_{0} asin \theta \cdot acos \theta d\theta

    I have more, but can some one check this up until this point?
    Thanks!! Molly
    From above **** x' = \frac{2a - 2y}{2\sqrt{a^2-y^2}} = \frac{a - y}{\sqrt{a^2-y^2}} \ne \frac{a-y}{a-y}.
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