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Math Help - Area integral in polar co-ords of a surface

  1. #1
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    Unhappy Area integral in polar co-ords of a surface

    Express as an area integral in polar coordinates the area of the surface
    z= y^2 +2xy -x^2 +2


    lying over the annulus defined by 8/3 x^2 + y^2 1.
    Hence evaluate this area in terms of π.

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  2. #2
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    Have you made any attempt at this at all?

    The area you are to integrate over, in polar coordinates, is just 8/3\le r^2\le 1 which, since r must be positive is exactly \sqrt{8/3}\le r\le 1. That makes the limits of integration easy. And don't forget that the "differential of area" in polar cordinates is rdrd\theta. Write z= y^2+ 2xy- x^2+ 2 in polar coordinates and integrate!

    Ouch! I just realized this is a surface integral, not a volume integral. But the "polar coordinates" part is the same. To get the surface area integrand you can think of the surface z= y^2+ 2xy- x^2+ 2 as given by the vector function \vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}=  x\vec{i}+ y\vec{j}+ (y^2- 2xy- x^2+ 2)\vec{k} with the x and y coordinates as parameters. The derivative vectors \vec{r}_x= \vec{i}- (2y+ 2x)\vec{k} and \vec{r}_y= \vec{j}+ (2y- 2x)\vec{k} are tangent to the surface and their cross product,
    \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -(2y+2x) \\ 0 & 1 & 2y- 2x\end{array}\right|= -(2y+ 2x)\vec{i}- (2y+ 2x)\vec{j}+ \vec{k}
    is perpendicular to the surface and gives the "vector differential of surface area". Its length, \sqrt{1+ 8x^2+ 8y^2} gives the "scalar differential of surface area", \sqrt{1+ 8x^2+ 8y^2}dx dy. I think you can see that will be much easier to integrate in polar coordinates!
    Last edited by HallsofIvy; September 18th 2010 at 04:28 AM.
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