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Math Help - Double Integral in Polar Coordinates

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    Double Integral in Polar Coordinates

    Use polar coordinates to find the volume of the solid above the cone z = (x^2+y^2)^1/2 and below the hemisphere
    z =( 8 − x^2 − y^2)^1/2.
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  2. #2
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    Quote Originally Posted by crwhd4 View Post
    Use polar coordinates to find the volume of the solid above the cone z = (x^2+y^2)^1/2 and below the hemisphere
    z =( 8 − x^2 − y^2)^1/2.
    You need to find

    V = \int \, \int_{R_{xy}} \sqrt{8 - x^2 - y^2} - \sqrt{x^2 + y^2} \, dx \, dy

    where R_{xy} is the region inside the circle x^2 + y^2 = 4.

    Note: x^2 + y^2 = 8 - x^2 - y^2 \Rightarrow x^2 + y^2 = 4.


    Make the switch to polar coordinates:

    V = \int_{\theta = 0}^{\theta = 2 \pi} \int_{r = 0}^{r = 2} (\sqrt{8 - r^2} - r) \, r \, dr \, d\theta


    = \int_{r = 0}^{r = 2} (\sqrt{8 - r^2} - r) \, r\, dr \cdot \int_{\theta = 0}^{\theta = 2 \pi} d \theta .


    Now do the simple integrations.
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    This is probably a stupid question, but how is it that an iterated integral becomes a product of integrals? Or is that the case in this one only because the inner integral will yield a constant?
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    Quote Originally Posted by rualin View Post
    This is probably a stupid question, but how is it that an iterated integral becomes a product of integrals? Or is that the case in this one only because the inner integral will yield a constant?
    The variables are seperable. In general:

    \int \int f(x) \, g(y) \, dx \, dy = \int f(x) \, dx \cdot \int g(y) \, dy
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