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Math Help - Help with finding Boundaries for Triple INtegrals

  1. #1
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    Help with finding Boundaries for Triple INtegrals

    For example:

    x^2 + y^2 + z^2 <= R^2

    z >= 0

    Can someone help me find the x,y, and z boundaries? Would x and y be -R and R? And would z be 0 and R^2 - x^2 -y^2?


    Also, for:

    x^2 + y^2 <= R^2 and 0<= z <= H

    I'm confused on the boundaries for this one as well. Anyone?

    Last one;

    If the question only says that a cub with edge length a, what would be the x,y,z boundaries for the triple integral?
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  2. #2
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    Here you will need strong visualization skills. The inequality x^2+y^2+z^2\leq R^2 describes a solid sphere with radius R centered on the origin. If z\geq 0, then we are only talking about the top half of the figure, the part above the xy-plane, a "semi-sphere". This shape is the domain D=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2\leq R^2\} for the function \to\mathbb{R}" alt="r\to\mathbb{R}" />, r(x,y,z) not defined in your problem. So we are actually finding the hypervolume of a 4-dimensional shape.

    The bounds on z go from 0 to R, as given in the problem, so that will be the innermost integral. So we are taking horizontal cross-sections of the semi-sphere, giving us circular disks of radius \sqrt{R^2-z^2}. So y must range from -\sqrt{R^2-z^2} to +\sqrt{R^2-z^2}. The intersection of these two planes give a line of length 2\sqrt{R^2-z^2-y^2}. Therefore the outermost integral will be x ranging from -\sqrt{R^2-z^2-y^2} to +\sqrt{R^2-z^2-y^2}. Hence,

    I=\int_{-\sqrt{R^2-z^2-y^2}}^{+\sqrt{R^2-z^2-y^2}}\int_{-\sqrt{R^2-z^2}}^{+\sqrt{R^2-z^2}}\int_{0}^{R} r(x,y,z) dzdydx

    (I suppose it would be inappropriate in this particular problem to convert to polar or cylindrical coordinates, which would incredibly simplify the problem.)
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  3. #3
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    I think that most (in cartesian coords although spherical polars is the way to go) would put

     <br />
\int_{_R}^{R} \int_{ -\sqrt{R^2-x^2} }^{\sqrt{R^2-x^2}} \int_0^{\sqrt{R^2-x^2-y^2}} f(x,y,z)dz\,dy\,dx<br />

    @media man - I think you want to reverse the order of your integrals.
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