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Thread: 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 $\displaystyle x^2+y^2+z^2\leq R^2$ describes a solid sphere with radius $\displaystyle R$ centered on the origin. If $\displaystyle z\geq 0$, 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 $\displaystyle D=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2\leq R^2\}$ for the function $\displaystyle r\to\mathbb{R}$, $\displaystyle r(x,y,z)$ not defined in your problem. So we are actually finding the hypervolume of a 4-dimensional shape.

    The bounds on $\displaystyle z$ go from $\displaystyle 0$ to $\displaystyle R$, 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 $\displaystyle \sqrt{R^2-z^2}$. So $\displaystyle y$ must range from $\displaystyle -\sqrt{R^2-z^2}$ to $\displaystyle +\sqrt{R^2-z^2}$. The intersection of these two planes give a line of length $\displaystyle 2\sqrt{R^2-z^2-y^2}$. Therefore the outermost integral will be $\displaystyle x$ ranging from $\displaystyle -\sqrt{R^2-z^2-y^2}$ to $\displaystyle +\sqrt{R^2-z^2-y^2}$. Hence,

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

    (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

    $\displaystyle
    \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
    $

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