Is there any way to change the limits of a 3-D function without sketching its graph? I mean, can I obtain the same results algebraically for 3-D? For example $$\int_{0}^{2}\int_{0}^{x}\int_{-\sqrt{4-2x}}^{\sqrt{4-2x}}f(x,y,z) dy dz dx$$ to $$\int_{?}^{?}\int_{?}^{?}\int_{?}^{?}f(x,y,z) dz dx dy$$ or $$\int_{?}^{?}\int_{?}^{?}\int_{?}^{?}f(x,y,z) dx dz dy$$ and I need to obtain the limits. Is there any algebraic method?
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I think you are almost always better off sketching the graph. – JP McCarthy Dec 05 '13 at 13:36
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$$ \int_0^2\int_0^x\int_{-\sqrt{4-2x}}^{\sqrt{4-2x}}f(x,y,z)\,\mathrm{d}y\,\mathrm{d}z \,\mathrm{d}x $$ Note that $0\le x\le2$, $0\le z\le x$, and $2x+y^2\le4$, so one change of order would be $$ \int_{-2}^2\int_0^{2-y^2/2}\int_z^{2-y^2/2}f(x,y,z)\,\mathrm{d}x\,\mathrm{d}z \,\mathrm{d}y $$ another would be $$ \int_{-2}^2\int_0^{2-y^2/2}\int_0^xf(x,y,z)\,\mathrm{d}z\,\mathrm{d}x \,\mathrm{d}y $$
robjohn
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1@Artemisia: it is about covering the same region in $\mathbb{R}^3$, and that region is defined by these inequalities. – robjohn Dec 06 '13 at 11:34
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You would just change the limits as per the order of the differentials on the right side.
$$\int_{-\sqrt{4-2x}}^{\sqrt{4-2x}}\int_{0}^{2}\int_{0}^{x}f(x,y,z) dz dx dy$$
That would not work however because the outermost limits of integration rely on $x$ and $x$ is integrated in an inner integral.
$$\int_{-\sqrt{4-2x}}^{\sqrt{4-2x}}\int_{0}^{x}\int_{0}^{2}f(x,y,z) dx dz dy$$
Once again, this will not work because of the above reason.
user1251385
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1I think the OP is asking how the limits need to be altered to change the order of integration. They seem to know that simply blindly changing order is not valid. – robjohn Dec 05 '13 at 13:14