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Consider: $$ \int_0^3 \int_4^{\sqrt{25-z^2}} \int_{-\sqrt{25-y^2-z^2}}^{\sqrt{25-y^2-z^2}} dxdydz$$

i. Clearly sketch the graph of the solid whose volume this triple integral determines.

ii. Present all other triple integrals in rectangular coordinates equivalent to the given triple integral but each of a different order than the others.<br>

Here is my graph and chart for changing order of integration any help at all in confirming my answers would be great! I also added my work if that helps as well.

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No, that is not correct.

Note that only $x$ can take negative values, the other two coordinates, whichever order you are doing it, must have $y\geq 4$ and $z\geq 0$ built in to the limit. The region is only in 2 of the 8 octants.

The $\mathrm{d}x\,\mathrm{d}z\,\mathrm{d}y$ integral should have limits $$ \int_{4}^{5}\int_{0}^{\sqrt{25-y^2}}\int_{-\sqrt{25-y^2-z^2}}^{\sqrt{25-y^2-z^2}}\mathrm{d}x\,\mathrm{d}z\,\mathrm{d}y $$ similarly the other four.

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  • $\begingroup$ ty! if i can ask a question, when given that first triple integral, can i assume that 4 is going to be the lowest for all y values? and same with z? and we get 5 for dy's highest value cause we are solving y from dz with respect to dy? Thanks again for answering. $\endgroup$ Jul 30 '20 at 14:58

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