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I have a triple integral $ \int_{0}^{2} \int_{0}^{y^3} \int_{0}^{y^2} f(x,y,z) \ dz \ dx \ dy $. I have to find five different iterated integrals equivalent to this integral. I know that order of integration involves projecting the graph onto the $xy,\ xz$, and $yz$ planes but I am not really sure what to do beyond that.

EDIT: I came with the following solution: $$ \int_{0}^{2} \int_{0}^{y^2} \int_{0}^{y^3} f(x,y,z) \ dx \ dz \ dy \\ \int_{0}^{8} \int_{\sqrt[3]{x}}^{2} \int_{0}^{y^2} f(x,y,z) \ dz \ dy \ dx \\ \int_{0}^{4} \int_{\sqrt{z}}^{2} \int_{0}^{y^3} f(x,y,z) \ dx \ dy \ dz \\ \int_{0}^{4} \int_{0}^{8} \int_{2}^{\sqrt[3]{x}} f(x,y,z) \ dy \ dx \ dz \\ \int_{0}^{8} \int_{0}^{4} \int_{\sqrt{z}}^{2} f(x,y,z) \ dy \ dz \ dx $$ Would this be correct?

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    $\begingroup$ There are 6 ways to order 3 things. You already have one ordering in your integral $dz\ dx\ dy$. The other 5 come from changing the order of integration. The tricky part will be correctly changing the limits of integration. $\endgroup$ – John Habert Jan 30 '14 at 2:01
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You have: $$ \int_0^2 \left( \int_0^{y^3} \left(\int_0^{y^2} f(x,y,z) \ dz\right) \ dx \right) \ dy. $$

Look at the outside integral: $$ \int_0^2 \cdots\cdots\cdots\,dy. $$ The variable $y$ goes from $0$ to $2$. Inside this integral you have $$ \int_0^{y^3} \cdots\cdots\cdots\,dx. $$ That means that, for each fixed value of $y$ between $0$ and $2$, the variable $x$ goes from $0$ to $y^3$. So $x$, being between $0$ and $y^3$, could be anywhere between $0$ and $2^3=8$. But $x$ cannot be more than $y^3$, or in other words $y^3$ must be more than $x$, or equivalently $y$ must be more than $\sqrt[3]{x}$. Thus one can write $$ \int_0^8\cdots\cdots\cdots\,dx, $$ and then say that for any value of $x$ between $0$ and $8$, the variable $y$ goes from $\sqrt[3]{x}$ to $2$. Thus we have $$ \int_0^8\left( \int_{\sqrt[3]{x}}^2 \cdots\cdots\cdots\,dy \right)\,dx. $$

In other words, when you change the order, you have to adjust the bounds on each integral accordingly.

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  • $\begingroup$ I came with the following solution $\int_{0}^{2} \int_{0}^{y^2} \int_{0}^{y^3} f(x,y,z) \ dx \ dz \ dy \\ \int_{0}^{8} \int_{\sqrt[3]{x}}^{2} \int_{0}^{y^2} f(x,y,z) \ dz \ dy \ dx \\ \int_{0}^{4} \int_{\sqrt{z}}^{2} \int_{0}^{y^3} f(x,y,z) \ dx \ dy \ dz \\ \int_{0}^{4} \int_{0}^{8} \int_{2}^{\sqrt[3]{x}} f(x,y,z) \ dy \ dx \ dz \\ \int_{0}^{8} \int_{0}^{4} \int_{\sqrt{z}}^{2} f(x,y,z) \ dy \ dz \ dx$. Would this be correct? $\endgroup$ – Jackie Jan 31 '14 at 0:31
  • $\begingroup$ I think you need $\displaystyle \int_0^8 \int_0^4 \int_{\max\{\sqrt{z},\sqrt[3]{x}\}}^2 f(x,y,z)\, dy\, dz \, dx$, and similarly for the other one with $\displaystyle \int\cdots\cdots \,dy$ as the innermost integral. $\endgroup$ – Michael Hardy Jan 31 '14 at 0:58
  • $\begingroup$ Sorry for the confusion but could you clarify a little? Thanks. $\endgroup$ – Jackie Jan 31 '14 at 1:41
  • $\begingroup$ The two constraints, that $z\le y^2$ and that $x\le y^3$, mean that $y\ge\text{both }\sqrt{z}\text{ and }\sqrt[3]{x}$. If $y\ge\text{both of two things}$ then $y\ge\max\text{ of those two things}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 31 '14 at 17:50

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