I find it insanely hard to imagine and change the order of integration for triple integrals. I have this problem for example. $\int_{0}^{2}\int_{0}^{y^3}\int_{0}^{y^2}f(x,y,z) dzdxdy$ So it is easy to change this to dzdydx because that only looking at the area at the bottom. However, changing it to anything else seems very hard, and requires sketching the volume. Could anyone tell me how to approach this problem, and problems like this in the future?
Thank you.
On
I'm going to try to expand on @SangchulLee's answer a little bit.
You think through the original problem by realizing that our 3-dimensional region projects down to the region in the $xy$-plane between the $y$-axis and $x=y^3$ [or, equivalently, $y=x^{1/3}$]. In $\Bbb R^3$, $x=y^3$ is a vertical cylinder over this plane curve. For each point in that plane region, a vertical line enters the region at $z=0$ and exits at $z=y^2$. So this surface is a parabolic cylinder, with lines parallel to the $x$-axis. Here's a sketch with the assistance of Mathematica:
Indeed, the region lies over the rectangle $R=[0,8]\times [0,4]$ in the $xz$-plane, and the two cylinders intersect along the crease $y=z^{1/2}=x^{1/3}$, which projects to the curve $C$ given by $x^2=z^3$ in the $xz$-plane. If we take a point in $R$, we want to know when a line parallel to the $y$-axis enters and exits our region. If we're below $C$ in $R$, then $z^3<x^2$ and the line enters at $y=x^{1/3}$ and exits at $y=2$. (As a check, if $y\ge x^{1/3}$, then $y\ge z^{1/2}$ as well in this range, so we're "inside" both cylinders.) If we're above $C$ in $R$, then $z^3>x^2$ and the line enters at $y=z^{1/2}$ and exits at $y=2$.
Thus, we end up with the following iterated integral in the order $dy\,dz\,dx$: $$\int_0^8\int_0^{x^{2/3}}\int_{x^{1/3}}^2 dy\,dz\,dx + \int_0^8\int_{x^{2/3}}^4\int_{z^{1/2}}^2 dy\,dz\,dx.$$
(By the way, you should be able to set up the limits in the orders $dx\,dz\,dy$ or $dx\,dy\,dz$ much more easily.)
It may help to convert the iterated integrals to a multiple integral. For instance,
$$ \int_{0}^{2}\int_{0}^{y^3}\int_{0}^{y^2} f(x,y,z)\,dzdxdy = \iiint_{\mathcal{D}} f(x,y,z) \, dxdydz $$
where $\mathcal{D}$ is a region in $\mathbb{R}^3$ given by
$$\mathcal{D} = \{ (x, y, z) \in \mathbb{R}^3 : 0 \leq y \leq 2, \ 0 \leq x \leq y^3, \ 0 \leq z \leq y^2 \}$$
Now, for instance, say you want to convert this to the iterated integrals in the order of $dydxdz$. This requires both geometric intuition and algebraic manipulation. The key idea here is to convert this set of inequalities, while not changing the region this set describes, to another so that
To do this, we first need to figure out the range of $z$. This is easily obtained from $0\leq z\leq y^2$ and $0 \leq y \leq 2$. These together yield $0 \leq z \leq 4$, and it is clear that $z$ can take any values in-between.
Next, given $0 \leq z \leq 4$, we want to figure out the range of $x$ possibly depending on $z$ but not on $y$. This can be done in a similar manner, yielding $0 \leq x \leq 8$. (The fact that the range of $x$ given $z$ is independent of $z$ is pure luck in this particular setting, and need not be true in general case.)
Finally, the range of $y$ follows from all three inequalities that constraint the range of $y$, and the result is $\max\{\sqrt{z}, \sqrt[3]{x}\} \leq y \leq 2$. Of course, you can perform sanity-check by recovering the original constraints from this new set of inequalities. As a result,
$$ \iiint_{\mathcal{D}} f(x,y,z) \, dxdydz = \int_{0}^{4} \int_{0}^{8} \int_{\max\{\sqrt{z}, \sqrt[3]{x}\}}^{2} f(x,y,z) \, dydxdz. $$