# Change order of integration for triple integral not possible?

Find the triple integral over the region $$E$$, where $$E$$ is bounded by $$x=3z^2$$ and the planes $$x=y, y=0$$,and $$x=12$$. I know that it is possible to integrate in the order $$dy$$, $$dx$$, and $$dz$$ and in the order $$dy$$, $$dz$$, and $$dx$$, but I am having no luck integrating in the order $$dx$$, $$dy$$ and $$dz$$. The biggest problem is that I am struggling to find a way to represent the domain of $$y$$ in terms of $$z$$ for the the second integral's limits of integration.

When I graph $$x=3z^2$$, I see that for any value of $$z$$, it does not map to a single $$y$$-value, but rather the entire $$y$$-axis. I'm lost on how to find these limits of integration. Any assistance would be greatly appreciated.

Look down onto the $$y,z$$ plane. In the plot below, the front wall of the bounding frame corresponds to $$x=0$$ and the back wall to $$x=12$$.

You'll notice that the two surfaces $$x=y$$ and $$x=3z^2$$ (resp. blue and orange) compete for "bottom surface" with respect to the $$x$$-axis. They intersect in a parabolic curve (red) along the wall of the parabolic cylinder. To one side of this parabola (below it, from this angle) the plane $$x=y$$ lies beneath the cylinder $$x=3z^2$$, i.e. $$x=y$$ is closer to $$x=0$$ and $$x=3z^2$$ is closer to $$x=12$$; in this case, the cylinder acts as the floor for the given region. Otherwise, to the other side of the parabola ("above" it) $$x=3z^2$$ lies below $$x=y$$, so the latter is closer to $$x=12$$ and acts as lower bound.

Over the whole rectangle $$(y,z)\in[0,12]\times[-2,2]$$, we can then say $$x$$ must be between $$12$$ and the larger of the two functions $$x=y$$ and $$x=3z^2$$, denoted $$\max\left\{y,3z^2\right\}$$.

So we can capture the whole region with the condensed triple integral,

$$\int_{z=-2}^2 \int_{y=0}^{12} \int_{x=\max\left\{y,3z^2\right\}}^{12} f(x,y,z) \, dx \, dy \, dz$$

The OP only mentions a generic integral, not a volume, so I assume $$f\neq1$$. We need to carefully split up the domain if we actually want to compute this.

For simplicity, let's suppose we just want the volume and that $$f=1$$. For comparison, using the much simpler order $$dy\,dx\,dz$$, we have a volume of

\begin{align*} I_{yxz} &= \int_{z=-2}^2 \int_{x=3z^2}^{12} \int_{y=0}^x dy \, dx \, dz \\ &= 2 \int_0^2 \int_{3z^2}^{12} \int_0^x dy \, dx \, dz \\ &= \frac{1152}5 \end{align*}

Now for our target order $$dz\,dy\,dx$$, we must split up the $$y$$ integral depending on which subset of the region we find ourselves in. For reference, consult the plot below showing a projection of the region's boundaries onto the $$y,z$$ plane.

\begin{align*} I_{xyz} &= \int_{z=-2}^2 \int_{y=0}^{12} \int_{x=\max\left\{y,3z^2\right\}}^{12} dx \, dy \, dz \\ &= 2 \int_0^2 \int_0^{12} \int_{\max\left\{y,3z^2\right\}}^{12} dx \, dy \, dz \\ &= 2 \left(\int_0^2 \int_0^{3z^2} \int_{3z^2}^{12} dx \, dy \, dz + \int_0^2 \int_{3z^2}^{12} \int_y^{12} dx \, dy \, dz\right) \\ &\stackrel{\checkmark}= \frac{1152}5 \end{align*}

• You cannot have $x$ as a limit of integration on your $dy$ integral …. Commented Aug 5 at 21:50
• I just had a massive brainfart for the last five hours reading this, but it clicked all of a sudden. Thank you! Commented Aug 5 at 22:27
• Does this mean that sometimes there are iterated integrals with a certain order of x, y, and z that cannot be written with a single integral? Commented Aug 5 at 22:30
• @TreeGuy Yes. As a simple example in 2D, consider the triangle with vertices $(0,0)$, $(0,2)$, and $(1,1)$. Setting up and computing the integral for area in $dy\,dx$ order is easy and doesn't require any additional surgery, while $dx\,dy$ requires splitting up the region along $y=1$ because the horizontal distance between the boundaries $x=0$ and $y=x$, and between $x=0$ and $y=2-x$, is not the same for all $y\in[0,2]$. In this example, we'd have$$\int_{x=0}^1\int_{y=x}^{2-x}dy\,dx=\int_{y=0}^2\int_{x=0}^{\min\{y,2-y\}}dx\,dy=\left(\int_0^1\int_0^y+\int_1^2\int_0^{2-y}\right)dx\,dy$$ Commented Aug 6 at 13:44