How do you switch the order of integration?

Question

Problem.^src) Re-iterate the integral $$ \int_{y=0}^{\sqrt{5}} \int_{z=0}^{\sqrt{5-y^2}} \int_{x=y^2+z^2}^{5} f(x, y, z) \, \mathrm{d}x\mathrm{d}z\mathrm{d}y $$ in the following order: $$ \int_{x=x_1}^{x_2} \int_{y=g_1(x)}^{g_2(x)} \int_{z=f_1(x,y)}^{f_2(x,y)} f(x, y, z) \, \mathrm{d}z\mathrm{d}y\mathrm{d}x $$

• $x = \boxed{\phantom{hi!}} \text{ to } \boxed{\phantom{hi!}} $
• $y = \boxed{\phantom{hi!}} \text{ to } \boxed{\phantom{hi!}} $
• $z = \boxed{\phantom{hi!}} \text{ to } \boxed{\phantom{hi!}} $

I have no clue how to approach this problem.

My first though was to just get $z$ in terms of $x$ and $y$. $x$ in terms of $z$ and an integer, and $y$ in terms of 2 integers, like so:

$$ \sqrt{x-y^2}\leq z \leq 1, \quad 0\leq y \leq \sqrt{5}, \quad\text{and}\quad 5 \leq z \leq 1$$

Buut this was wrong apart from the lower limits. This was the only way I knew how to approach this, so I don't actually know what to do here. Can anybody point me in the right direction?

Answer 1

When you convert the first iterated integral into a triple integral, the domain of integration is specified by the set of inequalities

$$ \begin{cases} 0 \leq y \leq \sqrt{5}, \\ 0 \leq z \leq \sqrt{5-y^2}, \\ y^2 + z^2 \leq x \leq 5. \end{cases} \tag{*}$$

Now we rephrase these inequalities to obtain the desired bounds:

\begin{align*} \text{(*)} &\quad\iff\quad \begin{cases} 0 \leq y \text{ and } \color{red}{y \leq \sqrt{5}}, \\ 0 \leq z \text{ and } \color{red}{y^2 + z^2 \leq 5}, \\ 0 \leq x \leq 5 \text{ and } y^2 + z^2 \leq x \end{cases} \tag{1} \\[0.667em] &\quad\iff\quad \begin{cases} 0 \leq y, \\ 0 \leq z, \\ 0 \leq x \leq 5 \text{ and } y^2 + z^2 \leq x \end{cases} \tag{2} \\[0.667em] &\quad\iff\quad \begin{cases} 0 \leq x \leq 5 \\ 0 \leq y \text{ and } \color{blue}{y^2 \leq x} \\ 0 \leq z \text{ and } y^2 + z^2 \leq x \end{cases} \tag{3} \\[0.667em] &\quad\iff\quad \begin{cases} 0 \leq x \leq 5 \\ 0 \leq y \leq \sqrt{x} \\ 0 \leq z \leq \sqrt{x - y^2} \end{cases} \tag{4} \end{align*}

Here are some explanation

• $\text{(1)} \iff \text{(2)}$ : The red-colored inequalities are redundant since they are already implied by the condition $y^2 + z^2 \leq x \leq 5$. So we can erase them without changing the region.

• $\text{(2)} \iff \text{(3)}$ : From the inequality $y^2 + z^2 \leq x$, we obtain the blue-colored inequality $y^2 \leq x$ for free, without changing the region the system describes.

Now $\text{(4)}$ gives the answer.

Answer 2

While this has got a good Answer & various hints via Comments , I wanted to share my thoughts on this.

Given a Volume (Domain of Integration) in terms of 3 Criteria , we should try to imagine (visualize) what that Volume looks like. Description given in 1 Order of $y,z,x$ can be converted to Single Set , which we can then give in some other Order $x,y,z$.

Hence , let us try to figure out what that Volume (Domain of Integration) is , visually.

Consider the 3 axes with $z$ Perpendicular to the $xy$ Plane. Consider arbitrary Point $X$ on the $x$ axis , where we are given the first Criteria $0<y<\sqrt{5}$ , with the Second Criteria $0<z<\sqrt{5-y^2}$ which is equivalent to $(y,z)$ within $z^2+y^2=5$ , which is a Circle in the $yz$ Plane , Perpendicular to the $x$ axis.

Here is the Image till now , where I have shown a Quarter Circle & the Blue lines are the Integration Points when $z$ goes from 0 to the Circle Edge.

Moving to the third Criteria , we see that when $(y,z)=(0,0)$ , $x$ is between $0^2+0^2=0$ & $5$ , hence our arbitrary Point was actually $X=5$. Every other value of $y,z$ gives a Positive Starting Point $x=y^2+z^2$ with the Ending Point $x=5$. These Starting Points are Circles of smaller radius , with $0$ radius at $(y,z)=(0,0)$ terminating with radius $\sqrt{5}$ at $(y,z)=(\sqrt{5},0)$.

It will look something like this :

Hence , the Volume is a Parabola $x=y^2+0^2$ [[ $y=\sqrt{x}$ ]] rotated around the $x$ axis , which gives a Paraboloid in 3D.

Here , I am showing the Parabola on the $xy$ Plane in Purple , the Circle in Black in the $yz$ Plane where the Circular Area gets the Blue $z$ lines :

Describing in Order $x,y,z$ can be attempted now.
We can see that $x$ is between $0$ & $5$ : $0<x<5$.
At each value of $x$ , $y$ is between $0$ & $\sqrt{x}$ : $0<y<\sqrt{x}$.
At each Point in that Area , $z$ is within the Circle of radius $\sqrt{x}$ , that is $z^2+y^2=x$ : $0<z<\sqrt{x-y^2}$.

Thus the Integration limits in Order $x,y,z$ are :
$0<x<5$.
$0<y<\sqrt{x}$.
$0<z<\sqrt{x-y^2}$

Observations :
(1) We know the Actual Volume Set (Domain of Integration) , hence we can easily convert to some other Order $x,z,y$.
(2) We can even try to convert back to Original Order $y,z,x$ to verify that we get the Correct Original Criteria , which will confirm that we are not making mistakes.
(3) Paraboloid in 3D is $x = y^2+z^2$ [[ $z = \sqrt{x-y^2}$ ]] where we are interested in the Volume where all variables are Non-Negative.