I need to find out if it is true or false that these integrals are the same

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1-x^2} f(x,y,z)dz dy dx $$

$$\int_{0}^{1}\int_{0}^{1}\int_{-\sqrt{1-z}}^{\sqrt{1-z}}f(x,y,z) dx dy dz$$

I know that $0\leq z\leq 1-x^2$

From just the $z\leq 1-x^2$ part, I know that $-\sqrt{1-z}\leq x \leq \sqrt{1-z}$

But I'm not sure where the bounds for z would come from i.e. how does the second integral shown have 0 and 1 as the limits of integration for z?

Also, isn't it technically that

$0\leq z \leq 1-x^2 \implies$

$x^2 \leq z+x^2 \leq 1 $

$x^2-z\leq x^2 \leq 1-z$

$\pm\sqrt{x^2-z}\leq x \leq \pm \sqrt{1-z}$

Do I just ignore the left side of the equation?

  • $\begingroup$ Uh, because the second integral explicitly gives $[0, 1]$ as the bounds for $z$. You do know that the leftmost integral sign corresponds to the rightmost d(variable), right? $\endgroup$
    – Dan
    May 27 at 22:32
  • $\begingroup$ Yes but I mean if I were to come up with the integral on my own in the order dx dy dz $\endgroup$ May 27 at 22:53

1 Answer 1


If they are equal, then they should be equal when $f=1$.

But when $f=1$ the first one is equal to $$ \int_0^1(1-x^2)\;dx=\frac23 $$ while the second one is $$ \int_0^12\sqrt{1-z}\;dz = \frac43. $$

To figure out the exact bounds when changing the order of the first iterated integral, it is helpful to draw a picture of the region.

Alternatively, observe that the region of the first integral is given by $$ E=\{(x,y,z)\mid 0\le x\le 1,\quad 0\le y\le 1,\quad 0\le z\le 1-x^2\} $$ which is a normal domain in $\mathbf{R}^3$: $$ \displaystyle E=\{(x,y,z)\mid 0\le y\le 1, (x,z)\in D\} $$ where $D:=\{(x,z)\mid 0\le x\le 1, 0\le z\le 1-x^2\}$ is a 2-dimensional region, which itself is a normal domain in $\mathbf{R}^2$.

Now we have already reduced the problem to a 2-dimensional one. Drawing a picture (see below), you can see that $D$ is nothing but a region bounded by the graph of the function $z=1-x^2$ ($0\le x\le 1$) and three other lines. It is equivalent to $$ D=\{(x,z)\mid 0\le z\le 1, 0\le x\le \sqrt{1-z}\}\tag{0} $$ If you want to find (0) algebraically, note that $0\le z\le 1-x^2$ implies $$ -\sqrt{1-z}\le x\le\sqrt{1-z},\quad z\ge 0\tag{1} $$ which together with the condition $0\le x\le 1$ gives $$ 0\le x\le \sqrt{1-z}\tag{2} $$

On the other hand, $0\le x\le 1$ and $0\le z\le 1-x^2$ together imply that $0\le z\le 1$.

enter image description here

  • $\begingroup$ Ohhh thank you, that makes sense for the bounds of x. Still wondering how to get the bounds for z as 0 and 1 though $\endgroup$ May 27 at 22:51
  • $\begingroup$ @user8290579: see the edited answer. If you draw a picture of D, you can clearly see the bounds for $z$. Let me know if you have further questions. $\endgroup$ May 27 at 22:56

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