3
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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

9
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.