# How do I know if these two integrals are the same?

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?

• 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?
– Dan
May 27 at 22:32
• Yes but I mean if I were to come up with the integral on my own in the order dx dy dz May 27 at 22:53

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 $$0\le z\le 1$$.

• 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 May 27 at 22:51
• @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. May 27 at 22:56