Integral $\int_{0}^{1} \int_{0}^{z^2} \int_{z}^{1} e^y dx\,dy\,dz \,\,+\,\, \int_{0}^{1} \int_{z^2}^{1} \int_{\sqrt{y}}^{1} e^y dx\,dy\,dz$
I am completely lost..
At first, I tried graphing it, which basically gives a square plus some euler curve on top of the square.
I understand the geometry of this however, I just cannot seem to determine the bounds for each sub-region. If someone can walk me through step by step or just briefly explain, I would very much appreciate it.
For example, I have two graphs:
Graph A - we know that 0 <= y <= z^2 and z <= x <= 1
Graph B - z^2<=y<=1 and root(y) <= x <= 1.
So graph A basically is a square (z, 0), (1,0), (z, z^2), (1, z^2).
Graph B is basically a curve y = x^2 so we have two points, (z, x^2) (1, 1)
From here, I honestly don't know how to manipulate the bounds and re-order the integral to solve it.
The inner integrals always tell the biggest story because they always give you concrete surfaces involved in the problem. We get
$$y = x^2$$ $$x=1$$ $$z=x$$
for free. Drawing these out we can see that we also get
$$y=0$$ $$z=0$$
Fortunately for us there is an integration order that will give us one integral. To find it, note that
$$0 \leq y \leq x^2$$
since all of the other boundaries are "invisible" to $y$ (i.e. they are parallel to the $y$ axis). We also have that
$$0 \leq z \leq x$$
for the same reason. Then we have
$$I = \int_0^1 \int_0^x \int_0^{x^2} e^y \:dy \:dz \:dx = \frac{e-3}{2}$$