Help me understand easy (not for me) concepts in volume integral 
Keep looking at the page for an hour.
Still not sure how I can get the sloping surface of $x+y+z=1$ and integration ranges for $x, y, z$. and why their range is different too.
The book keeps throwing things at me without much explanation. Help me.
 A: For the sloping surface, this is the plane through the points $ ( 0 , 0 , 1 ) $, $ ( 0 , 1 , 0 ) $, and $ ( 1 , 0 , 0 ) $.  Since none of these is the origin, this plane has an equation of the form $ A x + B y + C z = 1 $, so you can put in the coordinates of these three points to get a system of linear equations to solve for $ A $, $ B $, and $ C $.  (But you might also be able to see, once you realize the form that the equation needs to take, that all three of these constants have to be $ 1 $.)  So that's how you get $ 1 x + 1 y + 1 z = 1 $, or $ x + y + z = 1 $ for short.  You should also realize that the other surfaces are the coordinate planes, with equations $ x = 0 $, $ y = 0 $, and $ z = 0 $.
Now you have to decide which order you want to do the variables.  This is arbitrary, and while sometimes some orders are more convenient than others, it makes no difference at all this time.  Choosing to do $ x $ first gives the result in your question.  You simply solve $ x + y + z = 1 $ for $ x $ to get $ x = 1 - y - z $, and you also have $ x = 0 $ for one of the coordinate planes, and there you go, $ x $ runs from $ 0 $ to $ 1 - y - z $.  (Or maybe from $ 1 - y - z $ to $ 0 $, but it should be clear from the picture that the sloping plane is in front of the plane $ x = 0 $.  I'll mention later how you can tell when you don't have a nice picture.)
We've done $ x $, so let's say that we want to do $ y $ next.  We can't solve $ x + y + z = 1 $ for $ y $; we've already used that equation, and we're not allowed to use $ x $ anymore anyway.  So we only have $ y = 0 $.  What's the other bound on $ y $?  We can find that by setting the two bounds on $ x $ equal: $ 0 = 1 - y - z $.  (Visually, $ 0 = 1 - y - z $ is the line in the $ ( y , z ) $-plane where the sloping surface $ x = 1 - y - z $ meets the back plane $ x = 0 $.)  Solve that for $ y $, and you get $ y = 1 - z $.  So $ y $ runs from $ 0 $ to $ 1 - z $.
Now only $ z $ is left.  And there's only one equation left, $ z = 0 $.  But again we can get another equation by setting the bounds on $ y $ equal: $ 0 = 1 - z $.  (Visually, this is the point on the $ z $-axis within the $ ( y , z ) $-plane where the line $ y = 1 - z $ meets the line $ y = 0 $.)  Solve this for $ z $ to get $ z = 1 $.  So $ z $ runs from $ 0 $ to $ 1 $.  (You can often get the last one easily from just the picture; you can see that $ 0 $ is the smallest value that $ z $ takes while $ 1 $ is the largest value.  But if you don't have a good picture, then these can still be calculated as I did here.)
We're done now, but even if you were unsure whether $ x $ runs from $ 0 $ to $ 1 - y - z $ or the reverse, it's clear that $ z $ must run from $ 0 $ to $ 1 $, since $ 0 < 1 $.  To be sure of the order for $ y $, pick a value of $ z $ in between, such as $ z = 1 / 2 $.  (Actually $ z = 0 $ would work too, but not $ z = 1 $, since this is the value that you got by setting the bounds on $ y $ equal.)  Then use this value of $ z $ to compare $ 0 $ and $ 1 - z $, the bounds on $ y $.  You get $ 0 $ and $ 1 - ( 1 / 2 ) = 1 / 2 $, and $ 0 < 1 / 2 $, so you know that $ 0 < 1 - z $ in general.  (Technically it's $ 0 \leq 1 - z $ since they may be equal at the bounds on $ z $, and indeed they are equal when $ z = 1 $, but that's not important here.)  Then keeping $ z = 1 / 2 $, pick a value of $ y $ between $ 0 $ and $ 1 / 2 $, say $ y = 1 / 4 $, to check the order of the bounds on $ x $.  You get $ 0 $ and $ 1 - ( 1 / 2 ) - ( 1 / 4 ) = 1 / 4 $.  Since $ 0 < 1 / 4 $, you know that $ 0 < 1 - y - z $ in general (or technically, $ 0 \leq 1 - y - z $).  So now the orders of the bounds are clear even without using the picture to help.
And now you can set up volume integrals as $$ \int _ { z = 0 } ^ 1 \int _ { y = 0 } ^ { 1 - z } \int _ { x = 0 } ^ { 1 - y - z } f ( x , y , z ) \, \mathrm d x \, \mathrm d y \, \mathrm d z \text . $$
A: First we can state that $z$ is the "free variable", so it doesn't depend on $x$ and $y$ but both of them will depend on $z$. We can see in the figure that it goes from $0$ to $1$.
Then, we will see the dependence for $y$. You can visualize it projecting the figure on the $YZ$ plane. To do this, let us make $x=0$ and we get the line $z= 1-y$. If $z$ varies from $0$ to $1$, then $y$ goes from $0$ to $1-z$.
Then, from the equation of the surface, we get that $x$ varies from $0$ to $1-y-z$.
Analogously, we could have chosen other dependences, like $x$ being from $0$ to $1$, projecting in the $XY$ plane to get that $y$ goes from $0$ to $1-x$ and obtaining from the equation that $z$ goes from $0$ to $1-x-y$ (it is a very symmetric figure, so no surprise that the equations look the same switching variables).
A: This is an attempt to capture how I tend to visualize a region when setting up a volume integral. Consider the following plots:

On the left, the forest of arrows join the $y,z$-plane (i.e. the plane $x=0$) and the given plane $x+y+z=1$. They represent the range that the variable $x$ can take on, $0 \le x \le 1 - y - z$.
On the right, there's a similar array of arrows that join the $z$-axis (i.e. the line $x=y=0$) to the line where the plane $x+y+z=1$ intersects $x=0$. This line has equation $y + z = 1$, or $y = 1 - z$. These arrows represent the range that $y$ can take on, $0 \le y \le 1 - z$.
It's not pictured, but the last step is recognizing that $z$ must range between $0$ and $1$.
Then the volume of the region is given by the triple integral
$$\int_0^1 \int_0^{1-z} \int_0^{1-y-z} dx \, dy \, dz$$
