I was going over my textbook and have trouble understanding this process:
Why does the innermost integral have the bounds $\sqrt{y}$ and $1$? The shape on the $xz$ plane looks the same as the one on the $xy$ plane so I figured that you could replace $\sqrt{y}$ with $\sqrt{z}$ in the bounds.
I did a quick calculation on my computer using $f(x,y,z)=xyz$ to check and the results were $0.0125$ and $0.02916667$ respectively. What am I missing here?
On
If we use Figure 13, there is a simpler way to compute the limits of integration for any given order.
Step 1: For the variable corresponding to the innermost integral ($x$ in this case), visualize a line parallel to the $x$-axis that passes through the solid. Then the surface where such a line would first touch the solid is given by $y= x^2$, and the surface where it would exit the solid is given by $x=1$.
Therefore, the lower and upper limits for $x$ are $\sqrt y$ and $1$ respectively.
Step 2: For the remaining two variables ($y$ and $z$ in this case), find the region $R$ in the $yz$-plane that includes all the $y$ and $z$ coordinates being used in the solid. Looking at Figure 13, this 'projection' on $yz$-plane can be intuitively seen as $R= \{(y,z):0\leq y \leq 1,0\leq z \leq y\}$.
Step 3: Now the problem has basically been reduced to computing limits for a double integral. Since $y$ is the variable corresponding to the outermost integral, express limits of $z$ in terms of $y$.
Therefore, the lower and upper limits for $z$ are $0$ and $y$ respectively. Whereas the lower and upper limits for $y$ are $0$ and $1$ respectively.
Why does the innermost integral have the bounds $\sqrt y$ and $1$? The shape on the xz plane looks the same as the one on the xy plane so I figured that you could replace $\sqrt y$ with $\sqrt z$ in the bounds.
Based on the above interpretation, replacing $\sqrt y$ with $\sqrt z$ would essentially mean that if an arbitrary line parallel to the $x$-axis were to pass through your solid, it would first touch the solid at $x=\sqrt z$. However, we know that is not the case, and thus doing so would change the solid we are dealing with.
The two integrals are the same: see here and here.
As for the limits, the whole "projection" argument sounds too complicated to me. You have $$\tag1 0\leq x\leq 1, \ \ 0\leq y\leq x^2,\ \ 0\leq z\leq y. $$ If you now want $y$ to be the "independent" variable, you get $0\leq y\leq 1$. You also have, from above, that $0\leq z\leq y$. And, finally, $\sqrt y\leq x\leq 1$ from the first two sets of inequalities in $(1)$.