Chapter 14.5, Problem 40ES

Express each integral as an equivalent integral in which the $z\text{-<x-custom-btb-me data-me-id="1678" class="microExplainerHighlight">integration</x-custom-btb-me>}$ is performed first, the $y\text{-integration}$ second, and the $x\text{-integration}$ last.

$\begin{array}{l}\left(\text{a}\right){\displaystyle {\int }_0^3{\displaystyle {\int }_0^{\sqrt{9-z^2}}{\displaystyle {\int }_0^{\sqrt{9-y^2-z^2}}f\left(x,y,z\right)dxdzdy}}}\\ \left(\text{b}\right){\displaystyle {\int }_0^4{\displaystyle {\int }_0^2{\displaystyle {\int }_0^{x/2}f\left(x,y,z\right)dydzdx}}}\\ \left(\text{c}\right){\displaystyle {\int }_0^4{\displaystyle {\int }_0^{4-y}{\displaystyle {\int }_0^{\sqrt{z}}f\left(x,y,z\right)}dxdzdy}}\end{array}$