Determine the appropriate order of integration to find the volume

Question

Find an appropriate order of integration bounded by the solid. Can someone help me figure out an appropriate order and maybe walk me through the integration? I am rather new to this. Thank you!

Answer 1

Re-arrange $z=1-y$ into $y=1-z.\;$ Now the volume is given by

\begin{align*} V &=\int_{0}^{1} \int_{0}^{1-z} \int_{0}^{1-z^{2}} dx \, dy \, dz \\ &= \int_{0}^{1} (1-z)(1-z^{2}) \, dz \\ &= \int_{0}^{1} (1-z-z^{2}+z^{3}) \, dz \\ &= \left[ z-\frac{z^{2}}{2}-\frac{z^{3}}{3}+\frac{z^{4}}{4} \right]_{0}^{1} \\ &= \frac{5}{12} \end{align*}

Answer 2

Rather than choosing a single order of integration, we will present all of them.

If we fix $z$, then we are choosing horizontal slices perpendicular to the $z$-axis. These will be rectangles parallel to the $xy$-plane with $y$ ranging from $0$ to $1-z$, and $x$ ranging from $0$ to $1-z^2$; thus $$V = \int_{z=0}^1 \int_{y=0}^{1-z} \int_{x=0}^{1-z^2} \, dx \, dy \, dz,$$ or equivalently, $$V = \int_{z=0}^1 \int_{x=0}^{1-z^2} \int_{y=0}^{1-z} \, dy \, dx \, dz.$$ Note that because the cross-sectional slices are rectangles in this case, the order of integration of $x$ and $y$ are free to be interchanged without impacting the limits of integration: we can also see this from the fact that the limits of integration for $x$ and $y$ are independent of $y$ and $x$, respectively.

If we fix $y$, then we are taking vertical slices perpendicular to the $y$-axis and parallel to the $xz$-plane. These will be oddly shaped, consisting of a piecewise parabolic boundary and a horizontal boundary. Specifically, if $z$ ranges from $0$ to $1-y$, then $x$ ranges from $0$ to $1-z^2$, and we get $$V = \int_{y=0}^1 \int_{z=0}^{1-y} \int_{x=0}^{1-z^2} \, dx \, dz \, dy.$$ This corresponds to integrating each cross-section in horizontal strips along the $z$-axis. If we choose to reverse the order of $x$ and $z$, the integral becomes more complicated. We would need to observe that $x$ would always range from $0$ to $1$ for a fixed $y$, but then $z$ would range from $0$ to $\sqrt{1-x}$ if $1-(1-y)^2 < x < 1$, and from $0$ to $1-y$ if $0 < x < 1-(1-y)^2$: $$V = \int_{y=0}^1 \left(\int_{x=0}^{1-(1-y)^2} \int_{z = 0}^{1-y} \, dz \, dy + \int_{x=1-(1-y)^2}^1 \int_{z=0}^{\sqrt{1-x}} \, dz \, dy \right) \, dx.$$ Finally, consider fixing $x$ and taking cross-sections perpendicular to the $x$-axis and parallel to the $yz$-plane. These are trapezoids. We can integrate in horizontal strips within each cross-section to give $$V = \int_{x=0}^1 \int_{z=0}^{\sqrt{1-x}} \int_{y=0}^{1-z} \, dy \, dz \, dx.$$ If we integrate in vertical strips, we need to go piecewise, observing that $z$ ranges from $0$ to $\sqrt{1-x}$ if $0 < y < 1 - \sqrt{1-x}$, and $z$ ranges from $0$ to $1-y$ if $1 - \sqrt{1-x} < y < 1$: $$V = \int_{x=0}^1 \left(\int_{y=0}^{1-\sqrt{1-x}} \int_{z=0}^{\sqrt{1-x}} \, dz \, dy + \int_{y=1-\sqrt{1-x}}^1 \int_{z=0}^{1-y} \, dz \, dy \right) \, dx.$$