analysisintegrationmultivariable-calculusvolume
I am just studying triple integrals in my calculus class and I have the following problem as homework and I am not sure how to start:
Let $P$ be a pyramid defined by the base [-2,2]x[-2,2] in XY plane and tip (0,0,5). Compute the volume in 2 ways using triple integrals:
once as $dzdxdy$ and then as $dxdydz$.
Since I am new in this type of problems I would appreciated some help. Thank you
A volume element is
$$dV = A(y) dy$$
where the sides in the square cross-sectional area $A(y)$ behaves linearly with height:
$$A(y) = \left[ b-\frac{y}{h} (b-a)\right ]^2 = b^2-2 b (b-a)\frac{y}{h} +\frac{(b-a)^2} {h^2} y^2$$
so the integral is
$$V = \int_0^h dy \, A(y) = b^2 h - b (b-a) h +\frac13 (b-a)^2 h$$
or, simplifying,
$$V = \frac13\frac{b^3-a^3}{b-a} h $$
I'm going to try to expand on @SangchulLee's answer a little bit.
You think through the original problem by realizing that our 3-dimensional region projects down to the region in the $xy$-plane between the $y$-axis and $x=y^3$ [or, equivalently, $y=x^{1/3}$]. In $\Bbb R^3$, $x=y^3$ is a vertical cylinder over this plane curve. For each point in that plane region, a vertical line enters the region at $z=0$ and exits at $z=y^2$. So this surface is a parabolic cylinder, with lines parallel to the $x$-axis. Here's a sketch with the assistance of Mathematica:
Indeed, the region lies over the rectangle $R=[0,8]\times [0,4]$ in the $xz$-plane, and the two cylinders intersect along the crease $y=z^{1/2}=x^{1/3}$, which projects to the curve $C$ given by $x^2=z^3$ in the $xz$-plane. If we take a point in $R$, we want to know when a line parallel to the $y$-axis enters and exits our region. If we're below $C$ in $R$, then $z^3<x^2$ and the line enters at $y=x^{1/3}$ and exits at $y=2$. (As a check, if $y\ge x^{1/3}$, then $y\ge z^{1/2}$ as well in this range, so we're "inside" both cylinders.) If we're above $C$ in $R$, then $z^3>x^2$ and the line enters at $y=z^{1/2}$ and exits at $y=2$.
Thus, we end up with the following iterated integral in the order $dy\,dz\,dx$: $$\int_0^8\int_0^{x^{2/3}}\int_{x^{1/3}}^2 dy\,dz\,dx + \int_0^8\int_{x^{2/3}}^4\int_{z^{1/2}}^2 dy\,dz\,dx.$$
(By the way, you should be able to set up the limits in the orders $dx\,dz\,dy$ or $dx\,dy\,dz$ much more easily.)
Best Answer
If you have a volume $A$, and want to express $\int_A f(x,y,z)\,d(x,y,z)$ as an iterated integral, i.e. as $$ \int_A f(x,y,z) \,d(x,y,z) = \int_{A_x} \int_{A_y(x)} \int_{A_z(x,y)} f(x,y,z) \,dz\,dy\,dx $$ you have to find $$\begin{eqnarray} A_x &\subset& \mathbb{R} \\ A_y &\,:\,& A_x \to \mathcal{P}({\mathbb{R}}) \\ A_z &\,:\,& \{(x,y) \,:\, x \in A_x, y \in A_y(x)\} \to \mathcal{P}(\mathbb{R}) \end{eqnarray}$$ such that $$ A = \left\{(x,y,z) \,:\, x \in A_x,\, y \in A_y(x),\, z \in A_z(x,y,z) \right\}. $$
Integration order $dz\,dx\,dy$
Let's look at the case of a pyramid with base $[-2,2]\times [-2,2]$ in the $x,y$-plane, tip at $(0,0,5)$, and integration order $dz\,dx\,dy$. We have $$ A = \left\{(x,y,z) \,:\, x \in [-2,2], y \in [-2,2], 0 \leq z \leq \tfrac{5}{2}\min \{2-|x|,2-|y|\}\right\} \text{.} $$ So $A_y$ doesn't actually depend on $x$, and $A_z(x,y) = [0, \tfrac{5}{2}\min \{2-|x|,2-|y|\}]$. Therefore, $$ \int_A 1 \,d(x,y,z) = \int_{-2}^2 \int_{-2}^2 \int_0^{\tfrac{5}{2}\min \{2-|x|,2-|y|\}} 1 \,dz\,dy\,dx \text{.} $$
Integration order $dx\,dy\,dz$
For integration order $dx\,dy\,dz$ we have to find $$\begin{eqnarray} A_z &\subset& \mathbb{R} \\ A_y &\,:\,& A_z \to \mathcal{P}({\mathbb{R}}) \\ A_x &\,:\,& \{(z,y) \,:\, z \in A_z, y \in A_y(z)\} \to \mathcal{P}(\mathbb{R}) \end{eqnarray}$$ such that $$ A = \left\{(x,y,z) \,:\, x \in A_x(z,y),\, y \in A_y(z),\, z \in A_z \right\}. $$ Obviously, $A_z = [0,5]$. For a fixed $z \in A_z$, $(x,y,z) \in A$ exactly if $$\begin{eqnarray} z &\leq& \tfrac{5}{2}(2 - |x|) &\Leftrightarrow& |x| \leq \tfrac{2}{5}z - 2 &\text{ and } \\ z &\leq& \tfrac{5}{2}(2 - |y|) &\Leftrightarrow& |y| \leq \tfrac{2}{5}z - 2 \text{,}\\ \end{eqnarray}$$ which means $$ \int_A 1 \,d(x,y,z) = \int_0^5 \int_{-\frac{2}{5}z + 2}^{\frac{2}{5}z - 2} \int_{-\frac{2}{5}z + 2}^{\frac{2}{5}z - 2} 1 \,dx\,dy\,dz \text{.} $$