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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

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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 $$ \int_A 1 \,d(x,y,z) = \int_0^5 \int_{-\frac{2}{5}z + 2}^{\frac{2}{5}z - 2} \int_{-\frac{2}{5}z + 2}^{\frac{2}{5}z - 2} 1 \,dx\,dy\,dz \text{.} $$

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  • $\begingroup$ could you explain a bit more how you get the $z\leq \frac{5}{2}min{2-|x|, 2-|y|}$? $\endgroup$ – user147744 May 4 '14 at 18:42
  • $\begingroup$ @user147744 By geometric insight, mostly. I start with a quadratic pyramid whose height and width are both $1$. The I looked at the projections to the $xz$ and $yz$-plane (assuming that the base square lies in the $xy$-plane). This yields $z \leq 1 - |x|$ and $z \leq 1 - |y|$. The just scale the result appropriately. The $\min$ simply is the result of combining both inequalities into one. $\endgroup$ – fgp May 4 '14 at 18:49

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