Volume of a pyramid, using an integral

Pyramid and equations of the lines situated on the planes $y=0$ and $z=0$.

$$y=0,\qquad\frac{x}{a}+\frac{z}{c}=1\Leftrightarrow z=\left( 1-\frac{x}{a}\right) c,$$

$$z=0,\qquad\frac{x}{a}+\frac{y}{c}=1\Leftrightarrow y=\left( 1-\frac{x}{a}\right) b.$$

The area $A(x)$ is given by

$$A(x)=\frac{1}{2}\left( 1-\frac{x}{a}\right) b\left( 1-\frac{x}{a}\right) c=% \frac{bc}{2}\left( 1-\frac{x}{a}\right) ^{2},$$

because the intersection of the solid with the plane perpendicular to the $x$-axis in $x$ is a right triangle with catheti $\left( 1-\frac{x}{a}\right) c$ and $\left( 1-\frac{x}{a}\right) b$. Hence

$$\begin{eqnarray*} V &=&\int_{0}^{a}A(x)dx \\ &=&\frac{bc}{2}\int_{0}^{a}\left( 1-\frac{x}{a}\right) ^{2}dx \\ &=&\frac{bc}{2}\int_{0}^{a}1-\frac{2x}{a}+\frac{x^{2}}{a^{2}}dx \\ &=&\frac{bc}{2}\left( \int_{0}^{a}1dx-\int_{0}^{a}\frac{2x}{a}dx+\int_{0}^{a}% \frac{x^{2}}{a^{2}}dx\right) \\ &=&\frac{bc}{2}\left( a-\frac{2}{a}\int_{0}^{a}xdx+\frac{1}{a^{2}}% \int_{0}^{a}x^{2}dx\right) \\ &=&\frac{bc}{2}\left( a-\frac{2}{a}\frac{a^{2}}{2}+\frac{1}{a^{2}}\frac{a^{3}% }{3}\right) \\ &=&\frac{bc}{2}\left( a-a+\frac{a}{3}\right) \\ &=&\frac{abc}{6} \end{eqnarray*}$$

To calculate the volume of some object, which runs from height $z=z_0$ to $z=z_1$, you can use e.g.

$$\int_{z=z_0}^{z_1} A(z) \ dz$$

where $A(z)$ is the area of the object (if we took a slice of it) at height $z$.

For the pyramid, you could integrate along the z-axis, so that the areas become right triangles. You can verify that at $z = 0$ the sides are $a,b$ while at $z=c$ the sides are $0,0$, and the sides' lengths decrease linearly in $z$. So at height $z$ the sides are $a(1 - \frac{z}{c})$ and $b(1 - \frac{z}{c})$ respectively, giving $A(z) = \frac{1}{2} a b (1 - \frac{z}{c})^2$. Filling this in in the integral above we get

$$\int_{z=0}^{c} A(z) \ dz = \frac{1}{2} a b \int_{z=0}^{c} (1 - \frac{z}{c})^2 \ dz = \frac{1}{2} a b \int_{z=0}^{c} (1 - \frac{2z}{c} + \frac{z^2}{c^2}) \ dz = \frac{1}{2} a b(c - c + \frac{1}{3} c) = \frac{1}{6}abc$$

Another way to calculate the volume would be to make a triple integral over the whole object (and integrate 1), and make sure the bounds are right. You have that $x,y,z \geq 0$ and furthermore requirements like $y \leq b(1 - \frac{z}{c})$ and $x \leq a(1 - \frac{z}{c} - \frac{y}{b})$. The right integral then gives

$$\int_{z=0}^{c} \int_{y=0}^{b(1 - \frac{z}{c})} \int_{x=0}^{a(1 - \frac{z}{c} - \frac{y}{b})} 1 \ dx \ dy \ dz = \frac{1}{6} abc$$

Hope this helps. This would be the way I'd do it, but there may be faster ways to do it.