<s>[INST] Find the volume using triple integrals Using triple integrals and Cartesian coordinates, find the volume of the solid bounded by
$$ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1  $$
and the coordinate planes $x=0, y=0,z=0$

My take
  I have set the parameters to $$ 0\le x \le a$$ $$0\le y \le b\left( 1 - \frac{x}{a} \right)$$  $$0\le z \le c \left( 1 - \frac{y}{b} -\frac{x}{a} \right)  $$ and evaluated $$ \int_0^{a} \int_0^{b\left( 1 - \frac{x}{a}\right)} \int_0^{c \left( 1 - \frac{y}{b} -\frac{x}{a} \right)} 1 dzdydx$$ and gotten $0$ as my final answer but the actal answer is $\frac{abc}{6}$
Never mind, I found the mistake I was making, just a simple integral mistake but it was the right procedure. Thank you for viewing! :)

 [/INST] Here is an alternative computation using a  single variable integral that confirms your result. The following figure represents the given pyramid. 

The equations of the lines situated on the planes $y=0$ and $z=0$ are:
$$y=0,\qquad\frac{x}{a}+\frac{z}{c}=1\Leftrightarrow z=\left( 1-\frac{x}{a}\right) c,$$
$$z=0,\qquad\frac{x}{a}+\frac{y}{c}=1\Leftrightarrow y=\left( 1-\frac{x}{a}\right) b.$$
The intersection of the pyramid with the plane perpendicular to the $x$-axis in $x$ is a right triangle with catheti $\left( 1-\frac{x}{a}\right) c$ and $\left( 1-\frac{x}{a}\right) b$, whose area $A(x)$  is given by
$$A(x)=\frac{1}{2}\left( 1-\frac{x}{a}\right) b\left( 1-\frac{x}{a}\right) c=\frac{bc}{2}\left( 1-\frac{x}{a}\right) ^{2}.$$
Hence the volume is given by the integration of the area $A(x)$ from $x=0$ to $x=a$ 
$$\begin{eqnarray*}
V &=&\int_{0}^{a}A(x)dx \\
&=&\frac{bc}{2}\int_{0}^{a}\left( 1-\frac{x}{a}\right) ^{2}dx \\
&=&\frac{bc}{2}\left( a-\frac{2}{a}\frac{a^{2}}{2}+\frac{1}{a^{2}}\frac{a^{3}
}{3}\right) \\
&=&\frac{abc}{6}.
\end{eqnarray*}$$
</s>