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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! :)

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  • $\begingroup$ Very nice. From basic three-dimensional geometry you also get the volume of this straight-angle triangular pyramid: it is the basis's area times the height divided by $\;3\;$....which is exactly what you got, of course. $\endgroup$ – DonAntonio Jun 14 '13 at 13:07
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    $\begingroup$ Thankyou for your feedback @DonAntonio $\endgroup$ – amanda Jun 14 '13 at 14:12
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Here is an alternative computation using a single variable integral that confirms your result. The following figure represents the given pyramid.

enter image description here

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 $x=a$

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

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This is entirely correct. Here is a geometric check. The unit simplex, bounded by the positive orthant and the plane $x + y + z = 1$ has volume 1/6. You stretch this along the three axes by factors $a$, $b$ and $c$, hence your result $abc/6$.

It is good to see you found the glych.

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