# Volume of a pyramid, using an integral

I have a calculus exam tomorrow and this is a possible question. However, I don't know how to handle this question.

Suppose you have 3 points in space: $p_1=(a,0,0)$, $p_2=(0,b,0)$ and $p_3=(0,0,c)$, $a,b,c \gt 0$. If we connect these points we get a pyramid in the first octant, with the origin as a top.

(i) Proove that the volume of this pyramid is given by $V = \frac{1}{6}abc$ by using a volume integral. Use the formula we made for non-revolution-solids. (Integrate $A(x)$ from $a$ to $b$, where $A(x)$ is the area of the intersection of the solid with the plane perpendicular to the $x$-axis in $x$. (Maybe $A(y)$ is better suitable in this problem.)) Hint: Calculate the surface of a slice at height $z = z_0$, with $z_0$ is a constant. (Maybe I should find $A(z)$?)

(ii) If you know the equation of the plane $V$ through these points is given by $V \leftrightarrow \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, and that $p=(1,2,3)$ is an element of $V$, find the minimal volume of the pyramid cut off from the first octant. Explain physically why this has to be a minimum.

Notes: Excuse my English, it's not my native language. The original question is in Dutch. We use the textbook Calculus 6E, metrical edition, by James Stewart. Which is fortunately in English, so I should understand most of your answers.

Thanks a lot already!

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## 2 Answers

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.

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Thanks, it is clear that I have a lack of insight. The first integral is the one I'm looking for. – Mats Aug 23 '11 at 19:44

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

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

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The image is wonderful, thanks! – Mats Aug 23 '11 at 20:03
@Mats: You are welcome! – Américo Tavares Aug 23 '11 at 20:06