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