calculusvectors
Use a triple integral to find the volume of the pyramid P whose base is the square with
vertices (1,0,0), (0,1,0), (−1,0,0), and (0,−1,0) and whose top vertex is (0,0,1).
I've been given the problem above however I'm not sure how to go about answering it, I understand how to triple integrate given a function however I'm quite stumped how to answer this.
Any help would be appreciated!
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
$$\begin{eqnarray*} V &=&\int_{0}^{a}A(x)dx \\ &=&\frac{bc}{2}\int_{0}^{a}\left( 1-\frac{x}{a}\right) ^{2}dx \\ &=&\frac{bc}{2}\int_{0}^{a}1-\frac{2x}{a}+\frac{x^{2}}{a^{2}}dx \\ &=&\frac{bc}{2}\left( \int_{0}^{a}1dx-\int_{0}^{a}\frac{2x}{a}dx+\int_{0}^{a}% \frac{x^{2}}{a^{2}}dx\right) \\ &=&\frac{bc}{2}\left( a-\frac{2}{a}\int_{0}^{a}xdx+\frac{1}{a^{2}}% \int_{0}^{a}x^{2}dx\right) \\ &=&\frac{bc}{2}\left( a-\frac{2}{a}\frac{a^{2}}{2}+\frac{1}{a^{2}}\frac{a^{3}% }{3}\right) \\ &=&\frac{bc}{2}\left( a-a+\frac{a}{3}\right) \\ &=&\frac{abc}{6} \end{eqnarray*}$$
Note that it depends upon the orientation you have used for the vectors, indeed
$$60°=180°-120°$$
then your calculation is almost correct but you need to fix orientations according to the convention adopted.
If we choose as direction of the normal vectors the outer direction, your calculation is correct, notably we have
normal to face $ADB$: $v=DA\times DB=(1,0,-1)\times(1,1,-1)=(1,0,1)$
normal to face $BCD$: $w=DB\times DC=(1,1,-1)\times(1,1,-1)=(0,1,1)$
then
$$\cos \theta =\frac{v\cdot w}{|v||w|}=\frac{1}{2}\implies \theta=60°$$
As observed by Jyrki Lahtonen, if we consider the dihedral angle between the two half-planes bordered by the shared edge, it can be easily shown that this angle is exactly $180°-60°=120°$.
Best Answer
The "function" which you are integrating is in this case plainly $$f(x,y,z) = 1$$
Important are the integration borders. First, formulate the borders for $z$ depending on $x$ and $y$. For the lower border you just get $0$, which is correspondent to the base area of the pyramid. The upper border, however, is a function of $x$ and $y$, which also depends on the quadrant you are looking at. For the quadrant where $x\geq 0, y\geq 0$ the upper integration border is $-x-y+1$, which is just inclinations of $x$ and $y$ ($-1$ each) and the $z$-axis intercept.
The other quadrants have upper borders deviating from this one in sign combination, but you can make use of the symmetry that the pyramid volume is exactly quartered by the $xy$-crossing. So you just set up the integral for one quadrant and multiply by $4$ after integration.
The $y$ lower border is again $0$, the upper border is a function of $x$ namely $1-x$. The borders for $x$ are plainly $0$ and $1$.
So we have the borders and are ready to set up our integral.
$$V=4\cdot\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{-x-y+1}dz\,dy\,dx$$
First integration over $z$ yields
$$V=4\int_{0}^{1}\int_{0}^{1-x}\left[z\right]_{0}^{-x-y+1}dy\,dx =4\int_{0}^{1}\int_{0}^{1-x}(-x-y+1)dy\,dx$$
And so on... You will find $V=2/3$