calculusdefinite integralsintegrationmultivariable-calculusvolume
$D$ is the tetrahedron bounded by the coordinate planes and the plane $3x+3y+z=3$, then express the volume of $D$ as a triple integral.
Mt Try:
The z-limits are $0\le z\le3-3x-3y$
If $y$ is the limit on the $xy$– plane then $z=0$ $0\le y\le1-x$
Similarly, $0\le x\le3$
Finally, the integral will be $\int_{0}^{3}\int_{0}^{1-x}\int_{0}^{3-3x-3y}dz\ dy\ dx$
Is my above integral correct?
Just to convert my comment to an answer.
If you want to use triple integral to find the volume. There are two ways to do this.
First method: Direct triple integral. You have to find the plane equations for all faces of this tetrahedron. Due to the piecewise nature, the limits of the integral are somewhat messy.
Second method: We can transform the points $ A(1,2,3)$, $B(-2,1,5)$, $C(3,7,1)$ to $A'(1,0,0)$, $B'(0,1,0)$, $C'(0,0,1)$. So that the volume of the new tetrahedron is easy to compute. The transformation matrix from $A',B',C'$ to $A,B,C$ is: $$ T = \begin{pmatrix} 1&-2 &3 \\ 2 &1 &7 \\ 3 &5 &1\end{pmatrix} . $$ Because the linearity this is also the Jacobian matrix, so $$ \mathrm{Volume} = \iiint_{OABC} 1 \,dxdydz = \iiint_{OA'B'C'} 1 \,{|\det (T)|}\,dx'dy'dz' = \frac{17}{2}. $$
Another tip with a formula to compute $n$-simplex, I took it from my computational geometry notes: $$ |V| = \frac{1}{3!}\left|\det \begin{pmatrix} x_1 & x_2 & x_3 & x_4\\ y_1 & y_2 & y_3 & y_4\\ z_1 & z_2 & z_3 & z_4\\ 1 & 1 & 1 & 1\\ \end{pmatrix} \right|, $$ where $(x_i,y_i,z_i)$ are the coordinates for the $i$-th vertex. It bears the same form for the area formula of a triangle with three vertices give $$ |T| = \frac{1}{2!}\left|\det \begin{pmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \\ \end{pmatrix} \right|. $$
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 $x=a$
$$\begin{eqnarray*} V &=&\int_{0}^{a}A(x)dx \\ &=&\frac{bc}{2}\int_{0}^{a}\left( 1-\frac{x}{a}\right) ^{2}dx \\ &=&\frac{bc}{2}\left( a-\frac{2}{a}\frac{a^{2}}{2}+\frac{1}{a^{2}}\frac{a^{3} }{3}\right) \\ &=&\frac{abc}{6}. \end{eqnarray*}$$
Best Answer
It is correct for $y$ and $z$ bounds but for $x$ we have
and therefore
$$V=\int_{0}^{1}dx\int_{0}^{1-x}dy\int_{0}^{3-3x-3y}dz$$