i have a triple integral ∭zdV over the tetrahedron with vertices: (0, 0, 1), (0, 1, 1), (1, 0, 0), and (0, 0, 0).
I calculated the plane and got the boundary of 0≦x≦1, 0≦y≦1-x, 0≦z≦y.
the correct answer is supposed tobe 1/16 ,however, i keep getting the wrong answer
Can somebody help me where i get wrong and what the correct answer is? Thank you!!!
(Your planes do not correspond to the given points.)
Draw a figure! You then will see that the tetrahedron $T$ is defined by the four inequalities $$x\geq0,\quad y\geq0,\quad z\geq y,\quad x+z\leq 1\ .\tag{1}$$ Replacing the $\geq$ signs with $=$ we obtain the equations of the four planes containing the faces of $T$. Any three of these four planes intersect in one of the given vertices.
From $(1)$ we infer that automatically $z\geq0$ in all points of $T$. We therefore choose the triangle $T':=\bigl\{(x,0,z)\bigm| x\geq0, \ z\geq 0,\ x+z\leq 1\bigr\}$ as domain for the two-dimensional outer integral. Note that for any point $(x,y,z)\in T$ one necessarily has $(x,0,z)\in T'$. Given a point $(x,0,z)\in T'$ we now have to determine the set of $y$ for which $(x,y,z)\in T$. Looking at $(1)$ we see that this is the $y$-interval $0\leq y\leq z$. We therefore obtain $$\eqalign{\int_T z\>{\rm d}(x,y,z)&=\int_{T'} z\left(\int_0^z dy\right)\>{\rm d}(x,z)=\int_{T'}z^2\> {\rm d}(x,z)\cr &=\int_0^1 z^2\left(\int_0^{1-z} dx\right)dz=\int_0^1(z^2-z^3)\>dz\cr &={1\over12}\ .\cr} $$