To write the integral \(\iiint_{T} f(x, y, z) \, dV\) in the specified form, we need to determine the limits of integration for \(x\), \(y\), and \(z\) based on the given vertices of the tetrahedron. The vertices are \((2,0,0)\), \((0,2,0)\), \((0,0,4)\), and the origin \((0,0,0)\).
- The outer integral with respect to \(x\) ranges from 0 to 2.
- For a fixed \(x\), the range of \(z\) depends on \(x\). The plane containing the vertices \((2,0,0)\), \((0,0,4)\), and the origin can be described by the equation \(z = 2 - x\).
- For fixed \(x\) and \(z\), the range of \(y\) depends on both \(x\) and \(z\). The plane containing the vertices \((2,0,0)\), \((0,2,0)\), and the origin can be described by the equation \(y = 2 - x - z\).
Thus, the limits of integration are:
- \(x\) ranges from 0 to 2.
- \(z\) ranges from 0 to \(2 - x\).
- \(y\) ranges from 0 to \(2 - x - z\).
- Determine the limits of integration for \(x\), \(y\), and \(z\) based on the vertices of the tetrahedron.
- Write the integral in the specified form using these limits.
The tetrahedron \( T \) has vertices at \((2,0,0)\), \((0,2,0)\), \((0,0,4)\), and the origin \((0,0,0)\). We need to express the volume integral \(\iiint_{T} f(x, y, z) \, dV\) in the form of iterated integrals.
The \( x \)-coordinate ranges from 0 to 2, as the tetrahedron extends from the origin to the vertex \((2,0,0)\).
\[
0 \leq x \leq 2
\]
For a fixed \( x \), the \( z \)-coordinate ranges from 0 to the plane that intersects the \( z \)-axis at \( z = 4 \) and the \( x \)-axis at \( x = 2 \). The equation of this plane can be found by noting that it passes through \((2,0,0)\) and \((0,0,4)\). The equation of the plane is:
\[
\frac{x}{2} + \frac{z}{4} = 1 \implies z = 4 - 2x
\]
Thus, for a fixed \( x \):
\[
0 \leq z \leq 4 - 2x
\]
For fixed \( x \) and \( z \), the \( y \)-coordinate ranges from 0 to the plane that intersects the \( y \)-axis at \( y = 2 \) and the \( z \)-axis at \( z = 4 \). The equation of this plane can be found by noting that it passes through \((0,2,0)\) and \((0,0,4)\). The equation of the plane is:
\[
\frac{y}{2} + \frac{z}{4} = 1 \implies y = 2 - \frac{z}{2}
\]
Thus, for fixed \( x \) and \( z \):
\[
0 \leq y \leq 2 - \frac{z}{2}
\]
Combining all these limits, the iterated integral can be written as:
\[
\iiint_{T} f(x, y, z) \, dV = \int_{0}^{2} \int_{0}^{4-2x} \int_{0}^{2 - \frac{z}{2}} f(x, y, z) \, dy \, dz \, dx
\]
Thus, the functions \( u(x) \) and \( v(x, z) \) are:
\[
u(x) = 4 - 2x
\]
\[
v(x, z) = 2 - \frac{z}{2}
\]
\[
\boxed{u(x) = 4 - 2x}
\]
\[
\boxed{v(x, z) = 2 - \frac{z}{2}}
\]
Answered
