Question

1. let $\mathcal{u}$ be the solid tetrahedron bounded by the planes $x = 0$, $y = 1$, $z = 0$, and $x + 2y+3z = 8$. (the vertices of this tetrahedron are $(0,1,0)$, $(0,1,2)$, $(6,1,0)$, and $(0,4,0)$). write the triple integral $\iiint_{\mathcal{u}}f(x,y,z)dv$ as an iterated integral.

Explanation:

Step1: Determine the range of \(y\)

The planes \(y = 1\) and by considering the intersection of the planes, the range of \(y\) is from \(y = 1\) to \(y=4\).

Step2: Determine the range of \(x\) for a given \(y\)

From the plane \(x + 2y+3z=8\) and \(z = 0\), we get \(x=8 - 2y\). Also, \(x\geq0\). So for a given \(y\), \(x\) ranges from \(x = 0\) to \(x=8 - 2y\).

Step3: Determine the range of \(z\) for given \(x\) and \(y\)

From the plane \(x + 2y+3z=8\), we can solve for \(z\) as \(z=\frac{8 - x - 2y}{3}\), and \(z\geq0\). So \(z\) ranges from \(z = 0\) to \(z=\frac{8 - x - 2y}{3}\).

Answer:

\(\int_{1}^{4}\int_{0}^{8 - 2y}\int_{0}^{\frac{8 - x - 2y}{3}}f(x,y,z)dzdxdy\)