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 plane \(y = 1\) and from the vertices, the lower - bound of \(y\) is \(y = 1\) and the upper - bound is \(y = 4\). So, \(1\leq y\leq4\).

Step2: Determine the range of \(x\) in terms of \(y\)

From the plane \(x+2y + 3z=8\) and \(z = 0\), we get \(x=8 - 2y\). Also, \(x\geq0\). So, \(0\leq x\leq8 - 2y\).

Step3: Determine the range of \(z\) in terms of \(x\) and \(y\)

From the plane \(x + 2y+3z=8\), we can solve for \(z\) as \(z=\frac{8 - x - 2y}{3}\). Since \(z\geq0\), \(0\leq z\leq\frac{8 - x - 2y}{3}\).

Step4: Write the iterated integral

The triple - integral \(\iiint_{U}f(x,y,z)dV\) can be written as \(\int_{1}^{4}\int_{0}^{8 - 2y}\int_{0}^{\frac{8 - x - 2y}{3}}f(x,y,z)dzdxdy\).

Answer:

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