2. let $\\mathcal{u}$ be the solid tetrahedron bounded by the planes $x = 0, y = 1, z = 0$, and $x + 2y+3z =…

2. 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.

2. 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.

Answer

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)