Questions: ∭E(x+y-5z) dV where x, y, z -4 ≤ y ≤ 0, 0 ≤ x ≤ y, 0 ≤ z ≤ x+y^2

Solution Steps

To solve the given triple integral, we need to evaluate the integral over the specified region \( E \). The region is defined by the inequalities: \(-4 \leq y \leq 0\), \(0 \leq x \leq y\), and \(0 \leq z \leq x + y^2\). We will integrate the function \(x + y - 5z\) with respect to \(z\), \(x\), and \(y\) in that order.

We need to evaluate the triple integral

\[ \iiint_{E} (x + y - 5z) \, dV \]

over the region defined by the inequalities:

\(-4 \leq y \leq 0\), \(0 \leq x \leq y\), and \(0 \leq z \leq x + y^2\).

The limits of integration for the variables are set as follows:

• For \(z\): from \(0\) to \(x + y^2\)
• For \(x\): from \(0\) to \(y\)
• For \(y\): from \(-4\) to \(0\)

After performing the integration, we find that the value of the integral is

\[ \frac{22528}{15} \]

Final Answer