Rewrite the following integral using the indicated order of integration and then evaluate the resulting integral.

Integral from 0 to 5 Integral from negative 1 to 0 Integral from 0 to 6 x plus 6 dy dx dz∫05∫−10∫06x+6 dy dx dz 

in the order dz dx dy

To rewrite and evaluate the integral in the order \( dz \, dx \, dy \), let’s first analyze the bounds given for the original order \( dy \, dx \, dz \).

### Step 1: Rewrite in terms of \( dz \, dx \, dy \)

The original integral is:

\[
\int_0^5 \int_{-1}^0 \int_0^{6x + 6} \, dy \, dx \, dz
\]

Looking at the inner integral \( \int_0^{6x + 6} \, dy \), we see that this limits \( y \) from \( 0 \) to \( 6x + 6 \).

Now, let’s change the order to \( dz \, dx \, dy \). We need to determine the limits for \( z \), \( x \), and \( y \) based on this new order:

1. **Outer integral in \( y \):** From the original limits of \( x \) (from \( -1 \) to \( 0 \)), we can express \( y \) bounds as they affect \( x \):
   - \( y \) is bounded by \( 0 \leq y \leq 6 \) (since \( y \) can reach a maximum of \( 6 \) when \( x = 0 \)).

2. **Middle integral in \( x \):** For each fixed \( y \), \( x \) is bounded by the range where \( y \leq 6x + 6 \), solving:
   - \( x \) is bounded by \( -1 \leq x \leq \frac{y}{6} - 1 \).

3. **Inner integral in \( z \):** \( z \) remains bounded from \( 0 \) to \( 5 \) in all cases.

Thus, rewriting the integral in the order \( dz \, dx \, dy \), we have:

\[
\int_0^6 \int_{-1}^{\frac{y}{6} - 1} \int_0^5 \, dz \, dx \, dy
\]

### Step 2: Evaluate the resulting integral

Now, we evaluate each integral step-by-step:

1. **Integrate with respect to \( z \):**

   \[
   \int_0^6 \int_{-1}^{\frac{y}{6} - 1} \int_0^5 \, dz \, dx \, dy = \int_0^6 \int_{-1}^{\frac{y}{6} - 1} \left[ z \right]_0^5 \, dx \, dy
   \]

   \[
   = \int_0^6 \int_{-1}^{\frac{y}{6} - 1} 5 \, dx \, dy
   \]

2. **Integrate with respect to \( x \):**

   \[
   = \int_0^6 \left[ 5x \right]_{-1}^{\frac{y}{6} - 1} \, dy
   \]

   \[
   = \int_0^6 \left( 5 \left( \frac{y}{6} - 1 \right) - 5(-1) \right) \, dy
   \]

   \[
   = \int_0^6 \left( \frac{5y}{6} - 5 + 5 \right) \, dy
   \]

   \[
   = \int_0^6 \frac{5y}{6} \, dy
   \]

3. **Integrate with respect to \( y \):**

   \[
   = \frac{5}{6} \int_0^6 y \, dy
   \]

   \[
   = \frac{5}{6} \cdot \frac{y^2}{2} \Big|_0^6
   \]

   \[
   = \frac{5}{6} \cdot \frac{36}{2} = \frac{5}{6} \cdot 18 = 15
   \]

### Final Answer:

\[
\int_0^5 \int_{-1}^0 \int_0^{6x + 6} \, dy \, dx \, dz = 15
\]

Would you like further clarification or steps on this process? 

Here are some related questions:
1. How do you determine the limits of integration when changing the order?
2. What are common methods to simplify nested integrals?
3. How does the geometry of the region affect the integration limits?
4. What is the importance of checking boundaries in multivariable integrals?
5. How can double integrals simplify solving triple integrals in some cases?

**Tip:** Carefully analyze the bounds of each variable when changing the order of integration; sketching the region can help visualize limits.

Rewrite the following integral using the indicated order of integration and then evaluate the resulting integral. Integral from 0 to 5 Integral from negative 1 to 0 Integral from 0 to 6 x plus 6 dy dx dz in the order dz dx dy.

This page provides a detailed solution for rewriting and evaluating a triple integral with changed order of integration. It covers key concepts in multivariable calculus, including how to determine new limits of integration and evaluate step-by-step. This example demonstrates the use of Fubini's theorem and is suitable for undergraduate-level students.

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