Rewrite the following integral using the indicated order of integration and then evaluate the resulting integral.
Integral from 0 to 9 Integral from negative 1 to 0 Integral from 0 to 2 x plus 2 dy dx dz in the order dz dx dy

We are given the triple integral:

\[
\int_0^9 \int_{-1}^0 \int_0^{2x+2} \, dy \, dx \, dz
\]

and asked to rewrite and evaluate it in the order \( dz \, dx \, dy \).

### Step 1: Analyze the original limits

The integral is currently in the order \( dy \, dx \, dz \). The limits for each variable are as follows:
- \( z \in [0, 9] \)
- \( x \in [-1, 0] \)
- \( y \in [0, 2x + 2] \)

### Step 2: Change the order of integration

We need to rewrite the integral in the order \( dz \, dx \, dy \). For that, let's consider the region described by the current limits.

- For \( y \), we have \( 0 \leq y \leq 2x + 2 \), which gives information about how \( y \) depends on \( x \).
- For \( x \), the limits are \( -1 \leq x \leq 0 \).
- \( z \) is independent, and its limits are constant \( z \in [0, 9] \).

### Step 3: Rewrite the integral in the desired order

In this case, since the \( z \) limits are independent of \( x \) and \( y \), they stay the same: \( z \in [0, 9] \).

For \( x \), the limits must now depend on \( y \), based on the relationship \( y = 2x + 2 \). Solving for \( x \) in terms of \( y \):

\[
y = 2x + 2 \implies x = \frac{y - 2}{2}
\]

The bounds for \( y \) are \( 0 \leq y \leq 2 \), corresponding to the original region where \( x \) ranged from \(-1\) to \( 0 \).

Thus, the new order of integration is:
- \( y \in [0, 2] \)
- \( x \in \left[ \frac{y-2}{2}, 0 \right] \)
- \( z \in [0, 9] \)

### Step 4: Evaluate the integral

Now, we rewrite the integral and evaluate it step by step:

\[
\int_0^2 \int_{\frac{y-2}{2}}^0 \int_0^9 \, dz \, dx \, dy
\]

#### Step 4.1: Integrate with respect to \( z \)

The innermost integral with respect to \( z \) is straightforward, as there are no variables of \( z \) in the integrand:

\[
\int_0^9 dz = 9
\]

Now the integral becomes:

\[
9 \int_0^2 \int_{\frac{y-2}{2}}^0 \, dx \, dy
\]

#### Step 4.2: Integrate with respect to \( x \)

The next step is to integrate with respect to \( x \):

\[
\int_{\frac{y-2}{2}}^0 dx = 0 - \frac{y-2}{2} = \frac{2 - y}{2}
\]

Now the integral becomes:

\[
9 \int_0^2 \frac{2 - y}{2} \, dy
\]

#### Step 4.3: Integrate with respect to \( y \)

Finally, integrate with respect to \( y \):

\[
9 \int_0^2 \frac{2 - y}{2} \, dy = 9 \left[ \frac{2y}{2} - \frac{y^2}{4} \right]_0^2
\]

Evaluate the expression at the bounds \( 0 \) and \( 2 \):

\[
= 9 \left[ \left( \frac{4}{2} - \frac{4}{4} \right) - (0 - 0) \right]
= 9 \left[ 2 - 1 \right]
= 9 \times 1 = 9
\]

### Final Answer:

The value of the integral is \( \boxed{9} \).

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How do you handle changing the order of integration for a triple integral?
2. How does the region of integration affect the limits when changing the order?
3. Can this method be generalized to non-rectangular regions?
4. What happens if the bounds of \( z \) depend on \( x \) and \( y \) as well?
5. How would the solution change if the function being integrated was more complex?

**Tip:** When changing the order of integration, always sketch the region to help identify the new limits correctly.

Rewrite the following integral using the indicated order of integration and then evaluate the resulting integral. Integral from 0 to 9 Integral from negative 1 to 0 Integral from 0 to 2 x plus 2 dy dx dz in the order dz dx dy.

Learn how to rewrite and evaluate a triple integral by changing the order of integration. This problem involves multivariable calculus concepts like triple integrals and the application of Fubini's theorem. Suitable for college-level calculus students.

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