Rewrite the following integral using the indicated order of integration and then evaluate the resulting integral.
Integral from 0 to 3 Integral from negative 1 to 0 Integral from 0 to 2 x plus 2 dy dx dz
 
in the order dz dx dy
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Part 1
Integral from 0 to 3 Integral from negative 1 to 0 Integral from 0 to 2 x plus 2 dy dx dz
equalsIntegral from 0 to nothing Integral from nothing to nothing Integral from nothing to nothing dz dx dy

We are asked to rewrite and evaluate the triple integral given in the order \( dz \, dx \, dy \). The original integral is:

\[
\int_0^3 \int_{-1}^0 \int_0^{2x+2} \, dy \, dx \, dz
\]

### Step 1: Rewriting the integral in the order \( dz \, dx \, dy \)

The given limits are as follows:
- For \( z \), the limits are from 0 to 3.
- For \( x \), the limits are from -1 to 0.
- For \( y \), the limits are from \( 0 \) to \( 2x + 2 \).

Now, we switch the order of integration to \( dz \, dx \, dy \).

1. For \( y \), the limits are from \( y = 0 \) to \( y = 2x + 2 \).
2. For \( x \), we need to express the range of \( x \) in terms of \( y \). From \( y = 2x + 2 \), we can solve for \( x \):
   \[
   x = \frac{y - 2}{2}
   \]
   So, for \( x \), the limits are from \( x = -1 \) to \( x = \frac{y-2}{2} \).
3. \( z \) remains between 0 and 3.

Thus, the integral becomes:

\[
\int_0^2 \int_{-1}^{\frac{y-2}{2}} \int_0^3 dz \, dx \, dy
\]

### Step 2: Evaluating the integral

We can now evaluate the integral step-by-step.

#### 1. Evaluate the innermost integral with respect to \( z \):
\[
\int_0^3 dz = 3
\]

So the integral becomes:
\[
3 \int_0^2 \int_{-1}^{\frac{y-2}{2}} dx \, dy
\]

#### 2. Evaluate the second integral with respect to \( x \):
\[
\int_{-1}^{\frac{y-2}{2}} dx = \left( \frac{y - 2}{2} - (-1) \right) = \frac{y - 2}{2} + 1 = \frac{y + 2}{2}
\]

Thus, the integral now becomes:
\[
3 \int_0^2 \frac{y + 2}{2} \, dy
\]

#### 3. Evaluate the outermost integral with respect to \( y \):
We split it into two parts:
\[
\int_0^2 \frac{y + 2}{2} \, dy = \frac{1}{2} \int_0^2 (y + 2) \, dy = \frac{1}{2} \left( \int_0^2 y \, dy + \int_0^2 2 \, dy \right)
\]

- First, evaluate \( \int_0^2 y \, dy \):
\[
\int_0^2 y \, dy = \left[ \frac{y^2}{2} \right]_0^2 = \frac{4}{2} = 2
\]

- Now, evaluate \( \int_0^2 2 \, dy \):
\[
\int_0^2 2 \, dy = 2 \times (2 - 0) = 4
\]

So:
\[
\frac{1}{2} \left( 2 + 4 \right) = \frac{1}{2} \times 6 = 3
\]

#### Final result:
Now multiply by the constant 3:
\[
3 \times 3 = 9
\]

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

---

Would you like more details on any part of the process, or further clarification?

### 5 related questions:
1. How would you solve this integral if the limits were different?
2. What changes if we swap the order to \( dx \, dy \, dz \)?
3. Can you explain how to compute a double integral using iterated integrals?
4. How do you find the bounds of integration when changing the order of integration?
5. What is the geometric interpretation of a triple integral?

### Tip:
When changing the order of integration, always carefully rewrite the bounds in terms of the new variables to ensure correctness.

Rewrite the following integral using the indicated order of integration and then evaluate the resulting integral. Integral from 0 to 3 Integral from negative 1 to 0 Integral from 0 to 2x + 2 dy dx dz in the order dz dx dy.

Explore how to rewrite and solve a triple integral with limits from 0 to 3, -1 to 0, and 0 to 2x + 2 by changing the order of integration to dz dx dy. This guide covers changing the bounds, applying Fubini's Theorem, and evaluating step-by-step integrals. Suitable for undergraduate students studying Calculus.

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