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To find the volume under the surface \( z = f(x, y) = 2x + 5y + 2 \) and above the given rectangular region with boundaries \( -1 \leq x \leq 3 \) and \( 3 \leq y \leq 4 \), we need to set up and evaluate a double integral.

The volume \( V \) under the surface is given by:

\[
V = \int_{y=3}^{4} \int_{x=-1}^{3} (2x + 5y + 2) \, dx \, dy
\]

Let's break this down step-by-step:

### 1. Inner integral with respect to \( x \):
First, compute the integral \( \int_{-1}^{3} (2x + 5y + 2) \, dx \).

\[
\int (2x + 5y + 2) \, dx = x^2 + (5y + 2)x
\]

Now evaluate this expression from \( x = -1 \) to \( x = 3 \):

\[
\left[ (3)^2 + (5y + 2)(3) \right] - \left[ (-1)^2 + (5y + 2)(-1) \right]
\]

\[
= \left[ 9 + 3(5y + 2) \right] - \left[ 1 - (5y + 2) \right]
\]

\[
= \left[ 9 + 15y + 6 \right] - \left[ 1 - 5y - 2 \right]
\]

\[
= (15 + 15y) - (-1 - 5y)
\]

\[
= 15 + 15y + 1 + 5y
\]

\[
= 16 + 20y
\]

So the result of the inner integral is \( 16 + 20y \).

### 2. Outer integral with respect to \( y \):
Now, integrate with respect to \( y \):

\[
\int_{3}^{4} (16 + 20y) \, dy
\]

\[
= \left[ 16y + 10y^2 \right]_{3}^{4}
\]

Now evaluate this expression from \( y = 3 \) to \( y = 4 \):

\[
= \left[ 16(4) + 10(4)^2 \right] - \left[ 16(3) + 10(3)^2 \right]
\]

\[
= \left[ 64 + 10(16) \right] - \left[ 48 + 10(9) \right]
\]

\[
= \left[ 64 + 160 \right] - \left[ 48 + 90 \right]
\]

\[
= 224 - 138 = 86
\]

### Final Answer:
The volume under the surface is \( 86 \) cubic units.

Would you like a further breakdown of any part of the solution?

Here are 5 related questions for deeper understanding:
1. How would the volume change if the surface equation were different, for example, \( z = 3x + 4y + 1 \)?
2. What is the geometric meaning of evaluating this double integral?
3. How can this problem be solved using numerical methods instead of symbolic integration?
4. What if the boundaries were shifted, such as \( 0 \leq x \leq 2 \) and \( 2 \leq y \leq 5 \)?
5. How can the concept of double integrals be applied in real-world problems?

**Tip:** For complicated regions or functions, breaking down double integrals into smaller steps often makes the computation easier!

Find the volume under the given surface z = f(x, y) and above the rectangle with boundaries z = 2x + 5y + 2, -1 ≤ x ≤ 3, 3 ≤ y ≤ 4.

Explore how to find the volume under the surface z = 2x + 5y + 2 over the rectangle defined by the boundaries -1 ≤ x ≤ 3 and 3 ≤ y ≤ 4. This problem utilizes double integrals and covers key multivariable calculus concepts. Suitable for university-level calculus students.

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