Math Problem Statement

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.

Solution

To find the volume under the surface z=f(x,y)=2x+5y+2z = f(x, y) = 2x + 5y + 2 and above the given rectangular region with boundaries 1x3-1 \leq x \leq 3 and 3y43 \leq y \leq 4, we need to set up and evaluate a double integral.

The volume VV under the surface is given by:

V=y=34x=13(2x+5y+2)dxdyV = \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 xx:

First, compute the integral 13(2x+5y+2)dx\int_{-1}^{3} (2x + 5y + 2) \, dx.

(2x+5y+2)dx=x2+(5y+2)x\int (2x + 5y + 2) \, dx = x^2 + (5y + 2)x

Now evaluate this expression from x=1x = -1 to x=3x = 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)(15y)= (15 + 15y) - (-1 - 5y)

=15+15y+1+5y= 15 + 15y + 1 + 5y

=16+20y= 16 + 20y

So the result of the inner integral is 16+20y16 + 20y.

2. Outer integral with respect to yy:

Now, integrate with respect to yy:

34(16+20y)dy\int_{3}^{4} (16 + 20y) \, dy

=[16y+10y2]34= \left[ 16y + 10y^2 \right]_{3}^{4}

Now evaluate this expression from y=3y = 3 to y=4y = 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] ]

=224138=86= 224 - 138 = 86

Final Answer:

The volume under the surface is 8686 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+1z = 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 0x20 \leq x \leq 2 and 2y52 \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!

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Math Problem Analysis

Mathematical Concepts

Double Integration
Volume under Surface
Multivariable Calculus

Formulas

V = \int_{y=3}^{4} \int_{x=-1}^{3} (2x + 5y + 2) dx dy
Inner integral: \int_{-1}^{3} (2x + 5y + 2) dx
Outer integral: \int_{3}^{4} (16 + 20y) dy

Theorems

Fubini's Theorem

Suitable Grade Level

University Level - Calculus II or Multivariable Calculus