Math Problem Statement

Find the average value of f(x, y) = x^2y over the rectangle R with vertices (-4,0), (-4,8), (2,8), (2,0).

Solution

To solve the problem of finding the average value of f(x,y)=x2yf(x, y) = x^2 y over the rectangle RR with vertices at (4,0)(-4,0), (4,8)(-4,8), (2,8)(2,8), and (2,0)(2,0), we proceed as follows:

Formula for the Average Value of a Function:

The average value of a function f(x,y)f(x, y) over a region RR is given by: Average Value=1Area of RRf(x,y)dA\text{Average Value} = \frac{1}{\text{Area of } R} \iint_R f(x, y) \, dA

Step 1: Compute the Area of RR

The vertices define a rectangle. The width is 2(4)=62 - (-4) = 6, and the height is 80=88 - 0 = 8. Therefore, the area is: Area of R=6×8=48\text{Area of } R = 6 \times 8 = 48

Step 2: Set Up the Double Integral

The function f(x,y)=x2yf(x, y) = x^2 y will be integrated over RR: Rf(x,y)dA=x=42y=08x2ydydx\iint_R f(x, y) \, dA = \int_{x=-4}^{2} \int_{y=0}^{8} x^2 y \, dy \, dx

Step 3: Solve the Inner Integral

First, integrate x2yx^2 y with respect to yy: y=08x2ydy=x2y=08ydy=x2[y22]08=x2(822022)=x232\int_{y=0}^{8} x^2 y \, dy = x^2 \int_{y=0}^{8} y \, dy = x^2 \left[ \frac{y^2}{2} \right]_{0}^{8} = x^2 \left( \frac{8^2}{2} - \frac{0^2}{2} \right) = x^2 \cdot 32

Thus, the inner integral evaluates to 32x232x^2.

Step 4: Solve the Outer Integral

Now, integrate 32x232x^2 with respect to xx: x=4232x2dx=32x=42x2dx=32[x33]42\int_{x=-4}^{2} 32x^2 \, dx = 32 \int_{x=-4}^{2} x^2 \, dx = 32 \left[ \frac{x^3}{3} \right]_{-4}^{2} =32(233(4)33)=32(83643)=32723=3224=768= 32 \left( \frac{2^3}{3} - \frac{(-4)^3}{3} \right) = 32 \left( \frac{8}{3} - \frac{-64}{3} \right) = 32 \cdot \frac{72}{3} = 32 \cdot 24 = 768

Step 5: Compute the Average Value

Divide the integral by the area: Average Value=Rf(x,y)dAArea of R=76848=16\text{Average Value} = \frac{\iint_R f(x, y) \, dA}{\text{Area of } R} = \frac{768}{48} = 16

Final Answer:

The average value of f(x,y)f(x, y) over RR is: 16\boxed{16}

Would you like a step-by-step explanation of any part?

Here are 5 related questions for further exploration:

  1. How does the formula for the average value of a function change for non-rectangular regions?
  2. Can the average value of a function over a region be negative? Why or why not?
  3. How would the process differ if the function had a more complex form, such as f(x,y)=sin(xy)f(x, y) = \sin(xy)?
  4. What is the significance of the average value of a function in real-world applications?
  5. How do you verify the bounds of integration for more complex regions?

Tip: Double-check the order of integration and bounds when evaluating multiple integrals.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Geometry
Double Integrals

Formulas

Average Value of a Function: (1/Area of R) * ∫∫_R f(x, y) dA
Area of a Rectangle: width * height

Theorems

Fundamental Theorem of Calculus in Multiple Dimensions

Suitable Grade Level

Grades 10-12