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$Use the midpoint rule to estimate- the value of ∬(x^(2)-y^(2))dA where R={(x,y)∣-4 <= x <= 2,0 <= y <= 2} using three rectangles of equal area$

Use the midpoint rule to estimate- the value of $\iint\left(x^{2}-y^{2}\right) d A$ where $R=\{(x, y) \mid-4 \leq x \leq 2,0 \leq y \leq 2\}$ using three rectangles of equal area

Full solution

Q. Use the midpoint rule to estimate- the value of

\iint\left(x^{2}-y^{2}\right) d A

where

R=\{(x, y) \mid-4 \leq x \leq 2,0 \leq y \leq 2\}

using three rectangles of equal area

Determine dimensions and subintervals: Determine the dimensions and subintervals for the rectangles. $\newline$ Since the region R is defined by $-4 \leq x \leq 2$ and $0 \leq y \leq 2$ , the width in x-direction is $6$ and the height in y-direction is $2$ . Dividing the x-interval into three equal parts gives each rectangle a width of $2$ ( $6/3$ ).
Calculate x-coordinates for midpoints: Calculate the x-coordinates for the midpoints of each rectangle. The midpoints in the x-direction for the rectangles are at $x = -3$ , $-1$ , and $1$ . These are calculated by taking the middle of each interval: $[-4, -2]$ , $[-2, 0]$ , and $[0, 2]$ .
Calculate y-coordinate for midpoint: Calculate the y-coordinate for the midpoint of the rectangles. $\newline$ Since the y-interval is from $0$ to $2$ , the midpoint in y-direction is $y = 1$ .
Evaluate function at midpoints: Evaluate the function at each midpoint. $\newline$ Using the function $f(x, y) = x^2 - y^2$ , calculate: $\newline$ For $(-3, 1)$ : $f(-3, 1) = (-3)^2 - 1^2 = 9 - 1 = 8$ $\newline$ For $(-1, 1)$ : $f(-1, 1) = (-1)^2 - 1^2 = 1 - 1 = 0$ $\newline$ For $(1, 1)$ : $f(1, 1) = 1^2 - 1^2 = 1 - 1 = 0$
Apply midpoint rule: Apply the midpoint rule to estimate the integral. $\newline$ The area of each rectangle is $4$ ( $2$ in $x$ -direction times $2$ in $y$ -direction). The integral estimate is: $\newline$ $\iint(x^2 - y^2) dA \approx (8 + 0 + 0) \times 4 = 8 \times 4 = 32$