consider the solid that lies below the surface ( z = x^{2}-9y + 18 ) and above the rectangle (…

Explanation:

Step1: Determine the dimensions of sub - rectangles

The rectangle (R=[-2,2]\times[0,2]). The length in the (x) - direction is (2 - (-2)=4), and in the (y) - direction is (2-0 = 2). Since we divide (R) into (4) rectangles of equal size, each twice as wide as high. Let the height of each sub - rectangle in the (y) - direction be (\Delta y) and the width in the (x) - direction be (\Delta x). We have (4) sub - rectangles. Let (m = 2) (number of sub - intervals in (x)) and (n=2) (number of sub - intervals in (y)). Then (\Delta x=\frac{2 - (-2)}{2}=2) and (\Delta y=\frac{2 - 0}{2}=1). The sub - rectangles are (R_{11}=[-2,0]\times[0,1]), (R_{12}=[-2,0]\times[1,2]), (R_{21}=[0,2]\times[0,1]), (R_{22}=[0,2]\times[1,2])

Step2: Find the sample points for the largest Riemann sum (a)

For the function (z = f(x,y)=x^{2}-9y + 18), we want to maximize (f(x,y)) in each sub - rectangle.

• In (R_{11}=[-2,0]\times[0,1]): (f(x,y)) is maximized at ((x_1,y_1)=(- 2,0)) (since (f(x,y)) is a quadratic in (x) and linear in (y), for (x\in[-2,0]) and (y\in[0,1]), (x=-2) and (y = 0) gives (f(-2,0)=(-2)^{2}-9\times0 + 18=4 + 18=22))
• In (R_{12}=[-2,0]\times[1,2]): (f(x,y)) is maximized at ((x_2,y_2)=(-2,1)) ( (f(-2,1)=(-2)^{2}-9\times1 + 18=4-9 + 18=13))
• In (R_{21}=[0,2]\times[0,1]): (f(x,y)) is maximized at ((x_3,y_3)=(2,0)) ((f(2,0)=2^{2}-9\times0+18=4 + 18=22))
• In (R_{22}=[0,2]\times[1,2]): (f(x,y)) is maximized at ((x_4,y_4)=(2,1)) ((f(2,1)=2^{2}-9\times1 + 18=4-9 + 18=13))

The Riemann sum (S=\sum_{i = 1}^{2}\sum_{j = 1}^{2}f(x_i,y_j)\Delta x\Delta y) Since (\Delta x = 2) and (\Delta y=1), (S=(22 + 13+22 + 13)\times2\times1=(70)\times2=140)

Step3: Find the sample points for the smallest Riemann sum (b)

• In (R_{11}=[-2,0]\times[0,1]): (f(x,y)) is minimized at ((x_1,y_1)=(0,1)) ((f(0,1)=0^{2}-9\times1 + 18=9))
• In (R_{12}=[-2,0]\times[1,2]): (f(x,y)) is minimized at ((x_2,y_2)=(0,2)) ((f(0,2)=0^{2}-9\times2+18=0))
• In (R_{21}=[0,2]\times[0,1]): (f(x,y)) is minimized at ((x_3,y_3)=(0,1)) ((f(0,1)=9))
• In (R_{22}=[0,2]\times[1,2]): (f(x,y)) is minimized at ((x_4,y_4)=(0,2)) ((f(0,2)=0))

The Riemann sum (S=\sum_{i = 1}^{2}\sum_{j = 1}^{2}f(x_i,y_j)\Delta x\Delta y=(9 + 0+9 + 0)\times2\times1=(18)\times2 = 36)

Step4: Calculate the exact volume using iterated integrals (c)

The volume (V=\iint_{R}(x^{2}-9y + 18)dA=\int_{x=-2}^{2}\int_{y = 0}^{2}(x^{2}-9y + 18)dydx) First, integrate with respect to (y): (\int_{x=-2}^{2}\left[\int_{y = 0}^{2}(x^{2}-9y + 18)dy\right]dx=\int_{x=-2}^{2}\left[x^{2}y-\frac{9y^{2}}{2}+18y\right]{y = 0}^{2}dx) (=\int{x=-2}^{2}\left(2x^{2}-18 + 36\right)dx=\int_{x=-2}^{2}(2x^{2}+18)dx) Integrate with respect to (x): (\left[\frac{2x^{3}}{3}+18x\right]_{-2}^{2}=\left(\frac{2\times2^{3}}{3}+18\times2\right)-\left(\frac{2\times(-2)^{3}}{3}+18\times(-2)\right)) (=\left(\frac{16}{3}+36\right)-\left(-\frac{16}{3}-36\right)=\frac{32}{3}+72=\frac{32 + 216}{3}=\frac{248}{3}\approx82.67)

Answer:

(a) (140) (b) (36) (c) (\frac{248}{3})