EXAMPLE 1 Estimate the volume of the solid that lies above the square R = [0, 2] x [0, 2] and below the elliptical paraboloid z = 27 - x2 - 5y². Divide R into 4 equal squares and choose the sample point to be the upper right corner of each square Rj. Sketch the solid and the approximating rectangular boxes. SOLUTION The squares are shown in the top figure. The paraboloid is the graph of f(x, y) = 27 – x - 5y? and the area of each square is ΔΑ - Approximating the volume of the Riemann sum with m = n = 2, we have 2 i = 1j = 1 f(1, 1)AA + f(1, 2)AA + f(2, 1)AA + f(2, 2)AA = 21 + 6(1) + (1) + 3(1) This is the volume of the approximating rectangular boxes shown in the bottom figure.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.4: Linear Programming
Problem 4E
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EXAMPLE 1
Estimate the volume of the solid that lies above the square
R = [0, 2] x [0, 2] and below the elliptical paraboloid
z = 27 - x2 - 5y². Divide R into 4 equal squares and choose the sample
point to be the upper right corner of each square Rj. Sketch the solid and
the approximating rectangular boxes.
SOLUTION The squares are shown in the top figure. The paraboloid is
the graph of f(x, y) = 27 – x - 5y? and the area of each square is
ΔΑ -
Approximating the volume of the Riemann sum with
m = n = 2, we have
2
i = 1j = 1
f(1, 1)AA + f(1, 2)AA + f(2, 1)AA + f(2, 2)AA
= 21
+ 6(1) +
(1) + 3(1)
This is the volume of the approximating rectangular boxes shown in the
bottom figure.
Transcribed Image Text:EXAMPLE 1 Estimate the volume of the solid that lies above the square R = [0, 2] x [0, 2] and below the elliptical paraboloid z = 27 - x2 - 5y². Divide R into 4 equal squares and choose the sample point to be the upper right corner of each square Rj. Sketch the solid and the approximating rectangular boxes. SOLUTION The squares are shown in the top figure. The paraboloid is the graph of f(x, y) = 27 – x - 5y? and the area of each square is ΔΑ - Approximating the volume of the Riemann sum with m = n = 2, we have 2 i = 1j = 1 f(1, 1)AA + f(1, 2)AA + f(2, 1)AA + f(2, 2)AA = 21 + 6(1) + (1) + 3(1) This is the volume of the approximating rectangular boxes shown in the bottom figure.
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