Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 846: 7

Answer

We choose $N=4$ and $M=2$. So, the grid has eight subrectangles. We compute two different Riemann sums to estimate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left( {x + y} \right){\rm{d}}A$ using: 1. lower-left vertices as sample points ${S_{4,2}} = 0.625$ 2. midpoints as sample points ${S_{4,2}} = 1$

Work Step by Step

We have $f\left( {x,y} \right) = x + y$ and ${\cal R} = \left[ {0,1} \right] \times \left[ {0,1} \right]$. We choose $N=4$ and $M=2$. So, the grid has eight subrectangles. Using the regular partition, we get the dimensions of the subrectangles: $\Delta x = \frac{{1 - 0}}{4} = \frac{1}{4}$, ${\ \ \ \ }$ $\Delta y = \frac{{1 - 0}}{2} = \frac{1}{2}$ The Riemann sum ${S_{4,2}}$ to estimate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \left( {x + y} \right){\rm{d}}A$ is given by ${S_{4,2}} = \mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^2 f\left( {{P_{ij}}} \right)\Delta {x_i}\Delta {y_j} = \frac{1}{8}\mathop \sum \limits_{i = 1}^4 \mathop \sum \limits_{j = 1}^2 f\left( {{P_{ij}}} \right)$ Method 1. Use lower-left vertices as sample points ${S_{4,2}} = \frac{1}{8}(f\left( {0,0} \right) + f\left( {\frac{1}{4},0} \right) + f\left( {\frac{1}{2},0} \right) + f\left( {\frac{3}{4},0} \right)$ ${\ \ \ \ \ \ \ \ \ \ \ }$ $ + f\left( {0,\frac{1}{2}} \right) + f\left( {\frac{1}{4},\frac{1}{2}} \right) + f\left( {\frac{1}{2},\frac{1}{2}} \right) + f\left( {\frac{3}{4},\frac{1}{2}} \right))$ $ = \frac{1}{8}\left( {0 + \frac{1}{4} + \frac{1}{2} + \frac{3}{4} + \frac{1}{2} + \frac{3}{4} + 1 + \frac{5}{4}} \right)$ ${S_{4,2}} = 0.625$ Method 2. Use midpoints as sample points ${S_{4,2}} = \frac{1}{8}(f\left( {\frac{1}{8},\frac{1}{4}} \right) + f\left( {\frac{3}{8},\frac{1}{4}} \right) + f\left( {\frac{5}{8},\frac{1}{4}} \right) + f\left( {\frac{7}{8},\frac{1}{4}} \right)$ ${\ \ \ \ \ \ \ \ \ \ \ }$ $ + f\left( {\frac{1}{8},\frac{3}{4}} \right) + f\left( {\frac{3}{8},\frac{3}{4}} \right) + f\left( {\frac{5}{8},\frac{3}{4}} \right) + f\left( {\frac{7}{8},\frac{3}{4}} \right))$ $ = \frac{1}{8}\left( {\frac{3}{8} + \frac{5}{8} + \frac{7}{8} + \frac{9}{8} + \frac{7}{8} + \frac{9}{8} + \frac{{11}}{8} + \frac{{13}}{8}} \right)$ ${S_{4,2}} = 1$
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