# 1431 evaluate rr d x 8 sin x y 9 1 da where d is the

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Chapter 3 / Exercise 27
Intermediate Algebra
Mckeague
Expert Verified
14.31. Evaluate RR D ( x 8 sin x + y 9 + 1 ) dA , where D is the disk x 2 + y 2 4. 115
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Chapter 3 / Exercise 27
Intermediate Algebra
Mckeague
Expert Verified
LECTURE 15 Double Integrals in Polar Coordinates 15.1. The formula T heorem 15.1. Let f ( x , y ) be piecewise continuous in a region D in the (Cartesian) xy-plane and let R be the same region in polar coordinates, that is, R = ( r , θ ) : ( r cos θ , r sin θ ) D . Then ZZ D f ( x , y ) dA = ZZ R f ( r cos θ , r sin θ ) r dA . (15.1) The formal proof of (15.1) lies beyond the scope of these notes, but we should give at least some motivation for this formula. Let a , b , α , β be non-negative numbers such that a < b and 0 α < β 2 π . Consider the case when D is the region bounded by the circles x 2 + y 2 = a 2 and x 2 + y 2 = b 2 and by the half-lines through the origin forming angles α and β with the positive x -axis (see Figure 15.1). Note that D contains the points ( x , y ) in the plane whose polar coordinates ( r , θ ) satisfy the inequalities a r b and α θ β . We recall that the area of D is given by the formula area( D ) = 1 2 ( β - α ) ( b 2 - a 2 ) = ( β - α )( b - a ) b + a 2 . (15.2) For each integer n 1, let r 0 = a , r 1 = a + h , r 2 = a + 2 h , . . . , r n = a + nh = b , h = ( b - a ) / n , be the points that divide the interval [ a , b ] into n subintervals of equal lengths; and let θ 0 = α , θ 1 = α + δ , θ 2 = α + 2 δ , . . . , θ n = α + n δ = β , δ = ( β - α ) / n , θ = α θ = β θ = θ j r = a r = r i r = b ( x * i j , y * i j ) α θ j β a r i b r θ ( r * i , θ * j ) F igure 15.1. The regions D and R , with partitions and sample points 117
be the points that divide the interval [ α , β ] into n subintervals of equal lengths. We now subdivide the region D into n 2 subregions of similar shapes bounded by the circles of radii a , r 1 , r 2 , . . . , r n - 1 , b centered at the origin and the half-lines through the origin forming angles α , θ 1 , θ 2 , . . . , θ n - 1 , β with the positive x -axis (see Figure 15.1). Let D i j denote the subregion bounded by the circles of radii r i - 1 and r i and by the half-lines at angles θ j - 1 and θ j with the positive x -axis: in polar coordinates, D i j is given by the inequalities r i - 1 r r i , θ j - 1 θ θ j . We now pick a sample point ( x * i j , y * i j ) in each subregion D i j : we choose ( x * i j , y * i j ) to be the point in the plane whose polar coordinates are r * i = ( r i - 1 + r i ) / 2 and θ * j = ( θ j - 1 + θ j ) / 2. Using the subregions D i j and the sample points ( x * i j , y * i j ), we form the sum n X i = 1 n X j = 1 f ( x * i j , y * i j ) | D i j | , | D i j | = area( D i j ) . Note that this sum bears a strong resemblance to a Riemann sum for f ( x , y ). Indeed, it can be proved that if f ( x , y ) is continuous, then ZZ D f ( x , y ) dA = lim n →∞ n X i = 1 n X j = 1 f ( x * i j , y * i j ) | D i j | . (15.3) We now express the double sum in (15.3) in terms of the r i ’s and the θ j ’s. By (15.2) with a = r i - 1 , b = r i , α = θ j - 1 and β = θ j , we have | D i j | = ( θ j - θ j - 1 )( r i - r i - 1 ) r i + r i - 1 2 = δ hr * i . Furthermore, since the polar coordinates of ( x * i j , y * i j ) are ( r * i , θ * j ), we have x * i j = r * i cos θ * j , y * i j = r * i sin θ * j , whence f ( x * i j , y * i j ) = f ( r * i cos θ * j , r * i sin θ * j ). Therefore, n X i = 1 n X j = 1 f ( x * i j , y * i j ) | D i j | = n X i = 1 n X j = 1 f ( r * i cos θ * j , r * i sin θ * j ) r * i δ h .