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5.1 Double integrals over rectangular regions  (Page 2/12)

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Setting up a double integral and approximating it by double sums

Consider the function z = f ( x , y ) = 3 x 2 − y over the rectangular region R = [ 0 , 2 ] × [ 0 , 2 ] ( [link] ).

  1. Set up a double integral for finding the value of the signed volume of the solid S that lies above R and “under” the graph of f .
  2. Divide R into four squares with m = n = 2 , and choose the sample point as the upper right corner point of each square ( 1 , 1 ) , ( 2 , 1 ) , ( 1 , 2 ) , and ( 2 , 2 ) ( [link] ) to approximate the signed volume of the solid S that lies above R and “under” the graph of f .
  3. Divide R into four squares with m = n = 2 , and choose the sample point as the midpoint of each square: ( 1 / 2 , 1 / 2 ) , ( 3 / 2 , 1 / 2 ) , ( 1 / 2 , 3 / 2 ) , and ( 3 / 2 , 3 / 2 ) to approximate the signed volume.
    In xyz space, there is a surface z = f(x, y) = 3x2 minus y. The corners of the surface are given as (0, 0, 0), (2, 0, 12), (0, 2, negative 2), and (2, 2, 10). The surface is parabolic along the x axis.
    The function z = f ( x , y ) graphed over the rectangular region R = [ 0 , 2 ] × [ 0 , 2 ] .
  1. As we can see, the function z = f ( x , y ) = 3 x 2 − y is above the plane. To find the signed volume of S , we need to divide the region R into small rectangles R i j , each with area Δ A and with sides Δ x and Δ y , and choose ( x i j * , y i j * ) as sample points in each R i j . Hence, a double integral is set up as
    V = ∬ R ( 3 x 2 − y ) d A = lim m , n → ∞ ∑ i = 1 m ∑ j = 1 n [ 3 ( x i j * ) 2 − y i j * ] Δ A .
  2. Approximating the signed volume using a Riemann sum with m = n = 2 we have Δ A = Δ x Δ y = 1 × 1 = 1 . Also, the sample points are (1, 1), (2, 1), (1, 2), and (2, 2) as shown in the following figure.
    On the xy plane, the points (1, 1), (1, 2), (2, 1), and (2, 2) are marked, and these form the upper right corners of four squares marked R11, R12, R21, and R22, respectively. Each square has area Delta A = 1.
    Subrectangles for the rectangular region R = [ 0 , 2 ] × [ 0 , 2 ] .

    Hence,
    V = ∑ i = 1 2 ∑ j = 1 2 f ( x i j * , y i j * ) Δ A = ∑ i = 1 2 ( f ( x i 1 * , y i 1 * ) + f ( x i 2 * , y i 2 * ) ) Δ A = f ( x 11 * , y 11 * ) Δ A + f ( x 21 * , y 21 * ) Δ A + f ( x 12 * , y 12 * ) Δ A + f ( x 22 * , y 22 * ) Δ A = f ( 1 , 1 ) ( 1 ) + f ( 2 , 1 ) ( 1 ) + f ( 1 , 2 ) ( 1 ) + f ( 2 , 2 ) ( 1 ) = ( 3 − 1 ) ( 1 ) + ( 12 − 1 ) ( 1 ) + ( 3 − 2 ) ( 1 ) + ( 12 − 2 ) ( 1 ) = 2 + 11 + 1 + 10 = 24.
  3. Approximating the signed volume using a Riemann sum with m = n = 2 , we have Δ A = Δ x Δ y = 1 × 1 = 1 . In this case the sample points are (1/2, 1/2), (3/2, 1/2), (1/2, 3/2),
    and (3/2, 3/2).
    Hence
    V = ∑ i = 1 2 ∑ j = 1 2 f ( x i j * , y i j * ) Δ A = f ( x 11 * , y 11 * ) Δ A + f ( x 21 * , y 21 * ) Δ A + f ( x 12 * , y 12 * ) Δ A + f ( x 22 * , y 22 * ) Δ A = f ( 1 / 2 , 1 / 2 ) ( 1 ) + f ( 3 / 2 , 1 / 2 ) ( 1 ) + f ( 1 / 2 , 3 / 2 ) ( 1 ) + f ( 3 / 2 , 3 / 2 ) ( 1 ) = ( 3 4 − 1 4 ) ( 1 ) + ( 27 4 − 1 2 ) ( 1 ) + ( 3 4 − 3 2 ) ( 1 ) + ( 27 4 − 3 2 ) ( 1 ) = 2 4 + 25 4 + ( − 3 4 ) + 21 4 = 45 4 = 11.
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Use the same function z = f ( x , y ) = 3 x 2 − y over the rectangular region R = [ 0 , 2 ] × [ 0 , 2 ] .

Divide R into the same four squares with m = n = 2 , and choose the sample points as the upper left corner point of each square ( 0 , 1 ) , ( 1 , 1 ) , ( 0 , 2 ) , and ( 1 , 2 ) ( [link] ) to approximate the signed volume of the solid S that lies above R and “under” the graph of f .

V = ∑ i = 1 2 ∑ j = 1 2 f ( x i j * , y i j * ) Δ A = 0

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Note that we developed the concept of double integral using a rectangular region R . This concept can be extended to any general region. However, when a region is not rectangular, the subrectangles may not all fit perfectly into R , particularly if the base area is curved. We examine this situation in more detail in the next section, where we study regions that are not always rectangular and subrectangles may not fit perfectly in the region R . Also, the heights may not be exact if the surface z = f ( x , y ) is curved. However, the errors on the sides and the height where the pieces may not fit perfectly within the solid S approach 0 as m and n approach infinity. Also, the double integral of the function z = f ( x , y ) exists provided that the function f is not too discontinuous. If the function is bounded and continuous over R except on a finite number of smooth curves, then the double integral exists and we say that f is integrable over R .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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