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A contour map is shown for a function f ( x , y ) on the rectangle R = [ −3 , 6 ] × [ −1 , 4 ] .

A contour map is shown with the highest point being about 18 and centered near (4, negative 1). From this point, the values decrease to 16, 14, 12, 10, 8, and 6 roughly every 0.5 to 1 distance. The lowest point is negative four near (negative 3, 4). There is a local minimum of 2 near (negative 1, 0).
  1. Use the midpoint rule with m = 3 and n = 2 to estimate the value of R f ( x , y ) d A .
  2. Estimate the average value of the function f ( x , y ) .

Answers to both parts a. and b. may vary.

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Key concepts

  • We can use a double Riemann sum to approximate the volume of a solid bounded above by a function of two variables over a rectangular region. By taking the limit, this becomes a double integral representing the volume of the solid.
  • Properties of double integral are useful to simplify computation and find bounds on their values.
  • We can use Fubini’s theorem to write and evaluate a double integral as an iterated integral.
  • Double integrals are used to calculate the area of a region, the volume under a surface, and the average value of a function of two variables over a rectangular region.

Key equations

  • Double integral
    R f ( x , y ) d A = lim m , n i = 1 m j = 1 n f ( x i j * , y i j * ) Δ A
  • Iterated integral
    a b c d f ( x , y ) d x d y = a b [ c d f ( x , y ) d y ] d x
    or
    c d b a f ( x , y ) d x d y = c d [ a b f ( x , y ) d x ] d y
  • Average value of a function of two variables
    f ave = 1 Area R R f ( x , y ) d x d y

In the following exercises, use the midpoint rule with m = 4 and n = 2 to estimate the volume of the solid bounded by the surface z = f ( x , y ) , the vertical planes x = 1 , x = 2 , y = 1 , and y = 2 , and the horizontal plane z = 0 .

f ( x , y ) = 4 x + 2 y + 8 x y

27.

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f ( x , y ) = 16 x 2 + y 2

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In the following exercises, estimate the volume of the solid under the surface z = f ( x , y ) and above the rectangular region R by using a Riemann sum with m = n = 2 and the sample points to be the lower left corners of the subrectangles of the partition.

f ( x , y ) = sin x cos y , R = [ 0 , π ] × [ 0 , π ]

0.

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f ( x , y ) = cos x + cos y , R = [ 0 , π ] × [ 0 , π 2 ]

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Use the midpoint rule with m = n = 2 to estimate R f ( x , y ) d A , where the values of the function f on R = [ 8 , 10 ] × [ 9 , 11 ] are given in the following table.

y
x 9 9.5 10 10.5 11
8 9.8 5 6.7 5 5.6
8.5 9.4 4.5 8 5.4 3.4
9 8.7 4.6 6 5.5 3.4
9.5 6.7 6 4.5 5.4 6.7
10 6.8 6.4 5.5 5.7 6.8

21.3.

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The values of the function f on the rectangle R = [ 0 , 2 ] × [ 7 , 9 ] are given in the following table. Estimate the double integral R f ( x , y ) d A by using a Riemann sum with m = n = 2 . Select the sample points to be the upper right corners of the subsquares of R .

y 0 = 7 y 1 = 8 y 2 = 9
x 0 = 0 10.22 10.21 9.85
x 1 = 1 6.73 9.75 9.63
x 2 = 2 5.62 7.83 8.21
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The depth of a children’s 4-ft by 4-ft swimming pool, measured at 1-ft intervals, is given in the following table.

  1. Estimate the volume of water in the swimming pool by using a Riemann sum with m = n = 2 . Select the sample points using the midpoint rule on R = [ 0 , 4 ] × [ 0 , 4 ] .
  2. Find the average depth of the swimming pool.
    y
    x 0 1 2 3 4
    0 1 1.5 2 2.5 3
    1 1 1.5 2 2.5 3
    2 1 1.5 1.5 2.5 3
    3 1 1 1.5 2 2.5
    4 1 1 1 1.5 2

a. 28 ft 3 b. 1.75 ft.

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The depth of a 3-ft by 3-ft hole in the ground, measured at 1-ft intervals, is given in the following table.

  1. Estimate the volume of the hole by using a Riemann sum with m = n = 3 and the sample points to be the upper left corners of the subsquares of R .
  2. Find the average depth of the hole.
    y
    x 0 1 2 3
    0 6 6.5 6.4 6
    1 6.5 7 7.5 6.5
    2 6.5 6.7 6.5 6
    3 6 6.5 5 5.6
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The level curves f ( x , y ) = k of the function f are given in the following graph, where k is a constant.

  1. Apply the midpoint rule with m = n = 2 to estimate the double integral R f ( x , y ) d A , where R = [ 0.2 , 1 ] × [ 0 , 0.8 ] .
  2. Estimate the average value of the function f on R .
    A series of curves marked k = negative 1, negative ½, negative ¼, negative 1/8, 0, 1/8, ¼, ½, and 1. The line marked k = 0 serves as an asymptote along the line y = x. The lines originate at (along the y axis) 1, 0.7, 0.5, 0.38, 0, (along the x axis) 0.38, 0.5, 0.7, and 1, with the further out lines curving less dramatically toward the asymptote.

a. 0.112 b. f ave 0.175 ; here f ( 0.4 , 0.2 ) 0.1 , f ( 0.2 , 0.6 ) −0.2 , f ( 0.8 , 0.2 ) 0.6 , and f ( 0.8 , 0.6 ) 0.2 .

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Practice Key Terms 4

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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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