Labware - MA35 Multivariable Calculus - Two Variable Calculus

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Volume Under a Function Graph (Page: 1 | 2 )

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For a functions of one variable, a definite integral represents the area underneath its two-dimensional function graph. Likewise, the definite integral of a function of two variables represents the volume underneath its three-dimensional function graph.

Recall that the procedure for evaluating single variable integrals is to
1. approximate the area under the graph using rectangles; and
2. take the limit of the sum of the area of these rectangles as the number of rectangles approaches infinity.

Similarly, the volume underneath the function graph for a double integral can be found by
1. dividing the domain R into rectangles;
2. erecting rectangular prisms over these rectangles using the value of the function graph at the bottom-left vertex of each rectangle as the height;
3. taking the limit of the sum of the prismsí volumes as the number of rectangles in the domain aproaches infinity.

Mathematically, this can be stated \[ \int_R\int f(x,y) dA = \lim_{\|\Delta\| \to 0} \sum_{i=1}^n f(x_i,y_i)\Delta x_i \Delta y_i .\]

Demos

Exercises

  • 1. For which of the following two functions is the lower sum a better approximation for the domain 0 ≤ x ≤ 1, 0 ≤ y ≤ 1? Explain your answer.
    • f(x, y) = x2 + y2
    • f(x, y) = -(x2 + y2)
  • 2. What would happen if there were a large number of subdivisions (approaches infinity) along the x-axis and a small number of subdivisions along the y-axis? How could you go about finding the sum of the volumes of the rectangular prisms? What if instead there were a large number of subdivisions along the y-axis and a small number of subdivisions along the x-axis?