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Unformatted text preview: EE 505 C, Autumn, 2010 Vector Calculus 1 1 Calculus Review 1.1 Integration over a plane (double integrals) Integration over a plane is a special case of integrating over a surface, in which the surface is a plane. The integral is integraldisplayintegraldisplay D f ( x , y ) dA where D is some region on the plane. Think of integration over a plane as follows: at each point in the plane, there is a scaling factor f ( x , y ) (well call this the scalar field). The Riemann sum approximation to this integral is integraldisplayintegraldisplay D f ( x , y ) dA i j f ( x i , y j ) A ij where A ij is a small piece of the region D . So, all we are doing is weighting (scaling) the a number A ij (the small piece of area) by the scalar field value at that point, and adding up all of the results. Most of the work in evaluating an integral over a plane is to describe the region D . Example 1.1. Evaluate integraltextintegraltext D f ( x , y ) dA where D is the region bounded by the parabolas y = 2 x 2 and y = 1 + x 2 . The scalar field is f ( x , y ) = x + 2 y . y = 2 x 2 y = 1 + x 2 2 1 1 2 2 x y Figure 1: Area D is bounded by the parabolas y = 2 x 2 and y = 1 + x 2 . The key to evaluating this integral is to describe the region of integration. For dA , we can either use dx dy or dy dx . One will usually be a good choice, while the other will wont. It depends on the region. e Choose the variable of integration for the outer dimension. In this case, x is the smart choice. e Now find the limits of y given x (see dashed line in Figure 1 ). EE 505 C, Autumn, 2010 Vector Calculus 2 y is bounded by the top curve: y 1 + x 2 y is bounded by the bottom curve: y 2 x 2 So, for a particular value of x , 2 x 2 y 1 + x 2 . e Now find the limits of x . We can see from the figure that x must be between the points of intersection of the curves: 1 + x 2 = 2 x 2 x = 1 The integral can now be evaluated: integraldisplayintegraldisplay D f ( x , y ) dA = integraldisplay 1 1 integraldisplay 1 + x 2 2 x 2 f ( x , y ) dy dx = integraldisplay 1 1 integraldisplay 1 + x 2 2 x 2 x + 2 y dy dx = integraldisplay 1 1 bracketleftBig xy + y 2 bracketrightBig y = 1 + x 2 y = 2 x 2 dx = integraldisplay 1 1 x parenleftBig 1 + x 2 parenrightBig + parenleftBig 1 + x 2 parenrightBig 2 x parenleftBig 2 x 2 parenrightBig parenleftBig 2 x 2 parenrightBig 2 dx = integraldisplay 1 1 3 x 4 x 3 + 2 x 2 + x + 1 dx = bracketleftbigg 3 x 5 5 x 4 4 + 2 x 3 3 + x 2 2 + x bracketrightbigg 1 1 = 32 15 Efficiently describing the region of integration is the key to solving these problems. Example 1.2. D is bounded by y = x 1 and y 2 = 2 x + 6. Figure 2 shows the region over which a function (scalar field) is integrated....
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This note was uploaded on 03/01/2012 for the course EE 101 taught by Professor Munk during the Spring '12 term at Alaska Anch.
- Spring '12