vector_calculus

# vector_calculus - EE 505 C Autumn 2010 Vector Calculus 1 1...

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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 ) (we’ll 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 won’t. It depends on the region. ØeÔ Choose the variable of integration for the outer dimension. In this case, x is the smart choice. ØeÔ 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 . ØeÔ 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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vector_calculus - EE 505 C Autumn 2010 Vector Calculus 1 1...

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