15 Multiple Integrals Copyright Cengage Learning All rights reserved Change of Variables in Double Integrals 15 10 Copyright Cengage Learning All rights reserved Change of Variables in Double Integrals Learning objective simplify double integrals by a change of variables and then evaluate a similar technique for triple integrals will be discussed later 3 Change of Variables in Double Integrals In one dimensional calculus we often use a change of variable a substitution to simplify an integral By reversing the roles of x and u we can write where x g u and a g c b g d Another way of writing Formula 1 is as follows 4 Change of Variables in Double Integrals A change of variables can also be useful in double integrals We have already seen one example of this conversion to polar coordinates The new variables r and are related to the old variables x and y by the equations x r cos y r sin and the change of variables formula can be written as f x y dA f r cos r sin r dr d where S is the region in the r plane that corresponds to the region R in the xy plane 5 Change of Variables in Double Integrals More generally we consider a change of variables that is given by a transformation T from the uv plane to the xy plane T u v x y where x and y are related to u and v by the equations x g u v y h u v or as we sometimes write x x u v y y u v 6 Change of Variables in Double Integrals We usually assume that T is a C1 transformation which means that g and h have continuous first order partial derivatives A transformation T is really just a function whose domain and range are both subsets of If T u1 v1 x1 y1 then the point x1 y1 is called the image of the point u1 v1 If no two points have the same image T is called one to one 7 Change of Variables in Double Integrals Figure 1 shows the effect of a transformation T on a region S in the uv plane T transforms S into a region R in the xy plane called the image of S consisting of the images of all points in S Figure 1 8 Change of Variables in Double Integrals If T is a one to one transformation then it has an inverse transformation T 1 from the xy plane to the uv plane and it may be possible to solve Equations 3 for u and v in terms of x and y u G x y v H x y You have seen the first examples of such transformations on the last worksheet Here is another example 9 Example A transformation is defined by the equations x u2 v2 y 2uv Find the image of the square S u v 0 u 1 0 v 1 Solution The transformation maps the boundary of S into the boundary of the image So we begin by finding the images of the sides of S 10 Example Solution continued The first side S1 is given by v 0 0 u 1 See Figure 2 Figure 2 11 Example Solution continued From the given equations we have x u2 y 0 and so 0 x 1 Thus S1 is mapped into the line segment from 0 0 to 1 0 in the xy plane The second side S2 is u 1 0 v 1 and putting u 1 in the given equations we get x 1 v2 y 2v 12 Example Solution continued Eliminating v we obtain x 1 0 x 1 which is part of a parabola Similarly S3 is given by v 1 0 u 1 whose image is the parabolic arc x 1 1 x 0 13 Example Solution continued Finally S4 is given by u 0 0 v 1 whose image is x v2 y 0 that is 1 x 0 Notice that as we move around the square in the counterclockwise direction we also move around the parabolic region in the counterclockwise direction The image of S is the region R shown in Figure 2 bounded by the x axis and the parabolas given by Equations 4 and 5 Figure 2 14 Change of Variables in Double Integrals Now let s see how a change of variables affects a double integral We start with a small rectangle S in the uv plane whose lower left corner is the point u0 v0 and whose dimensions are u and v See Figure 3 Figure 3 15 Change of Variables in Double Integrals The image of S is a region R in the xy plane one of whose boundary points is x0 y0 T u0 v0 The vector r u v g u v i h u v j is the position vector of the image of the point u v The equation of the lower side of S is v v0 whose image curve is given by the vector function r u v0 16 Change of Variables in Double Integrals The tangent vector at x0 y0 to this image curve is ru gu u0 v0 i hu u0 v0 j Similarly the tangent vector at x0 y0 to the image curve of the left side of S namely u u0 is rv gv u0 v0 i hv u0 v0 j 17 Change of Variables in Double Integrals We can approximate the image region R T S by a parallelogram determined by the secant vectors a r u0 u v0 r u0 v0 b r u0 v0 v r u0 v0 shown in Figure 4 Figure 4 18 Change of Variables in Double Integrals But and so r u0 u v0 r u0 v0 u ru Similarly r u0 v0 v r u0 v0 v rv This means that we can approximate R by a parallelogram determined by the vectors u ru and v rv See Figure 5 Figure 5 19 Change of Variables in Double Integrals Therefore we can approximate the area of R by the area of this parallelogram which is u ru v rv ru rv u v Computing the cross product we obtain 20 Change of Variables in Double Integrals The determinant that arises in this calculation is called the Jacobian of the transformation and is given a special notation With this notation we can use Equation 6 to give an approximation to the area A of R where the Jacobian is evaluated at u0 v0 21 Change of Variables in Double Integrals Next we divide a region S in the uv plane into rectangles Sij and call their images in the xy plane Rij See Figure 6 Figure 6 22 Change of Variables in Double Integrals Applying the approximation 8 to each Rij we approximate the double integral of f over R as follows where the Jacobian is evaluated at ui vj Notice that this double sum is a Riemann sum for the integral 23 Change of Variables in Double Integrals The foregoing argument suggests that the following theorem is …
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