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Unformatted text preview: Mathematics 218: Analysis in Several Variables Professor Robert Gunning Fall Term 2009 Fifth Problem Set Rik Sengupta November 18, 2009 Evaluate the following integrals: 1 Group 1 1. R E x 2 1 where E is the set x 1 0, 1 x 2 1 + x 2 2 2. Solution. 3 8 . Consider E , the annulus around the origin whose inner radius is 1 and outer radius is 2. Now, E is given by the equation E = { ( x 1 ,x 2 ) : 1 x 2 1 + x 2 2 2 } . So we immediately see that E is given simply by the right half of this annulus. In fact, by symmetry, R E x 2 1 = 1 2 R E x 2 1 . So we will calculate R E x 2 1 . But in order to do that, we introduce a coor dinate change to polar coordinates. Note that the change in coordinates necessitates dx 1 dx 2 = rdrd , where r goes from 1 to 2 for the annulus, and x 1 = r cos , so that the integral simply becomes 1 Z E x 2 1 = Z 2 Z 2 1 r 2 cos 2 rdrd = Z 2 cos 2 d [ r 4 4 ]  2 1 = Z 2 cos 2 d 3 4 = 3 4 Z 2 1 + cos2 2 d = 3 8 [ Z 2 d + Z 2 cos2 d ] = 3 8 [2 + 1 2 Z 4 cos xdx ] = 3 8 [2 + 1 2 (sin x )  4 ] = 3 8 2 = 3 4 . So, from our discussion above, we get our required integral as 1 2 3 4 = 3 8 . 2. R E x 2 1 x 2 , where E is the region between the parabola x 1 = x 2 2 and the line x 1 = 2 x 2 . Solution. Using Fubinis Theorem, we can write this integral as an iter ated integral. We easily get Z E x 2 1 x 2 = Z 2 Z 2 x 2 x 2 2 ( x 2 1 x 2 ) dx 1 dx 2 = Z 2 ( 8 3 x 3 2 2 x 3 2 2 1 3 x 6 2 + x 5 2 2 ) dx 2 = 32 7 32 35 2 . 3. R E x 1 e x 2 , where E R 2 is the triangle with vertices (0 , 0) , (0 , 2) , (2 , 2). Solution. We may use Fubinis Theorem on this as well. We get 2 Z E x 1 e x 2 = Z 2 Z 2 x 1 x 1 e x 2 dx 2 dx 1 = Z 2 ( x 1 e 2 x 1 e x 1 ) dx 1 = e 2 1 . 4. R E x 2 e x 1 x 2 where E R 2 is the set defined by the conditions 1 x 1 2, x 2 1, x 1 x 2 1. Solution. We use Fubinis Theorem on this, once again. We readily obtain Z E x 2 e x 1 x 2 = Z 1 1 2 Z 2 1 x 2 x 2 e x 1 x 2 dx 1 dx 2 = Z 1 1 2 ( e 2 x 2 e ) dx 2 = [ 1 2 e 2 x 2 x 2 e ]  1 1 2 = 1 2 e ( e 2) . 5. R R 2 x 1 x 2 e x 2 1 x 2 2 , and show that this integral converges. Solution. We know the integral is independent of the choice of approx imating the Jordan domain, from page 62 of the notes. Therefore, we choose according to our convenience, to simplify calculations to a huge ex tent. Let R be the cell with vertices ( R,R ) , ( R,R...
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 '09
 RobertC.Gunning
 Integrals

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