HW 05 - Mathematics 218: Analysis in Several Variables...

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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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HW 05 - Mathematics 218: Analysis in Several Variables...

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