HW 06 - Mathematics 218 Analysis in Several Variables...

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Mathematics 218: Analysis in Several Variables Professor Robert Gunning Fall Term 2009 Sixth Problem Set Rik Sengupta November 25, 2009 1 Group 1 Evaluate the following integrals, using a change of variables to simplify the calculation. 1. R E ( x 1 + x 2 ) 2 + ( x 1 - x 2 ) 4 where E is the set - 1 x 1 + x 2 1, 1 x 1 - x 2 3. Solution. We make the obvious substitution u = x 1 + x 2 , and v = x 1 - x 2 . Then we know u v = x 1 + x 2 x 1 - x 2 = 1 1 1 - 1 x 1 x 2 whence dudv = 1 1 1 - 1 dx 1 dx 2 = 2 dx 1 dx 2 . So, we get our desired solution as 1
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Z E ( x 1 + x 2 ) 2 + ( x 1 - x 2 ) 4 = Z 3 1 Z 1 - 1 ( u 2 + v 4 ) 1 2 dudv = 1 2 ( Z 3 1 ( u 3 3 + v 4 u ) 1 - 1 dv ) = 1 2 ( Z 3 1 ( 1 3 + v 4 ) - ( - 1 3 - v 4 ) dv ) = 1 2 Z 3 1 ( 2 3 + 2 v 4 ) dv = Z 3 1 ( 1 3 + v 4 ) dv = v 3 + v 5 5 3 1 = ( 3 3 + 243 5 ) - ( 1 3 + 1 5 ) = 15 + 729 - 5 - 3 15 = 736 15 . So the desired answer is 736 15 . 2. R E ( x 2 1 + x 2 2 ) where E is the set x 1 0, x 2 0, 1 x 1 x 2 3, 1 ( x 2 1 - x 2 2 ) 4. Solution. As before, we make the obvious substitution u = x 1 x 2 , and v = x 2 1 - x 2 2 . Then, we get ( u, v ) ( x 1 , x 2 ) = x 1 u x 2 u x 1 v x 2 v = x 2 x 1 2 x 1 - 2 x 2 = - 2 x 2 2 - 2 x 2 1 = 2( x 2 1 + x 2 2 ) . This immediately gives us dudv = 2( x 2 1 + x 2 2 ) dx 1 dx 2 , and hence the integral becomes 2
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Z E ( x 2 1 + x 2 2 ) = Z x 1 Z x 2 ( x 2 1 + x 2 2 ) dx 1 dx 2 = 1 2 Z x 1 Z x 2 2( x 2 1 + x 2 2 ) dx 1 dx 2 = 1 2 Z u Z v dudv = 1 2 Z 4 1 Z 3 1 dudv = 1 2 Z 4 1 (3 - 1) dv = 1 2 Z 4 1 2 dv = Z 4 1 dv = 4 - 1 = 3 . So our required integral is simply 3. 3. Find the arc length of the plane curve described by the parametrization f ( t ) = ( 1 3 t 3 - t, t 2 ) where 0 t 2. Solution. We get the arc length directly from the formula L = Z r ( dx dt ) 2 + ( dy dt ) 2 dt where the integral ranges over all possible values of t . So we get 3
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L = Z 2 0 r ( dx dt ) 2 + ( dy dt ) 2 dt = Z 2 0 p ( t 2 - 1) 2 + (2 t ) 2 dt = Z 2 0 p t 4 - 2 t 2 + 1 + 4 t 2 dt = Z 2 0 p ( t 2 + 1) 2 dt = Z 2 0 ( t 2 + 1) dt = t 3 3 + t 2 0 = 2 3 3 + 2 = 14 3 . 2 Group 2 4. Evaluate the integral R E (1+ x 2 1 + x 2 2 ) - 2 where E is the triangle with vertices (0 , 0) , (2 , 0) , (1 , 3). Solution. The region E is simple the equilateral triangle with side 2, lying entirely on the first quadrant, with one side lying along the x -axis, and one vertex on the origin. So, if we change to polar coordinates, the angle θ must run from 0 to π 3 . This much is obvious. The problem is to find the limits on r . To this end, we have to find the equation of the line joining the two other vertices (i.e. the ones not on the origin). This line passes through the points (2 , 0) and (1 , 3). Suppose its equation is given by y = mx + c .
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