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Unformatted text preview: 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 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 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 L = Z 2 r ( dx dt ) 2 + ( dy dt ) 2 dt = Z 2 p ( t 2 1) 2 + (2 t ) 2 dt = Z 2 p t 4 2 t 2 + 1 + 4 t 2 dt = Z 2 p ( t 2 + 1) 2 dt = Z 2 ( t 2 + 1) dt = t 3 3 + t 2 = 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 xaxis, 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 ....
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This note was uploaded on 05/03/2010 for the course MAT 218 taught by Professor Robertc.gunning during the Fall '09 term at Princeton.
 Fall '09
 RobertC.Gunning
 Integrals

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