HW 06 - 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 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 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 ....
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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.

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

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