MIT18_024S11_soln-pset8

MIT18_024S11_soln-pset8 - Solutions for PSet 8 1. (10.5:11)...

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± Solutions for PSet 8 1. (10.5:11) Parameterize the sides of the square C by maps s i :[0 , 1] R 2 by s 1 ( t )=( 1 t, t ); s 2 ( t t, 1 t ); s 3 ( t t 1 , t ); s 4 ( t t, t 1) . With this parametrization: ² ² 1 ² 1 ² 1 ² 1 d x +d y 1+1 1 1 1 1 = d t + d t + d t + d t | x | + | y | (1 t )+ t t +(1 t ) (1 t t t t ) C 0 0 0 0 The first and the third summands are 0, and the second and fourth terms cancel each other, giving: ² d x y =0 | x | + | y | C 2. (10.9:6) Writing the equation of the cylinder in complete square form: 2 a a ( x ) 2 + y 2 = 2 4 Thus looking from high above the xy -plane the picture looks like: The parametrization of the cylinders’ intersection with the xy -plane is: a a a s ³ ( t )= cos t + , sin t, 0 2 2 2 1
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We need to lift it up to sit on the sphere: )= a a a s ( t cos t + , sin t, z ( t ) 2 2 2 where z ( t ) 0 and ± , a a ± 2 a 2 a a cos t + + sin t ± + z ( t ) 2 = a cos t + ± + z ( t ) 2 = a 2 2 2 2 2 2 This means, that a z ( t 1 2 cos t ² Now ( y 2 ,z 2 ,x 2 ) · d( x, y, z ) C ² 2 π a 3 ³ sin t = sin 2 t, 2(1 cos t ) , (cos t +1) 2 ) ´ · µ sin
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This note was uploaded on 01/18/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

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MIT18_024S11_soln-pset8 - Solutions for PSet 8 1. (10.5:11)...

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