# x4_sols - Fall 2010 Math 251 Exam 4: Solutions Tue, 07/Dec...

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Unformatted text preview: Fall 2010 Math 251 Exam 4: Solutions Tue, 07/Dec c 2010 Art Belmonte 1. Use a line integral to find the area of the lateral surface over the circle C : x 2 + y 2 = 4 in the xy-plane and under the surface z = x 2- y 2 + 4. • Let z = f ( x , y ) = x 2- y 2 + 4. • Parameterize the circle in the conventional manner. g ( t ) = [ 2cos t , 2sin t ] , ≤ t ≤ 2 π • The lateral surface area is A = Z C f ds = Z b a f ( g ( t )) g ( t ) dt = ?. • The composition is f ( g ( t )) = 4cos 2 t- 4sin 2 t + 4 or 8cos 2 t . • The vector derivative is g ( t ) = [- 2sin t , 2cos t ] . • Its magnitude is k g ( t ) k = 2. • Therefore, the area is ? = Z 2 π 16cos 2 t dt = 16 π ≈ 50 . 27 cm 2 . • NOTE: Alternatively compute the area via a surface integral of a scalar field. Let f = 1 and parameterize the surface as s = [ 2cos θ , 2sin θ , z ] , 0 ≤ θ ≤ 2 π , ≤ z ≤ 8cos 2 θ . (See composition above.) Then render the needful with the TAMUCALC command x.sisp (f,s,z,0,8cos ( θ ) 2 , θ , , 2 π ) to see all the steps. 2. Calculate the work done by the force field w = x 2 , y 2 , z 2 on a particle moving along the space curve g ( t ) = 2sin t , 2cos t , 1 2 t 2 , 0 ≤ t ≤ π . • Work is given by the line integral W = Z w · d g = Z b a w ( g ( t )) · g ( t ) dt . • The functional composition is w ( g ( t )) = 4sin 2 t , 4cos 2 t , 1 4 t 4 . • The vector derivative of g is g ( t ) = [ 2cos t ,- 2sin t , t ] . • The dot product is w ( g ( t )) · g ( t ) = 8sin 2 t cos t- 8cos 2 t sin t + 1 4 t 5 ....
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## This note was uploaded on 10/31/2011 for the course CHEN MATH 251 taught by Professor Dr.belmonte during the Fall '11 term at Texas A&M University-Galveston.

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x4_sols - Fall 2010 Math 251 Exam 4: Solutions Tue, 07/Dec...

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