MAT2122 HW4 sol - MAT2122 Multivariable Calculus(Fall 2016...

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MAT2122 Multivariable Calculus (Fall 2016) Assignment 4 solutions (the total is 85 points) 6.3.14 (9 points). Find the center of mass of the cylinder W : x 2 + y 2 1 , 1 z 2 if the density is δ ( x, y, z ) = ( x 2 + y 2 ) z 2 . Solution: Both W and the density function δ are rotationally symmetric with respect to the z axis, whence x = y = 0, c and it remains to Fnd the z coordinate z = iii W z δ ( x, y, z ) dx dy dz iii W δ ( x, y, z ) dx dy dz c of the center of mass only. In the cylindrical coordinates c δ ( r, θ, z ) = r 2 z 2 , c and with W = { ( r, θ, z ) : 0 r 1 , 0 θ 2 π, 1 z 2 } c the denominator takes the form III W * r 3 z 2 dr dθ dz = bI 1 0 r 3 dr B · bI 2 π 0 B · bI 2 1 z 2 dz B = 1 4 · 2 π · 7 3 = 7 π 6 . c In the same way, the numerator is W * r 3 z 3 dr dθ dz = bI 1 0 r 3 dr B · bI 2 π 0 B · bI 2 1 z 3 dz B = 1 4 · 2 π · 15 4 = 15 π 8 . c ±inally, z = 15 π 8 7 π 6 = 45 28 . c Absence of minor mistakes. c 6.3.16 (8 points). Find the average value of f ( x, y, z ) = e - z over the ball x 2 + y 2 + z 2 1 . Solution: The average value of the function f is f = iii W e z dx dy dz vol W , c where W is the unit ball centered at (0 , 0 , 0). In spherical coordinates c the numerator takes the form W * e r cos φ r 2 sin φ dr dθ dφ , c where W = { ( r, θ, φ ) : 0 r 1 , 0 θ 2 π, 0 φ π } , c so that this integral splits as bII D * e r cos φ r 2 sin φ dr dφ B · bI 2 π 0 B = 2 π II D * e r cos φ r 2 sin φ dr dφ , c where D = { ( r, φ ) : 0 r 1 , 0 φ π } .
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2 Now, ii D * e r cos φ r 2 sin φ dr dφ = i 1 0 i π 0 e r cos φ r 2 sin φ dφ dr = i 1 0 r i π 0 e r cos φ d ( r cos φ ) dr = i 1 0 r i r r e t dt dr = i 1 0 r ( e r e r ) dr = b ( r 1) e r + ( r + 1) e r B v v v r =1 r =0 = 2 e , c so that fnally f = 2 π · 2 e 4 3 π = 3 e . c Absence oF minor mistakes. c 7.1.12b (3 points). Evaluate the integral I of the function f ( x, y, z ) = ( x + y ) / ( y + z ) along the path c : t m→ ( t, 2 3 t 3 / 2 , t ) , t [1 , 2] . Solution: I = i 2 1 f ( x ( t ) , y ( t ) , z ( t )) · b ( x ( t ) , y ( t ) , z ( t )) b dt c = i 2 1 2 + t dt = 2 3 (2 + t ) 3 / 2 v v v t =2 t =1 = 2 3 (8 3 3) . c Absence oF minor mistakes. c 7.2.4a (3 points). Find the line integral I = I c ω , where ω = x dy y dx , and c ( t ) = (cos t, sin t ) for t [0 , 2 π ] Solution: I = i 2 π 0 ω ( c ( t )) · ( x ( t ) , y ( t )) dt c = i 2 π 0 ( sin t, cos t ) · ( sin t, cos t ) dt = 2 π .
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