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Vector Calculus Solutions Revised

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Vector calculus
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  Solutions to Vector Calculus Practice Problems 1. Let  R  be the region in  R 2 determined by the inequalities  x 2 +  y 2 ≤  4and  y 2 ≤ x 2 . Evaluate the following integral.   R sin( x 2 +  y 2 ) dA Answer:  The region looks like   xyy  = xy  =—  x We use polar coordinates:    π/ 4 − π/ 4    20 r sin( r 2 ) drdθ  +    5 π/ 43 π/ 4    20 r sin( r 2 ) drdθ =    π/ 4 − π/ 4  − 12 cos(4) + 12   dθ  +    5 π/ 43 π/ 4  − 12 cos(4) + 12   dθ =  − π 2 cos(4) +  π 21  2. Evaluate   R z dV   , where  R  is the following solid region:   xyz (1,  1,  0)(2,  0,  0)(1,  0,  1)(0,  0,  1)(0,  1,  1)(0,  1,  0) Answer:  If we integrate with respect to  x  first, we will not need tobreak the region into more than one piece (that is, we will only needto use one integral). The plane containing the front parallelogram hasequation  x + y + z   = 2. (You should be able to find this by the “guess-and-check” method.) So, the bounds for  x  are 0 ≤ x ≤ 2 − y − z  . Weare integrating over a square in the  yz  -plane, so the  y  and  z   boundsare 0 ≤ y  ≤ 1 and 0 ≤ z   ≤ 1. Thus:   R z dV   =    10    10    2 − y − z 0 z dxdydz   =    10    10  2 z  − yz  − z  2   dydz  =    10  23  −  12 y   dz   = 5123. Let  R  be the region in  R 3 determined by the inequalities  r  ≤  1 and0 ≤ z   ≤ 4 − r 2 , where  r  =   x 2 +  y 2 . Evaluate   R rdV   . Answer:  R  is the region under a paraboloid, inside a cylinder, andabove the  xy -plane. We will use cylindrical coordinates.   R rdV   =    2 π 0    10    4 − r 2 0 r 2 dz drdθ  =    2 π 0    10 r 2  4 − r 2   drdθ =    2 π 0  43 r 3 −  15 r 4  10 =    2 π 0 1715  dθ  = 34 π 152  4. Let  R  be the region in  R 3 defined by 4 ≤ x 2 +  y 2 +  z  2 ≤ 9 and  z   ≥ 0.Evaluate the following integral:   R  x 2 +  y 2   dV  Answer:  R  is the region inside a sphere of radius 3, outside a sphereof radius 2, and above the  xy -plane. We will use spherical coordinates,because the region is very easy to describe in spherical coordinates:2  ≤  ρ  ≤  3, 0  ≤  φ  ≤  π 2, 0  ≤  θ  ≤  2 π . Since  r  =  ρ sin φ , we areintegrating  x 2 +  y 2 =  r 2 =  ρ 2 sin 2 φ . Thus:   R  x 2 +  y 2   dV   =    2 π 0    π/ 20    32  ρ 2 sin 2 φ  ρ 2 sin φdρdφdθ =    2 π 0    π/ 20    32 ρ 4 sin 3 φdρdφdθ =    2 π 0    π/ 20 2115 sin 3 φdρdφdθ = 2115    2 π 0    π/ 20  1 − cos 2 φ  sin φdφdθ We use the substitution  u  = cos φ ,  du  = − sin φdφ :   R  x 2 +  y 2   dV   = 2115    2 π 0    01 (1 − u 2 ) dudθ  = 844 π 155. Let  C   be the curve  x  = 1 − y 2 from (0 , − 1) to (0 , 1). Evaluate   C  y 3 dx  +  x 2 dy Answer:  We parameterize the curve using  t  =  y : x  = 1 − t 2 y  =  t  − 1 ≤ t ≤ 13  Then: dx  =  − 2 tdtdy  =  dt Thus:   C  y 3 dx  +  x 2 dy  =    1 − 1  t 3 ( − 2 t ) +  1 − t 2  2   dt  =    1 − 1  − t 4 − 2 t 2 + 1   dt =  − 15 t 5 −  23 t 3 +  t  1 1 = 4156. Let  C   be the curve  x  = √  t ,  y  = 1 +  t 3 for 0 ≤ t ≤ 1. Evaluate   C  x 3 y 4 dx  +  x 4 y 3 dy Answer:  The rotation of this vector field equals 0:  ∂ ∂x∂ ∂yx 3 y 4 x 4 y 3  = 4 x 3 y 3 − 4 x 3 y 3 = 0This means that the vector field is conservative, so there is some func-tion  f   with ∇ f   =  x 3 y 4 i + x 4 y 3  j . It is fairly easy to see that  f   =  14 x 4 y 4 +  C  works. The endpoints of the curve  C   are (0 , 1) (when  t  = 0) and (1 , 2)(when  t  = 1). Thus:   C  x 3 y 4 dx  +  x 4 y 3 dy  =  14 x 4 y 4  (1 , 2)(0 , 1) = 4 − 0 = 4 Note:  This problem can also be done as a standard vector line integral,but the calculations are somewhat tedious.4
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