# Vector Calculus Solutions Revised

Description
Vector calculus
Categories
Published

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
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

Jul 23, 2017

#### Analysis and Design of Inclined Columns

Jul 23, 2017
Search
Similar documents

View more...
Tags

Related Search