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Calc 3

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Calc 3 Example Problems
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  Written Lesson Utkarsh NarayanMarch 19, 2014 For the vector field  F  ( x,y,z  ) to be incompressible, the divergence of   F  ,  ∇· F   = 0. Wehave: ∇· F  ( x,y,z  ) =  ∂f  ( y,z  ) ∂x  +  ∂g ( x,z  ) ∂y  +  ∂h ( x,y ) ∂z   = 0This is because the partials are being taken with respect to the variables the functions donot depend on.For the problem with the paraboloid, let’s picture it first. The paraboloid inside thecylinder will look something like this: Notice that the surfaces intersect when  x  = 9. Let  D be a disk with center at (9 , 0 , 0) with radius 3. Notice that  D  can be seen as the ”shadow”of S on the yz plane. We can calculate: dS   =   1 +  ∂x∂z   2 +  ∂x∂y  2 dA  =   1 + (2 z  ) 2 + (2 y ) 2 dA  =   1 + 4 z  2 + 4 y 2 dA Integrating:    S  dS   =    D   1 + 4 z  2 + 4 y 2 dA  =    2 π 0    30 √  1 + 4 r 2 rdrdθ 1  We converted to polar coordinates here by setting  z   =  r cos( θ ) and  y  =  r sin( θ ). In polarcoordinates,  dA  =  rdrdθ . We integrate through the disk, so the radius goes from 0 to 3.Next, we can substitute  u  = 1 + 4 r 2 and we will get    2 π 0 dθ    371 18 √  udu  = 18 [ θ ] θ =2 πθ =0  2 u 3 / 2 3  u =37 u =1 = 16 π (37 √  37 − 1)(a) Surface integral    S  ( x 2 + y 2 ) dS  . So let’s start by finding −→ r u  and −→ r v .  −→ r u  =  2 v, 2 u, 2 u  .Over here, we’re just taking the partial derivative of each component of   r ( u,v ) with respectto  u . Similarly for  v ,  −→ r v  =   2 u, − 2 v, 2 v  . Let  u  =  r cos( θ ) and  v  =  r sin( θ ). Then we have −→ r u  ×−→ r v  = 4 r 2  sin(2 θ ) ,cos (2 θ ) , − 1   and  |−→ r u  ×−→ r v |  = 4 r 2 √  3. Now we can start doing theintegral:    S  x 2 +  y 2 dS   =    D ( x 2 +  y 2 ) |−→ r u  ×−→ r v | dA =    D r 2 (4 r 2 √  3) rdrdθ = 4 √  3    2 π 0    10 r 5 drdθ = 4 √  3(2 π )  16 r 6  10 = 4 √  3 π 3(b) dS   =   1 +  ∂x∂y  2 +  ∂x∂z   2 dA  = √  1 + 1 + 4 z  2 dA where  dA  =  dydz  . So now we just need to integrate    S   zdS   =   10   10  z  √  2 + 4 z  2 dydz   whichshould be easy to evaluate.For the last problem with the surface integral of the vector field  F   =   5 , 5 , 5  , we firstwant to find the unit normal vector to the surface. Because this is a plane  x + y  + z   = 1, wecan write the unit normal vector to this was  −→ n  =  1 √  3  1 , 1 , 1  . So we get −→ F   · d −→ A  = −→ F   ·−→ n dA  = 1 √  3  5 , 5 , 5 · 1 , 1 , 1  = 1 √  3(5 + 5 + 5) dA  = 15 √  3 dA Then we have    S  −→ F   · d −→ A  = 15 √  3    S  dA  = 15 √  3[Area of the Disk] = 15 √  3 π 3 2 = 45 √  3 π Let me know if you have any questions about any of the above.2
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