# Calc 3

Description
Calc 3 Example Problems
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
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

Jul 23, 2017

#### Health Service Systems You Decide Assignment

Jul 23, 2017
Search
Similar documents

View more...
Tags

## Differential Topology

Related Search
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks