16.5

Surface Integrals Review : Theorem : If in the region between CC12 and we have Q P , then Pdx  Qdy  Pdx  Qdy xy  CC12

with CC12 and both traversed counter clockwise

Parametrization of a surface: r uv,  xuvyuvzuv (,),(,),(,) where (u , v ) lie in some region in the uv plane

area element of a surface d rruv dudv

area( S ) d = rr dudv   uv SS

Ex: Parametrization of a sphere   a : r ,   a sin(  )cos(  ), a sin(  )sin(  ), a cos(  ) ,

2 2 area element: darr   sin( ) ( recall volume element is dV   sin( ) )

Area of a sphere of radius aa is 4 2 Example : Compute the surface area of that portion of the the sphere x2 y 2  z 2  2 cut out by the cone z 2  x 2  y 2 , z  0.

Surface area : 1drr  d  d   a2 sin(  ) d  d     SS

2 Recall: rra sin( )   a2 sin( ) d  d 

2  a2 sin( ) d  d  a2  2 0 /4  2  2sin( )dd    2   2cos     /4 0  4

2 2 2  2   4  2 2  2 Surface area of the graph of a function: z f ( x , y ).

r u, v   u , v , f ( u , v )

ruu 1, 0, f 

rvv 0, 1, f  i j k

rruv10 fu  ffuvi  j  k

01 fv

22 Surface area element of z f ( x , y ) : d  1  fxy  f dxdy

area ( S ) 1  f22  f dxdy  xy Example : Find the surface area of the portion of the cone z2 x 2  y 2 lying above the disc ( x  1) 2  y 2  1.

z f ( x , y )  x22  y

1 x y f y  fxx  2  22 2 x2 y 2 x 2 y 2 xy

22 xy222 xy  11ff22     2 xy x2 y 2 x 2 y 2 xy22

area ( S ) 1  f22  f dxdy  xy

area ( S )  2 dxdy  2  area ( R )  2 R Surface area of a level surface : F ( x , y , z ) c assume that we can solve for z, i.e. z h( x , y ) with F ( x , y , h ( x , y )) c h differentiate implicity with respect to x : F + F  0 xzx Fx Fy hhxy= , = FFzz 22 Surface area element of z h ( x , y ) : d  1  hxy  h dxdy 221 F Fx  Fy  222 1   FFFx  y  z =     FF FFzz    zz F Surface area element of F ( x , y , z ) c : d dxdy Fz

F area ( S ) d  dxdy SRFz where R is the projection of S onto the xy plane. (we need to assume S projects uniquely, i.e. no folding) Example : Find the surface area of a spherical cap of height h in a sphere of radius a: x2 y 2  z 2  a 2 , a  h  z  a . F x2  y 2  z 2  a 2 F2 x ,2 y ,2 z

F 4 x2  4 y 2  4 z 2 4x2  y 2  z 2  F 2aa 42 aa2  Fa2  Fz 2 z z F 1 1 area ( S ) dA  a dA  a dA    2 2 2 Fz z R R R a x y 

R is the projection of the cap into the xy plane. Intersection between A disc whose boundary is a circle. x2 y 2  z 2  a 2 and z  a  h x2 y 2  a 2  a  h2  2 ah  h 2 1 Surface Area  a dA 2 2 2  2 2 2 R: x y  2 ah  h R a x y 

2 22 ah h r  a drd  22 00 ar 2 2ah h2 a d   a22  r  0 0

22 a  a22  ah  h  a  

22 a  a22  ah  h  a 

2 a  a  h  a  2 ah Surface area of a spherical cap of height h (if ha 2 , we get the whole 2 in a sphere of radius a is equal to 2 ah sphere with area 2ah 4 a ) General surface integral G x , y , z d where S is a surface in 3-space. S

Gd  G 1  f22  f dxdy if z  f ( x , y )   xy SR and R is the projection of S onto the xy plane

Gd = Gr r dudv where r u , v x ( u , v ), y ( u , v ), z ( u , v )   uv   SR describes S in parametric form.

F Gd =  G dxdy where F ( x , y , z ) c describes S implicitly, SRFz and R is the projection of S onto the xy plane Example : The surface S: yx 2 lies in the first quadrant and is bounded by y0, y  4, z  0, z  3. It has mass density equal to the distance to the yz plane. What is its total mass?

mass density   x total mass = xd S cannot project into the xy plane (why not?)

instead project into the xz plane (or yz plane): y f ( x , z ) x2

f2 x , f 0 2 2 2 xz d 1  fxz  f dxdz  1  4 x dxdz

total mass  x 1 4 x2 dxdz if 0 y  4, then 0 x  2 S 32 x1 4 x2 dxdz 00 3 2 12 3/2 1 3/2  dz 14  x2  17 1 0 0 83 4 Recall: Outward flux across a Flux through a surface curve C in the plane is  Fn ds C Let v x , y , z be the velocity field of a fluid. Volume of fluid flowing through an element of surface area  per unit time is approximated by

compn v  v  nd

The total volume of fluid flowing through the surface S per unit time is called the flux through S flux =  vn  d S

if the vector field is denoted by F , then flux =  Fn  d S ( or other common notation  Fn  dS ) S In a flux integral,  F nd we need to choose a normal n. S A surface S is orientable if there exists a continuous unit normal vector function n defined at each point on the surface

orientable upward downward non-orientable surface with orientation orientation surface since nn and  n  n nn becomes 

1 To find nn , define the surface S by g x , y , z  c , then   g g 11 Example : Let F  x2 , y 2 ,z  and S the paraboloid z  4  x 2  y 2 ,0  z  4. 22 Compute the flux of Fn through S, where is the upward pointing normal. flux = Fn  d 22  g x  y  z  4 R 33 (upward pointing normal, x y z 22  1 fxy  f dxdy g2 x ,2 y ,1 positive z coordinate)  22 R 4xy 4 1 33 22  (x  y  z ) dxdy g 4 x  4 y  1 R 3 3 2 2 11  (x  y  4  x  y ) dxdy n  g  2 x ,2 y ,1 R  22 22 g 4xy 4 1 z04  x  y  0 r  2,0   2 1122 2xy ,2 ,1 22 Fn  xy , ,z   3 3 3 2 22 4xy22 4 1 rcos  sin   4  r rdrd  00 x33 y z 2 2 Fn 54 22 rr3 3 2 54cos  sin   2rd   4xy 4 1    0 0 22 2 d 1  fxy  f dxdy 32 cos33  sin   4 d    5     8 0 22 22 33 1  4x  4 y dxdy why is  cosdd  sin   0? 00