surface integrals - university of pennsylvaniawziller/math114s14/ch16-5-6.pdf2 2 2 2 2 find the...
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![Page 1: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/1.jpg)
16.5
Surface Integrals
![Page 2: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/2.jpg)
( ) = u v
S S
area S d dudv r r
Parametrization of a surface: , ( , ), ( , ), ( , )u v x u v y u v z u vr
where ( , ) lie in some region in the planeu v uv
Review :
area element of a surface u vd dudv r r
Ex: Parametrization of a sphere :
, sin( )cos( ), sin( )sin( ), cos( ) ,
a
a a a
r
2area element: sin( )d a r r 2( recall volume element is sin( ) )dV
1 2
1 2
1 2
If in the region and
we have , then
with and both traversed counter clockwise
x y
C C
C C
Q P Pdx Qdy Pdx Qdy
C C
Theorem : between
2Area of a sphere of radius is 4a a
![Page 3: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/3.jpg)
2 2 2 2 2 2
Compute the surface area of that portion of the
the sphere 2 cut out by the cone , 0.x y z z x y z
Example :
2Recall: sin( )a r r
2Surface area : 1 sin( ) S S
d d d a d d r r
2 sin( )a d d
2
2
0 /4
sin( )a d d
2
/4 04
2sin( ) 2 2cosd d
22 2 2 4 2 2
2
2 2a
![Page 4: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/4.jpg)
Surface area of the graph of a function: ( , ).z f x y
, , , ( , )u v u v f u vr
1, 0, u ufr
0, 1, v vfr
u v r r 1 0
0 1
u
v
f
f
i j k
u vf f i j k
2 2Surface area element of ( , ) : 1 x yz f x y d f f dxdy
2 2 ( ) 1 x yarea S f f dxdy
![Page 5: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/5.jpg)
2 2 2 2 2
Find the surface area of the portion of the cone
lying above the disc ( 1) 1.z x y x y
Example :
2 2 ( , )z f x y x y
2 2 2 2
1 2
2x
xf x
x y x y
2 2
y
yf
x y
2 22 2
2 2 2 21 1x y
x yf f
x y x y
2 2
2 2
22
x y
x y
2 2 ( ) 1 x yarea S f f dxdy
( ) 2 2 ( ) 2R
area S dxdy area R
2 2 ( ) 1 x yarea S f f dxdy
![Page 6: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/6.jpg)
Surface area of a level surface : ( , , ) cF x y z
assume that we can solve for z, i.e. ( , ) with ( , , ( , )) cz h x y F x y h x y
differentiate implicity with respect to : + 0x z
hx F F
x
= , = yx
x y
z z
FFh h
F F
2 2Surface area element of ( , ) : 1 x yz h x y d h h dxdy
2 2
1 yx
z z
FF
F F
2 2 21= x y z
z z
FF F F
F F
Surface area element of ( , , ) c : z
FF x y z d dxdy
F
( ) zS R
Farea S d dxdy
F
where R is the projection of S onto the xy plane.
(we need to assume S projects uniquely, i.e. no folding)
![Page 7: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/7.jpg)
2 2 2 2F x y z a
2 ,2 ,2F x y z
2 2 24 4 4F x y z 2 2 24 x y z
24 2a a 2F a 2
2z
F a a
F z z
2 2 2 2
22 2 2 2
Intersection between
and
2
x y z a z a h
x y a a h ah h
1
R
a dAz
2 2 2
1
R
a dAa x y
2 2 2 2
Find the surface area of a spherical cap of height h
in a sphere of radius a: , . x y z a a h z a
Example :
( ) zR
Farea S dA
F
R is the projection of the cap into the xy plane.
A disc whose boundary is a circle.
![Page 8: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/8.jpg)
2 2 2
1Surface Area
R
a dAa x y
22 2
2 20 0
ah hr
a drda r
22 2
2 2
00
ah h
a d a r
2 22 2a a ah h a
Surface area of a spherical cap of height h
in a sphere of radius a is equal to 2 ah
2 2 2 R: 2x y ah h
2 22 2a a ah h a
2 a a h a 2 ah
2
(if 2 , we get the whole
sphere with area 2 4 )
h a
ah a
![Page 9: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/9.jpg)
2 2 1 if ( , )
and R is the projection of S onto the xy plane
x y
S R
Gd G f f dxdy z f x y
= where , ( , ), ( , ), ( , )
describes S in parametric form.
u v
S R
Gd G dudv u v x u v y u v z u v r r r
= where ( , , ) c describes S implicitly,
and R is the projection of S onto the xy plane
zS R
FGd G dxdy F x y z
F
General surface integral , , where is a surface in 3-space.S
G x y z d S
![Page 10: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/10.jpg)
2 The surface S: lies in the first quadrant and is bounded by
0, 4, 0, 3. It has mass density equal to the distance to the yz plane.
What is its total mass?
y x
y y z z
Example :
mass density x total mass =S
xd
cannot project into the xy plane (why not?)
2 , 0x zf x f
2instead project into the xz plane (or yz plane): ( , )y f x z x
2 2 21 1 4 x zd f f dxdz x dxdz
2total mass 1 4 S
x x dxdz 3 2
2
0 0
1 4 x x dxdz
3 2
3/22
00
1 21 4
8 3dz x
3/2117 1
4
if 0 y 4, then 0 2 x
![Page 11: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/11.jpg)
Let , , be the velocity field of a fluid.x y zv
Volume of fluid flowing through
an element of surface area per unit time
is approximated by
comp d nv v n
Flux through a surfaceRecall: Outward flux across a
curve C in the plane is C
ds F n
The total volume of fluid flowing through
the surface S per unit time is called the
flux through S
if the vector field is denoted by ,
then flux =
( or other common notation )
S
S
d
dS
F
F n
F n
flux = S
d v n
![Page 12: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/12.jpg)
A surface is if there exists a continuous unit normal vector
function defined at each point on the surface
S orientable
n
orientable
surface with
and n n
upward
orientation
n
downward
orientation
n
non-orientable
surface since
becomes n n
1
To find , define the surface by , , , then S g x y z c gg
n n
In a flux integral, we need to choose a normal S
d F n n.
![Page 13: Surface Integrals - University of Pennsylvaniawziller/math114s14/Ch16-5-6.pdf2 2 2 2 2 Find the surface area of the portion of the cone lying above the disz x y x y dc ( 1) 1. Example:](https://reader033.vdocuments.us/reader033/viewer/2022060915/60a88548cf848539181cdb5d/html5/thumbnails/13.jpg)
2 2 4g x y z
2 ,2 ,1g x y
2 24 4 1g x y
2 2
1 12 ,2 ,1
4 4 1g x y
g x y
n2 20 4
0 2,0 2
z x y
r
2 2
3 3 3 2
0 0
cos sin 4r r rdrd
5 4
2 23 3 2
5 40
0
cos sin 2r rr d
2
3 3325
0
cos sin 4 d
8
2 2 2 21 1
2 2 Let , ,z and S the paraboloid 4 ,0 4.
Compute the flux of through S, where is the upward pointing normal.
x y z x y z Example : F
F n
(upward pointing normal,
positive z coordinate)
2 2
3 3
0 0
why is cos sin 0?d d
2 2
2 2
1 1
2 2
2 ,2 ,1 , ,z
4 4 1
x yx y
x y
F n
3 3
2 2
4 4 1
x y z
x y
F n
2 21 x yd f f dxdy
2 21 4 4x y dxdy
3 32 2
2 2
flux =
14 4 1
R
x y
R
d
x y zf f dxdy
x y
F n
3 3 ( ) R
x y z dxdy
3 3 2 2 ( 4 ) R
x y x y dxdy