Math252CalculusIII:DivergenceandStokesTheorem by: javier
Math 252 Calculus III: Divergence and Stokes Theorem
by: javier
Quick Review the FLUX across a surface ∫
∫ F · dS = S
F · ndS S
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal STEP 2:√ Find the (warped) area dS, recall
dS =
(gx )2 + (gy )2 + 1dA
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal STEP 2:√ Find the (warped) area dS, recall
dS =
(gx )2 + (gy )2 + 1dA
STEP 3: Find the differential FLUX through the surface F · ndS
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal STEP 2:√ Find the (warped) area dS, recall
dS =
(gx )2 + (gy )2 + 1dA
STEP 3: Find the differential FLUX through the surface F · ndS
STEP 4: compute the double integral as usual
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal STEP 2:√ Find the (warped) area dS, recall
dS =
(gx )2 + (gy )2 + 1dA
STEP 3: Find the differential FLUX through the surface F · ndS
STEP 4: compute the double integral as usual
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
Quick Review the FLUX across closed vs open surfaces ∫
∫ F · dS = S
F · ndS S
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS F = ⟨2, 1, 5⟩ and S is the unit cube as shown with outward normals.
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS F = ⟨2, 1, 5⟩ and S is the unit cube as shown with outward normals. STEP 1: Suppose we desire the flux with respect to the outward normal on each side of the surface. Then we must compute 6 different calculations.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS F = ⟨2, 1, 5⟩ and S is the unit cube as shown with outward normals. STEP 1: Suppose we desire the flux with respect to the outward normal on each side of the surface. Then we must compute 6 different calculations.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP 1: We will compute 2 different calculations, the upward normal, top half, then the downward normal, bottom half.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP 1: We will compute 2 different calculations, the upward normal, top half, then the downward normal, bottom half.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP 1: We will compute 2 different calculations, the upward normal, top half, then the downward normal, bottom half.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP 1: We will compute 2 different calculations, the upward normal, top half, then the downward normal, bottom half.
∫
THE FLUX = F · ndS
but wait... for FLUX over CLOSED SURFACES
but wait... for FLUX over CLOSED SURFACES nice things happen when its a closed surface,
but wait... for FLUX over CLOSED SURFACES nice things happen when its a closed surface, so nice are closed surfaces, we call them ”Gauss surfaces”
but wait... for FLUX over CLOSED SURFACES nice things happen when its a closed surface, so nice are closed surfaces, we call them ”Gauss surfaces” and there is this beautiful trick to compute them.. ...
but wait... for FLUX over CLOSED SURFACES nice things happen when its a closed surface, so nice are closed surfaces, we call them ”Gauss surfaces” and there is this beautiful trick to compute them.. ... THE DIVERGENCE THEOREM aka Gauss Theorem
but first... a closer look at the FLUX
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
THE FLUX = 0
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
THE FLUX = 0
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
THE FLUX =0
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
THE FLUX = 0
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
∫
THE FLUX = ∇ · FdA
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary.
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary.
H Si
F · ndS ≈ div(F) · dV
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary.
H Si
F · ndS ≈ div(F) · dV
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary.
H
SF
· ndS =
∫
S div(F)
· dV
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary. GREEN’s THEOREM
H
SF
· ndS =
∫
S div(F)
· dV
back to 3D The Divergence Theorem ∫ ∫ ∫
I F · ndS = S
div(F)dV R
Examples of The Divergence Theorem ∫ ∫ ∫
I F · ndS = S
div(F)dV R
The Divergence Theorem ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1
The Divergence Theorem ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
S F · ndS =
∫∫∫
R div(F)dV
The Divergence Theorem ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
S F · ndS =
∫∫∫
R div(F)dV
The Divergence Theorem Evaluate ∫∫
S ((x
+ y)i + (y + z)j + (z + x)k) · ndS
where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals.
The Divergence Theorem Evaluate ∫∫
S ((x
+ y)i + (y + z)j + (z + x)k) · ndS
where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals. STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
SF
· ndS =
∫∫∫
R div(F)dV
The Divergence Theorem Evaluate ∫∫
S ((x
+ y)i + (y + z)j + (z + x)k) · ndS
where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals. STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
SF
· ndS =
∫∫∫
R div(F)dV
The Divergence Theorem ∫∫
Evaluate S (3xi + 2yj) · ndS where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals.
The Divergence Theorem ∫∫
Evaluate S (3xi + 2yj) · ndS where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals. STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
SF
· ndS =
∫∫∫
R div(F)dV
The Divergence Theorem ∫∫
Evaluate S (3xi + 2yj) · ndS where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals. STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
SF
· ndS =
∫∫∫
R div(F)dV
Finally, back to CLOSED PATH integrals Stoke’s Theorem
Finally, back to CLOSED PATH integrals Stoke’s Theorem recall GREENS THEOREM
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals.
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals. I F · dr = C
∑I
F · dr
Ci
H
CF
· dr ≈ curlk (F) · dA
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals. I F · dr =
∑I
C
I F · dr = C
F · dr
Ci
∑
curlk (F) · dAi
H Ci
F · dr ≈ curlk (F) · dAi
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals. I F · dr =
∑I
C
I
∫ ∫ F · dr =
C
F · dr
Ci
curlk (F) · dA R
H Ci
F · dr ≈ curlk (F) · dAi
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals. I F · dr =
∑I
C
I
∫ ∫ F · dr =
curlk (F) · dA
C
R
GREEN’s THEOREM I
∫ ∫
F · dr = C
F · dr
Ci
(Nx − My )dA R
H Ci
F · dr ≈ curlk (F) · dAi
and now... Stoke’s Theorem
Stoke’s Theorem Greens Theorem
Stokes Theorem
1 20 0
0 5 −1
−4
−2
0 0
2
4
I
−4
0
−2
0
−5
∫ ∫
2
I
F · dr = C
5
−20
−5
∫ ∫ F · dr =
curlk (F)dA S
4
C
curl(F) · ndS S
Stokes Theorem Example I
∫ ∫ F · dr =
C
curl(F) · ndS S
Example
The plane z = x + 4 and the cylinder x2 + y2 = 4 intersect in a curve C. Suppose C is oriented counterclockwise when viewed from above. Evaluate H C F · dr where
F = ⟨x3 + 2y, sin(y) + z, x + sin(z2 )⟩
Example
The plane z = x + 4 and the cylinder x2 + y2 = 4 intersect in a curve C. Suppose C is oriented counterclockwise when viewed from above. Evaluate H F · dr where C
F = ⟨x3 + 2y, sin(y) + z, x + sin(z2 )⟩
STEP H 1: Use Stokes ∫ ∫ Theorem CF
· dr =
S curl(F)
· ndS
Example The plane z = 8 − x2 − y and the cylinder x2 + y2 = 1 intersect in a curve C. Suppose C is oriented Hcounterclockwise when viewed from above. Evaluate C F · dr where
F = ⟨z2 − y2 , −2xy, e
√
z cos(z)⟩
Example
The plane z = 8 − x2 − y and the cylinder x2 + y2 = 1 intersect in a curve C. Suppose C is oriented Hcounterclockwise when viewed from above. Evaluate C F · dr where √ F = ⟨z2 − y2 , −2xy, e
z cos(z)⟩
STEP 1: Use Stokes Theorem
H
CF
· dr =
∫∫
S curl(F)
· ndS compute the curl...
Example
The plane z = 8 − x2 − y and the cylinder x2 + y2 = 1 intersect in a curve C. Suppose C is oriented Hcounterclockwise when viewed from above. Evaluate C F · dr where √ F = ⟨z2 − y2 , −2xy, e
z cos(z)⟩
STEP H 1: Use Stokes ∫ ∫ Theorem CF
· dr =
S curl(F)
· ndS compute the curl...
curl(F) = ⟨0, 2z, 2y − 2y2 ⟩
by: javier
Quick Review the FLUX across a surface ∫
∫ F · dS = S
F · ndS S
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal STEP 2:√ Find the (warped) area dS, recall
dS =
(gx )2 + (gy )2 + 1dA
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal STEP 2:√ Find the (warped) area dS, recall
dS =
(gx )2 + (gy )2 + 1dA
STEP 3: Find the differential FLUX through the surface F · ndS
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal STEP 2:√ Find the (warped) area dS, recall
dS =
(gx )2 + (gy )2 + 1dA
STEP 3: Find the differential FLUX through the surface F · ndS
STEP 4: compute the double integral as usual
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
the FLUX across a surface ∫
Evaluate S F · ndS F = ⟨x4 , 2y2 , z⟩ and S is the half of the surface x2 /4 + y2 /9 + z2 = 1 where z > 0 STEP 1: Find the normal to the surface ±⟨gx , gy , −1⟩, ⟨gx ,gy ,−1⟩ and get unit vector, n = ||⟨g with upward x ,gy ,−1⟩|| normal STEP 2:√ Find the (warped) area dS, recall
dS =
(gx )2 + (gy )2 + 1dA
STEP 3: Find the differential FLUX through the surface F · ndS
STEP 4: compute the double integral as usual
∫
THE FLUX = F · ndS SAGE FLUX HELPER TOOL
Quick Review the FLUX across closed vs open surfaces ∫
∫ F · dS = S
F · ndS S
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS F = ⟨2, 1, 5⟩ and S is the unit cube as shown with outward normals.
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS F = ⟨2, 1, 5⟩ and S is the unit cube as shown with outward normals. STEP 1: Suppose we desire the flux with respect to the outward normal on each side of the surface. Then we must compute 6 different calculations.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS F = ⟨2, 1, 5⟩ and S is the unit cube as shown with outward normals. STEP 1: Suppose we desire the flux with respect to the outward normal on each side of the surface. Then we must compute 6 different calculations.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP 1: We will compute 2 different calculations, the upward normal, top half, then the downward normal, bottom half.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP 1: We will compute 2 different calculations, the upward normal, top half, then the downward normal, bottom half.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP 1: We will compute 2 different calculations, the upward normal, top half, then the downward normal, bottom half.
∫
THE FLUX = F · ndS
the FLUX across closed vs open surfaces ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP 1: We will compute 2 different calculations, the upward normal, top half, then the downward normal, bottom half.
∫
THE FLUX = F · ndS
but wait... for FLUX over CLOSED SURFACES
but wait... for FLUX over CLOSED SURFACES nice things happen when its a closed surface,
but wait... for FLUX over CLOSED SURFACES nice things happen when its a closed surface, so nice are closed surfaces, we call them ”Gauss surfaces”
but wait... for FLUX over CLOSED SURFACES nice things happen when its a closed surface, so nice are closed surfaces, we call them ”Gauss surfaces” and there is this beautiful trick to compute them.. ...
but wait... for FLUX over CLOSED SURFACES nice things happen when its a closed surface, so nice are closed surfaces, we call them ”Gauss surfaces” and there is this beautiful trick to compute them.. ... THE DIVERGENCE THEOREM aka Gauss Theorem
but first... a closer look at the FLUX
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
THE FLUX = 0
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
THE FLUX = 0
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
THE FLUX =0
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
THE FLUX = 0
a closer look at the FLUX For simplicity let us consider a few examples of 2 dimensional cases of flux across a unit square with
outward normals.
∫
THE FLUX = ∇ · FdA
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary.
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary.
H Si
F · ndS ≈ div(F) · dV
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary.
H Si
F · ndS ≈ div(F) · dV
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary.
H
SF
· ndS =
∫
S div(F)
· dV
a closer look at the FLUX
Just as before the flux over small differentials will add up to the flux over entire boundary. GREEN’s THEOREM
H
SF
· ndS =
∫
S div(F)
· dV
back to 3D The Divergence Theorem ∫ ∫ ∫
I F · ndS = S
div(F)dV R
Examples of The Divergence Theorem ∫ ∫ ∫
I F · ndS = S
div(F)dV R
The Divergence Theorem ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1
The Divergence Theorem ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
S F · ndS =
∫∫∫
R div(F)dV
The Divergence Theorem ∫
Evaluate S F · ndS where F = ⟨2, 3, 5⟩ and S is the the surface x2 /4 + y2 /9 + z2 = 1 STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
S F · ndS =
∫∫∫
R div(F)dV
The Divergence Theorem Evaluate ∫∫
S ((x
+ y)i + (y + z)j + (z + x)k) · ndS
where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals.
The Divergence Theorem Evaluate ∫∫
S ((x
+ y)i + (y + z)j + (z + x)k) · ndS
where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals. STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
SF
· ndS =
∫∫∫
R div(F)dV
The Divergence Theorem Evaluate ∫∫
S ((x
+ y)i + (y + z)j + (z + x)k) · ndS
where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals. STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
SF
· ndS =
∫∫∫
R div(F)dV
The Divergence Theorem ∫∫
Evaluate S (3xi + 2yj) · ndS where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals.
The Divergence Theorem ∫∫
Evaluate S (3xi + 2yj) · ndS where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals. STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
SF
· ndS =
∫∫∫
R div(F)dV
The Divergence Theorem ∫∫
Evaluate S (3xi + 2yj) · ndS where S is the the surface x2 + y2 + z2 = 9 oriented with outward normals. STEP theorem H 1: Use The ∫ DIVERGENCE ∫∫ SF
· ndS =
R div(F)dV
THE FLUX =
H
SF
· ndS =
∫∫∫
R div(F)dV
Finally, back to CLOSED PATH integrals Stoke’s Theorem
Finally, back to CLOSED PATH integrals Stoke’s Theorem recall GREENS THEOREM
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals.
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals. I F · dr = C
∑I
F · dr
Ci
H
CF
· dr ≈ curlk (F) · dA
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals. I F · dr =
∑I
C
I F · dr = C
F · dr
Ci
∑
curlk (F) · dAi
H Ci
F · dr ≈ curlk (F) · dAi
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals. I F · dr =
∑I
C
I
∫ ∫ F · dr =
C
F · dr
Ci
curlk (F) · dA R
H Ci
F · dr ≈ curlk (F) · dAi
Stoke’s Theorem H
Suppose F = ⟨M, N⟩. Observe C F · dr can be computed as the sum of many smaller path integrals. I F · dr =
∑I
C
I
∫ ∫ F · dr =
curlk (F) · dA
C
R
GREEN’s THEOREM I
∫ ∫
F · dr = C
F · dr
Ci
(Nx − My )dA R
H Ci
F · dr ≈ curlk (F) · dAi
and now... Stoke’s Theorem
Stoke’s Theorem Greens Theorem
Stokes Theorem
1 20 0
0 5 −1
−4
−2
0 0
2
4
I
−4
0
−2
0
−5
∫ ∫
2
I
F · dr = C
5
−20
−5
∫ ∫ F · dr =
curlk (F)dA S
4
C
curl(F) · ndS S
Stokes Theorem Example I
∫ ∫ F · dr =
C
curl(F) · ndS S
Example
The plane z = x + 4 and the cylinder x2 + y2 = 4 intersect in a curve C. Suppose C is oriented counterclockwise when viewed from above. Evaluate H C F · dr where
F = ⟨x3 + 2y, sin(y) + z, x + sin(z2 )⟩
Example
The plane z = x + 4 and the cylinder x2 + y2 = 4 intersect in a curve C. Suppose C is oriented counterclockwise when viewed from above. Evaluate H F · dr where C
F = ⟨x3 + 2y, sin(y) + z, x + sin(z2 )⟩
STEP H 1: Use Stokes ∫ ∫ Theorem CF
· dr =
S curl(F)
· ndS
Example The plane z = 8 − x2 − y and the cylinder x2 + y2 = 1 intersect in a curve C. Suppose C is oriented Hcounterclockwise when viewed from above. Evaluate C F · dr where
F = ⟨z2 − y2 , −2xy, e
√
z cos(z)⟩
Example
The plane z = 8 − x2 − y and the cylinder x2 + y2 = 1 intersect in a curve C. Suppose C is oriented Hcounterclockwise when viewed from above. Evaluate C F · dr where √ F = ⟨z2 − y2 , −2xy, e
z cos(z)⟩
STEP 1: Use Stokes Theorem
H
CF
· dr =
∫∫
S curl(F)
· ndS compute the curl...
Example
The plane z = 8 − x2 − y and the cylinder x2 + y2 = 1 intersect in a curve C. Suppose C is oriented Hcounterclockwise when viewed from above. Evaluate C F · dr where √ F = ⟨z2 − y2 , −2xy, e
z cos(z)⟩
STEP H 1: Use Stokes ∫ ∫ Theorem CF
· dr =
S curl(F)
· ndS compute the curl...
curl(F) = ⟨0, 2z, 2y − 2y2 ⟩
