Stokes’ and Gauss’ Theorems Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

Stokes’ and Gauss’ Theorems Math 240 — Calculus III Summer 2013, Session II

Monday, July 8, 2013

Stokes’ and Gauss’ Theorems

Agenda

Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

1. Stokes’ theorem

2. Gauss’ theorem Calculating volume with Gauss’ theorem

Stokes’ and Gauss’ Theorems

Stokes’ theorem

Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

Theorem (Green’s theorem) Let D be a closed, bounded region in R2 with boundary C = ∂D. If F = M i + N j is a C 1 vector field on D then  I ZZ  ∂M ∂N − dx dy. M dx + N dy = ∂x ∂y C D   ∂M k = ∇ × F. − Notice that ∂N ∂x ∂y

Theorem (Stokes’ theorem) Let S be a smooth, bounded, oriented surface in R3 and suppose that ∂S consists of finitely many C 1 simple, closed curves. If F is a C 1 vector field whose domain includes S, then I ZZ ∇ × F · dS. F · ds = ∂S

S

Stokes’ and Gauss’ Theorems

Stokes’ theorem and orientation

Math 240 Stokes’ theorem

Definition

Gauss’ theorem

A smooth, connected surface, S is orientable if a nonzero normal vector can be chosen continuously at each point.

Calculating volume

Examples Orientable planes, spheres, cylinders, most familiar surfaces Nonorientable M¨ obius band To apply Stokes’ theorem, ∂S must be correctly oriented. Right hand rule: thumb points in chosen normal direction, fingers curl in direction of orientation of ∂S. Alternatively, when looking down from the normal direction, ∂S should be oriented so that S is on the left.

Stokes’ and Gauss’ Theorems

Stokes’ theorem

Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

z

Example Let S be the paraboloid z = 9 − x2 − y 2 defined over the disk in the xy-plane with radius 3 (i.e. for z ≥ 0). Verify Stokes’ theorem for the vector field F = (2z − y) i + (x + z) j + (3x − 2y) k. We calculate ∇ × F = − 3 i − j + 2 k and N = 2x i +

n S

x

C= S

Figure 7.31 The paraboloid z = 9 − x 2 − y2 with upward 2yoriented j + k. normal n. Note that the boundary circle C is oriented consistently with S.

Therefore, ZZ ZZ (−6x − 2y + 2) dx dy = 18π. ∇ × F · dS = S

D

y

Stokes’ and Gauss’ Theorems

Stokes’ theorem

Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

z

Example Let S be the paraboloid z = 9 − x2 − y 2 defined over the disk in the xy-plane with radius 3 (i.e. for z ≥ 0). Verify Stokes’ theorem for the vector field

n S

C= S

F = (2z − y) i + (x + z) j + (3x − 2y) k. x ZZ ZZ Figure 7.31 The paraboloid ∇ × F · dS = (−6x − 2y + 2) dx dyz ==9 18π. − x 2 − y2 S

D

oriented with upward

normal n. Note that Using Stokes’ theorem, we can do instead the boundary circle I I C is oriented consistently with S. F · ds = −y dx + x dy ∂S C Z 2π = (−3 sin t)2 + (3 cos t)2 dt = 18π.

0

y

Stokes’ and Gauss’ Theorems

Stokes’ theorem

Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

z

Example Let S be the paraboloid z = 9 − x2 − y 2 defined over the disk in the xy-plane with radius 3 (i.e. for z ≥ 0). Verify Stokes’ theorem for the vector field

n S

C= S

F = (2z − y) i + (x + z) j + (3x − 2y) k. x ZZ ZZ Figure 7.31 The paraboloid ∇ × F · dS = (−6x − 2y + 2) dx dyz ==9 18π. − x 2 − y2 S

D

oriented with upward

n. Note that Applying Stokes’ theorem a second time yields normal the boundary circle ZZ I I Z ZC is oriented S. ∇ × F · dS = F · ds = F · ds = consistently ∇ × Fwith · dS S ∂D D Z Z∂S = 2 dS = 2 (area of D) = 18π.

D

y

Stokes’ and Gauss’ Theorems

Gauss’ theorem

Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

Theorem (Gauss’ theorem, divergence theorem) Let D be a solid region in R3 whose boundary ∂D consists of finitely many smooth, closed, orientable surfaces. Orient these surfaces with the normal pointing away from D. If F is a C 1 vector field whose domain includes D then ZZ ZZZ F · dS = ∇ · F dV. ∂D

D

Stokes’ and Gauss’ Theorems

Gauss’ theorem 494 Chapter 7 Surface

Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

n2

Example Let F be the radial vector field x i + y j + z k and let D the be solid cylinder of radius a and height b with axis on the z-axis and faces at z = 0 and z = b. Let’s verify Gauss’ theorem. Let S1 and S2 be the bottom and top faces, respectively, and let S3 be the lateral face.

z S2

S3 n3 S1 x

y n1

Figure 7.35 The

To orient ∂D for Gauss’ theorem, choose normalssolid cylinder D of Example 3.

n1 = −k for S1 , n2 = k for S2 , and n3 = a1 (x i + y j) for S3 . Now we integrate over the surface ZZ ZZ ZZ F · dS = b dS + a ∂D

S2

S3

dS = 3πa2 b.

Stokes’ and Gauss’ Theorems

Gauss’ theorem 494 Chapter 7 Surface

Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

n2

Example

z S2

Let F be the radial vector field x i + y j + z k and let D the be solid cylinder of radius a and S3 height b with axis on the z-axis and faces at n3 z = 0 and z = b. Let’s verify Gauss’ theorem. y S1 Let S1 and S2 be the bottom and top faces, n 1 x respectively, and let S3 be the lateral face. ZZ ZZ ZZ Figure 7.35 The solid 2cylinder D of F · dS = b dS + a dS = 3πa b. Example 3. ∂D

S2

S3

On the other hand, ∇ · F = 3. Then ZZ ZZZ ZZZ F · dS = ∇ · F dV = 3 dV = 3πa2 b. ∂D

D

D

Stokes’ and Gauss’ Theorems Math 240 Stokes’ theorem Gauss’ theorem Calculating volume

Calculating volume Recall how we used Green’s theorem to calculate the area of a plane region via a line integral around its boundary.

Theorem Suppose D is a solid region in R3 to which Gauss’ theorem applies and F is a C 1 vector field such that ∇ · F is identically 1 on D. Then the volume of D is given by ZZ ∂D

F · dS

where ∂D is oriented as in Gauss’ theorem. Some examples are RR  (x i) · dS   RR∂D (y j) · dS Volume of D =  RR∂D   (z k) · dS ∂D

.

Stokes’ and Gauss’ Theorems

Calculating volume

Math 240

Example Stokes’ theorem Gauss’ theorem

Let’s calculate the volume of a truncated cone via an integral over its surface. Let D be the solid bounded by the cone

Calculating volume

x2 + y 2 = (2 − z)2 and the planes z RR = 1 and z = 0. Let’s use the vector field F = x i, so that S F · dS = 0 when S is the top or bottom face. Then we just need to calculate i j k sin θ −1 = x i + y j + r k N = cos θ −r sin θ r cos θ 0 and the volume of D is ZZ Z (x i) · dS = S

0

2π Z 2 1

(r cos θ)2 dr dθ = 37 π.