The Divergence Theorem

The Divergence Theorem MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Summer 2011

Green s Theorem Revisited Green s Theorem: M(x, y) dx + N(x, y) dy = C R ( N x M ) da y y x

Green s Theorem Vector Form (1 of 3) Simple closed curve C is described by the vector-valued function r(t) = x(t), y(t) for a t b. The unit tangent vector and unit (outward) normal vector to C are respectively T(t) = 1 r (t) x (t), y (t) and n(t) = 1 r (t) y (t), x (t).

Green s Theorem Vector Form (2 of 3) If the vector field F(x, y) = M(x, y)i + N(x, y)j, then along the simple closed curve C: F n = M(x(t), y(t)), N(x(t), y(t)) 1 r (t) y (t), x (t) = ( M(x(t), y(t))y (t) N(x(t), y(t))x (t) ) 1 r (t).

Green s Theorem Vector Form (2 of 3) If the vector field F(x, y) = M(x, y)i + N(x, y)j, then along the simple closed curve C: F n = M(x(t), y(t)), N(x(t), y(t)) 1 r (t) y (t), x (t) = ( M(x(t), y(t))y (t) N(x(t), y(t))x (t) ) 1 r (t). Now consider the line integral F n ds. C

Green s Theorem Vector Form (2 of 3) If the vector field F(x, y) = M(x, y)i + N(x, y)j, then along the simple closed curve C: F n = M(x(t), y(t)), N(x(t), y(t)) 1 r (t) y (t), x (t) = ( M(x(t), y(t))y (t) N(x(t), y(t))x (t) ) 1 r (t). Now consider the line integral F n ds. Note: this is a line integral with respect to arc length. C

Green s Theorem Vector Form (3 of 3) C F n ds = = = = = = b a b (F n)(t) r (t) dt ( M(x(t), y(t))y (t) N(x(t), y(t))x (t) ) r (t) a r (t) dt b ( M(x(t), y(t))y (t) N(x(t), y(t))x (t) ) dt a M(x, y) dy N(x, y) dx C ( M R x + N ) da (by Green s Theorem) y F da R

Summary and Objective Green s Theorem in vector form states F n ds = F(x, y) da. C A double integral of the divergence of a two-dimensional vector field over a region R equals a line integral around the closed boundary C of R. R

Summary and Objective Green s Theorem in vector form states F n ds = F(x, y) da. C A double integral of the divergence of a two-dimensional vector field over a region R equals a line integral around the closed boundary C of R. The Divergence Theorem (also called Gauss s Theorem) will extend this result to three-dimensional vector fields. R

Divergence Theorem Remark: the Divergence Theorem equates surface integrals and volume integrals. Theorem (Divergence Theorem) Let R 3 be a region bounded by a closed surface and let n be the unit outward normal to. If F is a vector function that has continuous first partial derivatives in, then F n ds = F dv.

Proof (1 of 7) Suppose F(x, y, z) = M(x, y, z)i + N(x, y, z)j + P(x, y, z)k, then the Divergence Theorem can be stated as F n ds = M(x, y, z)i n ds + N(x, y, z)j n ds + P(x, y, z)k n ds = = M x dv + F(x, y, z) dv. N y dv + P z dv

Proof (2 of 7) Thus the theorem will be proved if we can show that M M(x, y, z)i n ds = x dv N N(x, y, z)j n ds = y dv P P(x, y, z)k n ds = z dv. All of the proofs are similar so we will focus only on the third.

Proof (3 of 7) Suppose region can be described as = {(x, y, z) g(x, y) z h(x, y), for (x, y) R} where R is a region in the xy-plane. Think of as being bounded by three surfaces S 1 (top), S 2 (bottom), and S 3 (side).

Proof (4 of 7) y S 1 : z h x,y z S 2 : z g x,y S 3 x On surface S 3 the the unit outward normal is parallel to the xy-plane and thus P(x, y, z) k }{{} n ds = 0 ds = 0. =0

Proof (5 of 7) Now we calculate the surface integral over S 1. S 1 = {(x, y, z) z h(x, y) = 0, for (x, y) R} Unit outward normal: n = (z h(x, y)) (z h(x, y)) = h x (x, y)i h y (x, y)j + k [ h x (x, y)] 2 + [ h y (x, y)] 2 + 1 and k n = 1 [h x (x, y)] 2 + [h y (x, y)] 2 + 1

Proof (6 of 7) S 1 P(x, y, z)k n ds = = P(x, y, z) ds S 1 [h x (x, y)] 2 + [h y (x, y)] 2 + 1 P(x, y, h(x, y)) da R In a similar way we can show the surface integral over S 2 is P(x, y, z)k n ds = P(x, y, g(x, y)) da. S 2 R

Proof (7 of 7) Finally, = = = = = P(x, y, z)k n ds P(x, y, z)k n ds + P(x, y, z)k n ds S 1 S 2 + P(x, y, z)k n ds S 3 P(x, y, h(x, y)) da P(x, y, g(x, y)) da R R [P(x, y, h(x, y)) P(x, y, g(x, y))] da R P(x, y, z) R h(x,y) R g(x,y) P z z=h(x,y) z=g(x,y) da dz da = P z dv.

Example (1 of 2) Let be the solid unit sphere centered at the origin. Use the Divergence Theorem to calculate the flux of the vector field F(x, y, z) = z, y, x over the surface of the unit sphere.

Example (2 of 2) F(x, y, z) = z, y, x F = 1 S = {(x, y, z) x 2 + y 2 + z 2 = 1} = {(x, y, z) x 2 + y 2 + z 2 1} According to the Divergence Theorem, F n ds = F dv = 1 dv = 4π 3. S

Example (1 of 3) Let be the solid region bounded by the parabolic cylinder z = 1 x 2 and the planes z = 0, y = 0, and y + z = 2. Calculate the flux of the vector field F(x, y, z) = xyi + (y 2 + e xz2 )j + sin(xy)k over the boundary of.

Example (2 of 3) Region : 1 x 1 0 y 2 z 0 z 1 x 2 x 0.0 0.5 1.0 0.5 1.0 1.0 z 0.5 0.0 0.0 0.5 1.0 y 1.5 2.0

Example (3 of 3) F(x, y, z) = xy, y 2 + e xz2, sin(xy) F = 3y S = {(x, y, z) z = 1 x 2, z = 0, y = 0, y + z = 2} = {(x, y, z) 0 z 1 x 2, 0 y 2 z} According to the Divergence Theorem, F n ds = F dv = 3y dv S

Example (3 of 3) F(x, y, z) = xy, y 2 + e xz2, sin(xy) F = 3y S = {(x, y, z) z = 1 x 2, z = 0, y = 0, y + z = 2} = {(x, y, z) 0 z 1 x 2, 0 y 2 z} According to the Divergence Theorem, F n ds = F dv = 3y dv S = 1 1 x 2 2 z 1 0 0 3y dy dz dx

Example (3 of 3) F(x, y, z) = xy, y 2 + e xz2, sin(xy) F = 3y S = {(x, y, z) z = 1 x 2, z = 0, y = 0, y + z = 2} = {(x, y, z) 0 z 1 x 2, 0 y 2 z} According to the Divergence Theorem, F n ds = F dv = S = 1 1 x 2 2 z 1 = 184 35 0 0 3y dv 3y dy dz dx

Identities (1 of 2) Show that S ( F) n ds = 0.

Identities (1 of 2) Show that S ( F) n ds = 0. By the Divergence Theorem ( F) n ds = S = = 0 ( F) dv 0 dv

Identities (2 of 2) Show that S D n f (x, y, z) ds = 2 f (x, y, z) dv.

Identities (2 of 2) Show that S D n f (x, y, z) ds = 2 f (x, y, z) dv. S D n f (x, y, z) ds = = = f (x, y, z) n ds S f (x, y, z) dv 2 f (x, y, z) dv (Divergence Th.)

Average Value of a Function During Calculus I you learned the Integral Mean Value Theorem for a continuous f (x) defined on [a, b] as for some a c b. f (c) = 1 b f (x) dx = f avg, b a a

Average Value of a Function During Calculus I you learned the Integral Mean Value Theorem for a continuous f (x) defined on [a, b] as for some a c b. f (c) = 1 b f (x) dx = f avg, b a a The analogous result for triple integrals is f (A) = 1 f (x, y, z) dv V where A is a point in and V is the volume of region.

Interpretation of Divergence of a Vector Field (1 of 3) [ F] A = 1 F dv V 1 = F n ds V }{{} flux per unit volume

Interpretation of Divergence of a Vector Field (2 of 3) Let P be an arbitrary point in the interior of (not on ) then we may center a sphere a of radius a > 0 at P so that the sphere lies entirely in the interior of.

Interpretation of Divergence of a Vector Field (2 of 3) Let P be an arbitrary point in the interior of (not on ) then we may center a sphere a of radius a > 0 at P so that the sphere lies entirely in the interior of. [ F] A = 1 F n ds V a a 1 = F n ds 4 3 πa3 a lim [ F] a 0 + A = lim 1 F n ds a 0 + 4 3 πa3 a

Interpretation of Divergence of a Vector Field (3 of 3) Conclusion: the divergence of a vector field at a point is the limiting value of the flux per unit volume over a sphere centered at the point as the radius of the sphere approaches zero.

Interpretation of Divergence of a Vector Field (3 of 3) Conclusion: the divergence of a vector field at a point is the limiting value of the flux per unit volume over a sphere centered at the point as the radius of the sphere approaches zero. Suppose F represents the flow of a fluid in three dimensions. If F < 0 then the divergence represents a net loss of fluid (a sink). If F > 0 then the divergence represents a net gain of fluid (a source).

Application (1 of 5) Suppose a point charge q is located at the origin. By Coulomb s Law F(x, y, z) = c q x, y, z x, y, z 3 where c is constant. Show the electric flux of F over any closed surface containing the origin is 4πc q.

Application (2 of 5) F is not continuous on any region containing the origin. Think of as the region between two surfaces: (1) S a a sphere centered at the origin of radius a > 0, and (2) S any surface containing the origin inside it. F dv = F n ds + F n ds S a S

Application (3 of 5) y 2 2 1 x 0 1 2 0 2 4 2 z 0 Sa 2 4 S

Application (4 of 5) F(x, y, z) = c q x, y, z x, y, z 3 F = 0 According to the Divergence Theorem 0 = F dv = F n ds + S a F n ds S = F n ds S a S F n ds

Application (5 of 5) On surface S a the unit outward normal points toward the origin. S a = {(x, y, z) x 2 + y 2 + z 2 = a 2 } n = 1 x, y, z a According to the Divergence Theorem F n ds = F n ds S S a = S a = c q a = c q a = c q a 2 c q x, y, z x, y, z 3 S a 1 ( 1 ) x, y, z ds a x, y, z 3 x, y, z 2 ds x, y, z ds = c q 1 a S a a ds S a 1 S a 1 ds = 4πc q

Homework Read Section 14.7. Exercises: 1 35 odd