Section 3.3: A CALCULUS REVIEW.

This chapter rapidly eviews some ideas from the multi-dimensional calculus. The reader should already be familiar with such concepts as the gradient, divergence, and curl, and if you are comofrtable with these, you may skip to the next section.

We recall the following notational conventions :

grad u = --u = {[[partialdiff]]u/[[partialdiff]]x, [[partialdiff]]u/[[partialdiff]]y},

div v =--*v= [[partialdiff]]v1/[[partialdiff]]x + [[partialdiff]]v2/[[partialdiff]]y,

--²u = --*--u = [[partialdiff]]²u/[[partialdiff]]x² + [[partialdiff]]²u/[[partialdiff]]y² in rectangular coordinates,

--²u = F(1,r) F([[partialdiff]] ,[[partialdiff]]r) (r F([[partialdiff]]u,[[partialdiff]]r)) + F(1,r²) F([[partialdiff]]²u,[[partialdiff]][[Theta]]²⁾ in polar coordinates,

[[partialdiff]]u/[[partialdiff]][[eta]] = < --u, [[eta]] > where [[eta]] is a vector, and

curl F = det B(ACO3(i,j,k, [[partialdiff]]( )/[[partialdiff]]x, [[partialdiff]]( )/[[partialdiff]]y, [[partialdiff]]( )/[[partialdiff]]z, F1, F2, F3))

={ [[partialdiff]]F3/[[partialdiff]]y - [[partialdiff]]F2/[[partialdiff]]z, [[partialdiff]]F1/[[partialdiff]]z - [[partialdiff]]F3/[[partialdiff]]x, [[partialdiff]]F2/[[partialdiff]]x - [[partialdiff]]F1/[[partialdiff]]y}.

There is also information from the integral calculus which we should recall. Some of the important ideas involve methods of calculating surface integrals. Recall that if D is an open, connected region in the plane and f is a function defined on D which has continuous partial derivatives, and if S is the surface which is the graph of f, then we can define a surface integral

òòS H(x,y,z) dA

where H is a continuous function with domain S. This integral over the surface S in R³ can be evaluated by changing it to an integral over the 2-dimensional region D as

òòD H(x,y,f(x,y)) R( [(df/dx)²+(df/dy)² +1] ) dx dy.

For such a surface, a unit normal is given by

[[eta]] = { - [[partialdiff]]f/[[partialdiff]]x, -[[partialdiff]]f/[[partialdiff]]y, 1} / R ([(df/dx)²+(df/dy)² +1] ).

The unit normal to a plane curve described by {x(t), y(t)} is

[[eta]] = {y'(t), -x'(t)} / R([ x'(t)² + y'(t)² ]).

The following are fundamental ideas used in vector analysis.

STOKES'S THEOREM IN 2D. Suppose that D is a region in the plane with a piece-wise smooth boundary and that F(x,y) ={P(x,y),Q(x,y)} has continuous second partial derivatives and is a function from R² to R². Then

ò[[partialdiff]]D Pdx + Qdy = òòD ([[partialdiff]]Q/[[partialdiff]]x - [[partialdiff]]P/[[partialdiff]]y) dx dy.

STOKES'S THEOREM IN 3D. Suppose that D is a smooth surface with unit normal [[eta]] and has a smooth boundary. Suppose, also, that F is a function from R³ to R³ which has continuous second partial derivatives. Then

ò[[partialdiff]]D< F, dR > = òòD< curl F, [[eta]] > dA.

GAUSS'S DIVERGENCE THEOREM. With proper suppositions on D and F

ò[[partialdiff]]D< F, [[eta]] >ds = òòD div F dA = òòD --.F dA.

in two dimensions, while

òò[[partialdiff]]D< F, [[eta]] >dA = òòòDdiv F dV = òòòD --.F dV.

in three dimensions.

REMARKS.

1. Here is a suggestion for the proof of the divergence theorem which uses Stokes's theorem: Take R(t) to be a parameterization of the boundary given by{ [[Phi]](t), [[Psi]](t)}. Then dR(t) = { [[Phi]]'(t), [[Psi]]'(t) } dt and [[eta]] ds = {[[Psi]]'(t), -[[Phi]]'(t)} dt.

Thus,

òòD div F dA = òòD [[[partialdiff]]F1/[[partialdiff]]x + [[partialdiff]]F2/[[partialdiff]]y] dA

⁼ òòD [F([[partialdiff]]F1,[[partialdiff]]x) - F([[partialdiff]]-F2,[[partialdiff]]y)] dx dy.

Recall that Stokes's theorem in 2-D says that this last two dimensional integral can be changed to a line integral:

òòD [F([[partialdiff]]F1,[[partialdiff]]x) - F([[partialdiff]]-F2,[[partialdiff]]y)] dx dy

⁼ ò[[partialdiff]]D [-F2 dx + F1 dy]

= ò[[partialdiff]]D < {F1 , F2}, {[[Psi]]', -[[Phi]]'} > dt

= ò[[partialdiff]]D< F , [[eta]] > ds . [[florin]]

2. The divergence theorem is a generalization of the fundamental theorem of integral calculus in the following sense: Let D be the rectangle (a,b)x(c,d) and F(x,y) ={u(x),0}. Then div F =u'(x) and

(d-c) I( a,b, )u'(x) dx = òòD div F dA.

By the divergence theorem, this latter is

ò[[partialdiff]]D < F , [[eta]] > ds = I(a,b, )< {u, 0}, {0,-1} > dx +I(c,d, )< {u(b),0} , {1,0} > dy

+I(a,b, )< {u,0} , {0,1} > |dx| + I(c,d, )< {u(a) , 0} , {-1,0} > |dy|

= u(b) (d-c) - u(a) (d-c) = [u(b) - u(a)] (d-c).

Thus, in one dimension, the divergence theorem specializes to

I(a,b, u'(x) dx) = u(b) - u(a). [[florin]]

EXERCISE:

1. Suppose that A = B(ACO2(1, 2,2, 3)), B = {5,7}, and C = 11. Show that

--.A--u + B.--u + Cu = F([[partialdiff]]²u,[[partialdiff]]x²) + 4 F([[partialdiff]]²u,[[partialdiff]]x[[partialdiff]]y) + 3 F([[partialdiff]]²u,[[partialdiff]]y^{2) + 5} F([[partialdiff]]u,[[partialdiff]]x) + 7 F([[partialdiff]]u,[[partialdiff]]y) + 11 u.

2.Suppose that L[u] = F([[partialdiff]]²u,[[partialdiff]]x²) + 2 F([[partialdiff]]²u,[[partialdiff]]x[[partialdiff]]y) + 3 F([[partialdiff]]²u,[[partialdiff]]y^{2) + 4} F([[partialdiff]]u,[[partialdiff]]x) + 5 F([[partialdiff]]u,[[partialdiff]]y) + 6 u. Find A, B, and C such that L[u] = --.A--u + B.--u + Cu.

3. Suppose the P(x,y) and Q(x,y) have continuous partial derivatives and that F is defined by F(x,y,z) = {P(x,y), Q(x,y), 0}. Show that

< curl F, {0,0,1)> = ([[partialdiff]]Q/[[partialdiff]]x - [[partialdiff]]P/[[partialdiff]]y).

4. Application of Stokes's theorem in the plane: Integrate < curl F, [[eta]] > over D.

a. F(x,y) = {x,y}, D = D1(0) (= the unit disk).

b. F(x,y) = {-y,x}, D = D1(0),

c. F(x,y) = [3y, 5x}, D = D1(0), ans:2[[pi]]

d. F(x,y) = {0,x²}, D is the rectangle with vertices at {0,0}, {a,0}, {a,b}, and {0,b}. ans: a²b

e. F(x,y) = {3xy + y², 2xy + 5x²}, D= D1({1,2}) (= the disk with radius 1 and center {1,2} ). ans:7[[pi]].

5. Application of Stokes's theorem in 3D: Integrate < curl F, [[eta]] > over D.

a. F(x,y,z) = {x,y,z}, D is the upper half of the unit sphere. ans: 0.

b. F(x,y,z) = {z², 2x, -y³}, D is as above. ans: 2[[pi]].

c. F(x,y,z) = {2z, -y, x}, D is the triangle with vertices at {2,0,0}, {0,2,0}, {0,0,2}. ans: 2.

d. F(x,y,z) = {x⁴, xy, z⁴}, D is the as above. ans: 4/3.

6. Application of the divergence theorem. Verify the divergence theorem on these regions.

a. F(x,y,z) = {x, y, z}, D = the unit sphere. ans: 4[[pi]].

b. F(x,y,z) = {x², y², z²}, D = the unit sphere. ans: 0.

c. F(x,y,z)= {x,y,z}, D = the unit cube in the 1st octant. ans: 3.

d. F(x,y,z) = {x², -xz, z²}, D = the unit cube in the 1st octant ans: 2.