Another example of Jacobian

1. Draw the integral area.
2. Change the variables from (x, y) to (u, v) by u = x, v = x + 2y.
   

   d x d y = ⎢ ⎢ ∂(x, y) ∂(u, v) ⎢ ⎢ d u d v.

   (2)

Implicit functions

Example

Application:Lagrange multiplier

Example 1

Example 2

x=1; y=1; lambda=1;

for i=0:10
 eqs=[ 2*(x-1)-3*x*x*lambda ; 2*y+4 *y *lambda ; -x^3+2*y*y];

 jacob=[2-6*x*lambda, 0, -3*x*x; 0, 2+4*lambda, 4*y; -3*x*x, 4*y, 0];

 right=inv(jacob)*eqs;

 x=x-right(1);
 y=y-right(2);
 lambda=lambda-right(3);
end;

fprintf('%f %f %f\n', x, y,lambda);
fprintf('%f', sqrt((x-1)^2+y^2));

Vector Analysis

Scalar product (inner product, dot product)

1. a ·b = b ·a
2. c a ·b = a ·c b = c (a ·b)
3. a ·(b + c ) = a ·b + a ·c

Vector product (cross product)

1. a ×b = −b ×a
2. c a ×b = a ×c b = c (a ×b)
3. a ×(b + c ) = a ×b + a ×c

Equation of plane (application of dot product)

Triple products

Scalar triple products, a·(b×c)

Vector triple products, a ×(b×c)

1. a×( b×c) +b×( c×a) +c×( a×b) = ?
2. (a×b)·(c×d) = ?

Curves, Surfaces and Volumes

Arc length

Sqrt[1 + 4 u^2] u 0 1

exactly (case sensitive, square bracket) in the boxes then press the query button.

Integrate with respect to from to .

Surface area element

To obtain the surface area expressed by two parameters, u and v, one can first compute the surface element spanned by the two vectors shown above:

R(u + d u, v)− R(u, v) = R(u, v) + ∂R ∂u du − R(u, v) = eu du, (56) R(u, v + dv) − R(u, v) = R(u, v) + ∂R ∂v dv − R(u, v) = ev dv, (57)

where

eu ≡ ∂R ∂u , (58) ev ≡ ∂R ∂v . (59)

Remembering that the area of a parallelogram spanned by two vectors, a and b, is

ds =   ⎛  ⎥ √ det ⎛ ⎜ ⎜ ⎝ eu ·eu , eu ·ev ev ·eu , ev ·ev ⎞ ⎟ ⎟ ⎠   du dv = √   E F − G2   du dv. (60)

where

E = eu ·eu F = ev ·ev G = eu ·ev.

Alternatively, recall that the magnitude of a ×b is the area of the parallegram spanned by a and b, Equation (60) can be also written as

d s = | eu ×ev | du dv.

Example of surface integral

As an example of surface integrals, we attempt to compute the surface area of a sphere x² + y² + z² = a². By setting x = u and y = v, a position vector on the sphere is expressed as

R(u, v) = ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ u v √   a2 − u2 − v2   ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ,

(61)

so

eu = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 0 −u √   [ˉ(a2 − u2 − v2)] ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ (62) ev = ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 0 1 −v √   [ˉ(a2 − u2 − v2)] ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ (63)

and

eu ·eu = 1+ u2 a2 − u2 − v2 ev ·ev = 1+ v2 a2 − u2 − v2 eu ·ev = uv a2 − u2 − v2 .

(64)

Hence,

dS = √   (eu ·eu) (ev ·ev) − (eu ·ev)2   du dv = a √ a2 − u2 − v2 du dv.

(65)

The surface area of a semi-sphere (radius = a) is

S = ⌠ ⌡ ⌠ ⌡ a du dv √ a2 − u2 − v2 = a ⌠ ⌡ ⌠ ⌡ r dr dθ √ a2 − r2 2 = a ⌠ ⌡ a 0  r dr √ a2 − r2 ⌠ ⌡ 2π 0  1 dθ = 2πa ⌠ ⌡ a 0  r dr √ a2 − r2 3 = 2πa2.

(66)

Alternative approach

A bit faster approach than the one above is to use the spherical coordinate system. Noting that the equation, x² + y² + z² = a² is automatically satisfied if

x = a sin ϕcos θ, y = a sin ϕsin θ, z = a cos ϕ.

(67)

Hence we can set

R = ⎛ ⎜ ⎜ ⎜ ⎝ a sin ϕcos θ a sin ϕsin θ a cos ϕ, ⎞ ⎟ ⎟ ⎟ ⎠ .

(68)

It follows

eϕ = ⎛ ⎜ ⎜ ⎜ ⎝ a cos ϕcos θ a cos ϕsin θ −a sin ϕ ⎞ ⎟ ⎟ ⎟ ⎠ (69) eθ = ⎛ ⎜ ⎜ ⎜ ⎝ −a sin ϕsin θ a sin ϕcos θ 0 ⎞ ⎟ ⎟ ⎟ ⎠ , (70)

and

E = a2 (71) F = a2 sin 2 ϕ (72) G = 0 (73) dS = √   EF−G2   dθdϕ (74) = a2 sin ϕdθdϕ, (75)

thus

S = ⌠ ⌡ π/2 0  ⌠ ⌡ 2π 0  a2 sin ϕdθdϕ = 2πa2 ⌠ ⌡ π/2 0  sin ϕdϕ = 2πa2. (76)

z = f(x, y)

If the equation of a curved surface is given as

z = f(x, y),

(77)

the area element can be expressed as

d S = √   1 + fx2 + fy2   dx dy,

(78)

where

fx ≡ ∂f ∂x , fy ≡ ∂f ∂y .

(79)

Proof

ex = ∂R ∂x = ⎛ ⎜ ⎜ ⎜ ⎝ 1 0 fx ⎞ ⎟ ⎟ ⎟ ⎠ ey = ∂R ∂y = ⎛ ⎜ ⎜ ⎜ ⎝ 0 1 fy ⎞ ⎟ ⎟ ⎟ ⎠ , E = ex ·ex = 1 + fx2, F = ey ·ey = 1 + fy2, G = ex ·ey = fx fy, EF−G2 = (1 + fx2)(1 + fy2) − (fx fy)2 = 1 + fx2 + fy2.

(80)

Volume

The volume element of the parallelepiped spanned by the three vectors above can be expressed as⁴

dv =   ⎛  ⎥  ⎥ √ ⎢ ⎢ ⎢ ⎢ ⎢ eu ·eu, eu ·ev, eu ·ew ev ·eu, ev ·ev, ev ·ew ew ·eu, ew ·ev, ew ·ew ⎢ ⎢ ⎢ ⎢ ⎢   du dv dw = |eu·(ev×ew)| du dv dw.

(81)

Example: Volume of sphere

R = ⎛ ⎜ ⎜ ⎜ ⎝ x y z ⎞ ⎟ ⎟ ⎟ ⎠ = ⎛ ⎜ ⎜ ⎜ ⎝ r sin ϕcos θ r sin ϕsin θ r cos ϕ ⎞ ⎟ ⎟ ⎟ ⎠ ,

(82)

so

er = ⎛ ⎜ ⎜ ⎜ ⎝ sinϕ cos θ sinϕ sin θ cos ϕ ⎞ ⎟ ⎟ ⎟ ⎠ ,   eϕ = ⎛ ⎜ ⎜ ⎜ ⎝ r cosϕ cos θ r cosϕ sin θ − r sin ϕ ⎞ ⎟ ⎟ ⎟ ⎠ ,   eθ = ⎛ ⎜ ⎜ ⎜ ⎝ − r sin ϕ sin θ r sin ϕ cos θ 0 ⎞ ⎟ ⎟ ⎟ ⎠

(83)

Noting that

eϕ ×eθ = ⎛ ⎜ ⎜ ⎜ ⎝ r2  sin2 ϕ cos θ r2  sin2 ϕ sin θ r2  sin ϕ cos ϕ ⎞ ⎟ ⎟ ⎟ ⎠

(84)

it follows

er ·eϕ ×eθ = r2 sin ϕ.

(85)

so

V = ⌠ ⌡ ⌠ ⌡ ⌠ ⌡ dV = ⌠ ⌡ 2π 0  ⌠ ⌡ π 0  ⌠ ⌡ a 0  r2 sin ϕdr dϕdθ = ⌠ ⌡ 2π 0  dθ ⌠ ⌡ π 0  sin ϕdϕ ⌠ ⌡ a 0  r2 dr = 4πa3 3 . (86)

Summary

1. Length of Arc
   
   

   dl = √   eu ·eu   du = |eu | du.
2. Area
   
   

   d s =   ⎛  ⎥ √ | eu ·eu, eu ·ev eu ·ev, ev ·ev |   du dv = |eu ×ev| du dv.
3. Volume
   
   

   d v =   ⎛  ⎥  ⎥ √ ⎢ ⎢ ⎢ ⎢ ⎢ eu ·eu, eu ·ev, eu ·ew ev ·eu, ev ·ev, ev ·ew ew ·eu, ew ·ev, ew ·ew ⎢ ⎢ ⎢ ⎢ ⎢   du dv dw = |eu ·(ev ×ew) | du dv dw.

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