THE JACOBIAN & CHANGE OF VARIABLES MATH 200 GOALS Be able to - - PowerPoint PPT Presentation

THE JACOBIAN & CHANGE OF VARIABLES

MATH 200 WEEK 10 - WEDNESDAY

MATH 200

GOALS

▸ Be able to convert integrals in rectangular coordinates to

integrals in alternate coordinate systems

MATH 200

DEFINITION

▸ Transformation: A transformation, T, from the uv-plane to

the xy-plane is a function that maps (u,v) points to (x,y) points.

▸ x = x(u,v); y = y(u,v)

MATH 200

EXAMPLE

▸ Consider the region in the

rθ-plane bounded between r=1, r=2, θ=0, and θ=π/2

▸ The transformation

T:x=rcosθ, y=rsinθ maps the region in the rθ-plane into this region in the xy-plane

▸ Transformations need to be

• ne-to-one and must have

continuous partial derivatives

MATH 200

JACOBIAN

▸ The area of a cross section in the xy-plane may not be

exactly the same as the area of a cross section in the uv-

• plane. We want to determine the relationship; that is, we

want to determine the scaling factor that is needed so that the areas are equal.

MATH 200

IMAGE OF S UNDER T

▸ Suppose that we start with a tiny rectangle as a cross section in

the uv-plane with dimensions u and v. The image will be roughly a parallelogram (as long as our partition is small enough).

▸ T: x=x(u,v), y=y(u,v) ▸ x and y are functions of u and v

MATH 200

▸ Let’s label the corners of S as follows ▸ A(u0,v0) ▸ B(u0 + Δu,v0): B is a a little to the right of A ▸ C(u0,v0 + Δv): C is a little higher that A ▸ D(u0 + Δu,v0 + Δv): D is a little higher and to the right of A

MATH 200

▸ What happens to these points under the transformation T? ▸ Well, we have transformation functions x(u,v) and y(u,v), so we can plug

the coordinates of the points A, B, C, and D to get the corresponding points in the xy-plane

▸ For the image of a point P under T, we’ll write T(P) = P’, so we get: ▸ A’(x(u0,v0),y(u0,v0)) ▸ B’(x(u0 + Δu,v0), y(u0 + Δu,v0)) ▸ C’(x(u0,v0 + Δv), y(u0,v0 + Δv)) ▸ D’(x(u0 + Δu,v0 + Δv), y(u0 + Δu,v0 + Δv))

MATH 200

▸ We want to know how the area of R compares to the area

• f S.

▸ R is a parallelogram, so its area is the cross-product of the

vectors a and b:||

a × b||

• a =
• AB

= x(u0 + ∆u, v0) x(u0, v0), y(u0 + ∆u, v0) y(u0, v0), 0

• b =
• AC

= x(u0, v0 + ∆v) x(u0, v0), y(u0, v0 + ∆v) y(u0, v0), 0

WE’VE ADDED THE Z- COMPONENT BECAUSE CROSS PRODUCTS ARE ONLY DEFINED FOR VECTORS IN 3-SPACE

MATH 200

▸ To simplify these vectors down to something more

manageable, we look to the limit definition of the partial derivative:

d f dx = lim

∆x→0

f(x + ∆x) − f(x) ∆x ∂f ∂x = lim

∆x→0

f(x + ∆x, y) − f(x, y) ∆x

FROM CALC 1

▸ Notice that the first component of the vector a looks like

the numerator of the limit definition of partial derivative

• f x with respect to u

x(u0 + ∆u, v0) − x(u0, v0) ≈ ∂x ∂u∆u

IT’S APPROXIMATE BECAUSE WE’RE MISSING THE LIMIT

MATH 200

▸ Applying this idea to both vectors, we get

• a = x(u0 + ∆u, v0) x(u0, v0), y(u0 + ∆u, v0) y(u0, v0), 0
• ∂x

∂u∆u, ∂y ∂u∆u, 0

• xu∆u, yu∆u, 0
• b = x(u0, v0 + ∆v) x(u0, v0), y(u0, v0 + ∆v) y(u0, v0), 0
• ∂x

∂v ∆v, ∂y ∂v ∆v, 0

• xv∆v, yv∆v, 0

MATH 200

▸ Now we can take the cross product − → a × − → b =

• −

→ i − → j − → k xu∆u yu∆u xv∆v yv∆v

• = [(xu∆u)(yv∆v) − (xv∆v)(yu∆u)] −

→ k = (xuyv − xvyu)∆u∆v− → k ▸ For the area of R, we get

• −

→ a × − → b

• = |xuyv − xvyu|∆u∆v

∆AR = |xuyv − xvyu|∆u∆v

MATH 200

CONCLUSIONS

▸ We’ve shown that the area of R is ∆AR = |xuyv − xvyu|∆u∆v ▸ The area of S is ∆AS = ∆u∆v ▸ So the scaling factor we were looking for is |xuyv - xvyu| ▸ We can write this as the determinate of a special matrix

called the Jacobian Matrix:

∂(x, y) ∂(u, v) = xu xv yu yv

MATH 200

EXAMPLE

▸ Compute the Jacobian (i.e., the determinate of the

Jacobian Matrix) for the transformation from the rθ-plane to the xy-plane

T :

• x(r, θ) = r cos θ

y(r, θ) = r sin θ

xr = cos θ xθ = −r sin θ yr = sin θ yθ = r cos θ

MATH 200

T :

• x(r, θ) = r cos θ

y(r, θ) = r sin θ ∂(x, y) ∂(r, θ) = xr xθ yr yvθ

• =

cos θ −r sin θ sin θ r cos θ

• = cos θ(r cos θ) − (−r sin θ)(sin θ)

= r cos2 θ + r2 sin2 θ = r(cos2 θ + sin2 θ) = r

MATH 200

THEOREM

▸ Given a transformation T from the uv-plane to the xy-plane,

if f is continuous on R and the Jacobian is nonzero, we have

• R

f(x, y) dAxy =

• S

f(x(u, v), y(u, v))

• ∂(x, y)

∂(u, v)

• dAuv

▸ For example, when converting from rectangular to polar,

we have

• R

f(x, y) dAxy =

• S

f(x(r, θ), y(r, θ))r dArθ

MATH 200

EXAMPLE

▸ Consider the following double integral:

• R

x − y x + y dA, R : region enclosed by y = x, y = x − 1, y = −x + 1, y = −x + 3

R

IT WOULD BE NICE IF WE COULD TRANSFORM THIS REGION INTO AN UPRIGHT RECTANGLE

MATH 200

▸ Let’s take the transformation T to be T :

• u = x − y

v = x + y ▸ To convert our xy-integral to a uv-integral, we want the

Jacobian

∂(x, y) ∂(u, v) =

• xu

xv yu yv

• ▸ We could solve for u and v in our transformations OR we

could use the convenient fact that

∂(x, y) ∂(u, v) = 1

∂(u,v) ∂(x,y)

MATH 200

▸ First, we need the partial derivatives: ux = 1, uy = −1, vx = 1, vy = 1 ▸ Plug it all in to the Jacobian Matrix ∂(u, v) ∂(x, y) = ux uy vx vyθ

• =

1 −1 1 1

• = 2

▸ Take the reciprocal to get ∂(x, y) ∂(u, v) = 1 2

MATH 200

▸ For the bounds, we have          y = x = ⇒ x − y = 0 = ⇒ u = 0 y = x − 1 = ⇒ x − y = 1 = ⇒ u = 1 y = −x + 1 = ⇒ x + y = 1 = ⇒ v = 1 y = −x + 3 = ⇒ x + y = 3 = ⇒ v = 3 ▸ Finally, our integral becomes

• R

x − y x + y dA = 3

1

1 u v

• 1

2

• dudv

MATH 200

• R

x − y x + y dA = 3

1

1 u v

• 1

2

• dudv

= 1 2 3

1

1 u v dudv = 1 2 3

1

1 2 u2 v

• 1

dv = 1 4 3

1

1 v dv = 1 4 ln |v|

• 3

1

= 1 4 ln 3

MATH 200

EXAMPLE 2

▸ Use the transformation x = v/u, y=v to evaluate the integral 1 x y2 x2 ey/x dydx ▸ Let’s first convert the

region from the xy-plane to the uv-plane

▸ The region extends

from the line y=0 to the line y=x

▸ From x=0 to x=1

MATH 200

▸ Apply the conversion

formulas x=v/u and y=v

▸ y=0 becomes v=0 ▸ y=x becomes v=v/u ▸ u=1 ▸ x=0 becomes v/u=0 ▸ v=0 ▸ x=1 becomes v/u=1 ▸ v=u ▸ In summary, the region

extends from the line v=0 to the line v=u

▸ Bounded by u=1

SO THE REGION LOOKS THE SAME IN THE UV-PLANE

MATH 200

▸ For the Jacobian, we need

the partials of x=v/u and y=v

▸ xu = -v/u2 ▸ xv = 1/u ▸ yu = 0 ▸ yv = 1 ∂(x, y) ∂(u, v) =

• xu

xv yu yv

• =
• −v/u2

1/u 1

• = −v/u2

▸ Lastly, we need to convert

the integrand (the function we’re integrating)

y2 x2 ey/x = u2eu ▸ Now we have all the

pieces we need to set up the integral

y2 x2 ey/x = (v)2 (v/u)2 ev/(v/u)

MATH 200

▸ At this point we need to do integration by parts twice to

finish the problem

1 x y2 x2 ey/x dydx = 1 u u2eu

• − v

u2

• dvdu

= 1 u veu dvdu = 1 1 2v2eu

• u

du = 1 1 2u2eu du

1 1 2u2eu du = 1 2u2eu

• 1

− 1 2 1 2ueu du = 1 2e − 1 2

• 2ueu
• 1

− 1 2eu du

• = 1

2e − 1 2

• 2e − 2eu
• 1
• = 1

2e − e + 1 2(2e − 2) = 1 2e − 1

MATH 200

MATH 200

EXAMPLE 3

▸ Let R be the region enclosed by xy = 1, xy = 2, xy2 = 1, and

xy2 = 2. Evaluate the following integral.

• R

y2dA

R

MATH 200

▸ Let u=xy and v=xy2 ▸ This gives us a nice rectangular

region in the uv-plane

▸ u=1 to u=2 and v=1 to v=2 ▸ To find the Jacobian we’ll need to

solve for x and y in terms of u and v

▸ Solve each equation for x to get

x = u/y and x = v/y2

▸ Setting those equal: u/y = v/y2 ▸ Solve for y: y = v/u ▸ Plug into the u-equation: u=x(v/u) ▸ x = u2/v

MATH 200

▸ Jacobian: ▸ Setup:

• R

y2dA = 2

1

2

1

v2 u2 1 v dvdu ∂(x, y) ∂(u, v) =

• xu

xv yu yv

• =
• 2u/v

−u2/v2 −v/u2 1/u

• = 2

v − 1 v = 1 v

• R

y2dA = 2

1

2

1

v2 u2 1 v dvdu = 2

1

2

1

v u2 dvdu = 2

1

v2 2u2

• 2

1

du = 2

1

4 2u2 − 1 2u2 du = 2

1

3 2u2 du = − 3 2u

• 2

1

= −3 4 + 3 2 = 3 4