What are vector fields in Calculus?

 Objective: Define vector field

Introduction

Vector fields are used to describe physical phenomena such as gravitation, electromagnetism, atmospheric storms and deep-sea ocean currents. Let's give some examples.

Example (a). Gravitational force exerted by two astronomical objects such as a planet and a star or a planet and its moon.

The figure (a) models the gravitational force exerted by these two objects:

At any point in the figure, a vector associated with that point gives the net gravitational force exerted by the two objects on a unit of mass of an object. The vectors of the greater magnitude are associated with the largest object. The greater the mass of an object, the stronger is its gravitational force. The gravitational force of the larger object is stronger than that of the smaller object. The vectors of smaller magnitude are associated with the smaller object. The mass of this object is smaller. Its gravitational force is less strong than the larger object. 

Example (b). Velocity of a river at points on its surface.

The vector field velocity of water on a river shows the varied speeds of water. Vectors in red are of greater magnitude. Therefore, the water flows quickly. Vectors in blue are smaller in magnitude. They show a smaller flow of water. 

As we can see in these two examples, a vector field is a map of vectors. In the following posts we will study vectors fields in R² and R³.

Definition

Example

Solution

Practice

Solving a triple integral using the change of variables

 In the previous posts we spent time establishing formulas to solve a triple integral using a change of variables. Let's now solve an example.

Example

Solution

Let's apply the formula:

Let's take the following steps to apply the formula:

1) Let's find x, y, z from the given values of u, v, w. This leads to solve the following system of equations:

u = 2x-y/2 (1)

v = y/2 (2)

w = z/3 (3)

Solving this system, we find x = (u + v)/2 y = 2v z = 3w

2) Let's find the new function H(u,v,w) to be integrated by substituting x, y, z in the function F(x,y,z) = x +z/3. 

3) Let's find the limits of integration of the new function i.e u.v.w:

Let's isolate x, y, z as limits of integration of the given integral. We have:

x = y/2 x = y/2 + 1

y = 0 y = 4

z = 0 z = 3

These equations represent the planes that bound the surface G in the space xyz.

From the equation x=y/2 we have 2x = y 2x-y = 0 ⇒2x-y/2 = 0 ⇒ u = 0 since u = 2x-y/2 (1)

From the equation x = y/2 + 1 we have 2x = y+2 ⇒ 2x-y = 2 ⇒ 2x-y/2 = 1  ⇒ u = 1.

Let's use the equation v = y/2:

For y = 0 we have v = 0. For y = 4 v =2 

Let's use the equation w = z/3

For z = 0 w = 0

For z =3 w = 1

Let's calculate the jacobian:

Change of variables for triple integrals

 Objective: Set the formula for the change of variables in triple integrals.

Planar transformation in tridimensional space

a.The change of variables in triple integrals works the same way as in double integrals. We consider a transformation from a tridimensional space (u,v.w) to a tridimensional space (x,y,z). We define the Jacobian in the (u,v,w) space. Then we establish the formula for the triple integral defined in (u,v,w).

 Suppose that G i s a region in the uvww space transformed in another region D in the xyz space by a C¹ transformation by the transformation T so that T(u,v,w) = (x,y,z) where x = g(u,v,w) y = h(u,v.w) z = k(u,v,w).

Any function F(x.y,z) defined on D can be thought of another function H(u,v. w) defined G such that:

F(x.y.z) = F(g(u,v,w), h(u,v,w), k(u,v,w)) = H(u,v,w), Now let's define the Jacobian in a tridimensional space (u,v,w

Jacobian for three variables. Definition

The jacobian J(u,v,w) for three variables is defined as follows:

This is the same as:

Theorem

Change of Variables for Triple Integrals

Let T(u, v, w) = (x, y, z) where x = g(u, v, w), y = h(u, v, w), and z = k(u, v, w), be a one-to-one C¹ transformation, with a nonzero Jacobian, that maps the region G in the uvw-space into the region R in the xyz-space. As in the two-dimensional case, if F is continuous on R, then

∫∫∫_R F(x,y,z) dV = ∫∫∫_G F(g(u,v,w), h(u,v,w), k(u,v,w))   | ∂(x,y,z) ∂(u,v,w) |  du dv dw
= ∫∫∫_G H(u,v,w) |J(u,v,w)| du dv dw.

Example 

  Obtaining Formulas in Triple Integrals for Cylindrical and Spherical Coordinates

Derive the formula in triple integrals for

1. cylindrical and
2. spherical coordinates.

Solution

Let's apply the formula for change of variables in triple integrals:

∫∫∫_R F(x,y,z) dV = ∫∫∫_G H(u,v,w) |J(u,v,w)| du dv dw

In cylindrical coordinates it becomes:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(r cosθ, r sinθ, z)  |J(r,θ,z)|  dr dθ dz

J(r, θ, z) =  |

∂x/∂r	∂x/∂θ	∂x/∂z
∂y/∂r	∂y/∂θ	∂y/∂z
∂z/∂r	∂z/∂θ	∂z/∂z

| = |

cos θ	−r sin θ	0
sin θ	r cos θ	0
0	0	1

| = r cos²θ + r sin²θ = r (cos²θ + sin²θ) = r.

We know that r ≥ 0, so |J(r,θ,z)| = r. Then the triple integral becomes by substituting the value of the jacobian: 
∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(r cos θ, r sin θ, z) r dr dθ dz.

b. Let's find the formula for the triple integral in spherical coordinates by applying the same process.

For spherical coordinates, the transformation is T(ρ, θ, φ) = (x, y, z) from the Cartesian pθφ-plane to the Cartesian xyz-plane . Here x = ρ sin φ cos θ, y = ρ sin φ sin θ, and z = ρ cos φ. The expression of the triple integral can be written as:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ)  |J(ρ,θ,φ)|  dρ dφ dθ. Let's calculate the jacobian.

First write it as a quotient of partial derivatives:

J(ρ,θ,φ) = ∂(x,y,z) ∂(ρ,θ,φ)

Now write it as a determinant:

J(ρ,θ,φ) = |

∂x/∂ρ	∂x/∂θ	∂x/∂φ
∂y/∂ρ	∂y/∂θ	∂y/∂φ
∂z/∂ρ	∂z/∂θ	∂z/∂φ

|

Substitute the partial derivatives:

J(ρ,θ,φ) = |

sinφ cosθ	-ρ sinφ sinθ	ρ cosφ cosθ
sinφ sinθ	ρ sinφ cosθ	ρ cosφ sinθ
cosφ	0	-ρ sinφ

|

Expand along the third row:

J = cosφ |

-ρ sinφ sinθ	ρ cosφ cosθ
ρ sinφ cosθ	ρ cosφ sinθ

| - ρ sinφ |

sinφ cosθ	-ρ sinφ sinθ
sinφ sinθ	ρ sinφ cosθ

|

Calculate the first determinant:

(-ρ sinφ sinθ)(ρ cosφ sinθ) - (ρ cosφ cosθ)(ρ sinφ cosθ)

= -ρ² sinφ cosφ sin²θ - ρ² sinφ cosφ cos²θ

= -ρ² sinφ cosφ

Calculate the second determinant:

(sinφ cosθ)(ρ sinφ cosθ) - (-ρ sinφ sinθ)(sinφ sinθ)

= ρ sin²φ cos²θ + ρ sin²φ sin²θ

= ρ sin²φ

Therefore,

J = cosφ(-ρ² sinφ cosφ) - ρ sinφ(ρ sin²φ)

J = -ρ² sinφ cos²φ - ρ² sin³φ

J = -ρ² sinφ(cos²φ + sin²φ)

J(ρ,θ,φ) = -ρ² sinφ

So,

|J(ρ,θ,φ)| = ρ² sinφ

Let's substitute the jacobian in the expression of the integral. Then the triple integral becomes:

∫∫∫_D f(x,y,z) dV = ∫∫∫_G f(ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ)  ρ² sinφ dρ dφ dθ.

Change of variables for double integrals

 We have seen that under the change of variables T(u,v) = (x,y) where x = g(u,v) and y = h(u,v), a small region ΔA in the xy plane is related to the area formed by the product 𝜟u 𝜟v in the uv plane by the approximation: 𝜟A ≃J(u,v)𝜟u𝜟

Remember that the double integral is defined as:

In the following figure we divide the region S in small rectangles Sᵢⱼ and the region R in small rectangles Rᵢⱼ. A small rectangle Rᵢⱼ is the image of a small rectangle Sᵢⱼ under the transformation T.

 

Let's substitute f(xᵢⱼ, yᵢⱼ) and ΔA in the definition of the integral:

∫∫_R f(x,y) dA = ∫∫_S f(g(u,v),h(u,v)) |J(u,v)| du dv

Let's substitute J(u,v) in in the expression on the right side of the equality sign, we have:

Theorem

Example

Solution

Let's do the change of variables in the expression √x² + y² . By substituting x = rcosθ and y = sinθ, the expression becomes equal to r.

The expression dydx becomes J(r,θ) drdθ.  In a previous example J(r,θ) was equal to r. The integral becomes when we substitute everything:

Practice

What are Jacobians in multiple integrals?

 Let's consider a one-to-one C¹ transformation defined by: 

We want to see how it transforms a small rectangle with sides Δu and Δv in the (u,v) plane.

Let's call ΔA the area of the curved region R of sides Δurᵤ and Δvrᵥ. This area is approximately equal to  the norm of the product of the two sides. It can be written as:

ΔA≃ ॥Δurᵤ॥.॥Δvrᵥ॥ = ॥Δu॥॥rᵤ॥. ॥Δv॥rᵥ॥ = Δu॥rᵤ॥.Δv॥rᵥ॥ = ॥rᵤ॥॥rᵥ॥ ΔuΔv 

Let's find rᵤ and rᵥ in order to calculate the norm of their product.

Let's calculate the product of the vectors rᵤ and rᵥ. This is the tricky part. The surface is sitting in a tridimensional space, which is R³.  We can rewrite rᵤ and rᵥ as: 

rᵤ = ∂x/∂ui + ∂x/∂uj + 0k

rᵥ = ∂x/∂vi + ∂x/∂vj + 0k

So, the coordinates of the tangent vectors rᵤ and rᵥ at (u₀, v₀) are:

𝐫ᵤ = ( ∂x/∂u, ∂y/∂u, 0 )

𝐫ᵥ(u₀,v₀) = ( ∂x/∂v, ∂y/∂v, 0 )

Let's calculate the product of the vectors now:

Let's take the norm of the product:

॥rᵤ 🗙 rᵥ॥ =∥ ∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u) ) 𝐤 ) ∥

= ॥ (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u)॥ ∥𝐤∥

=  (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u)  since ∥𝐤∥ = 1

Let's substitute in the expression of ΔA. We have:

 . 

The jacobian can finally be written as:

J(u,v) = 

Example. Find the jacobian of the transformation given in the following example:

 

 

Solution

       

Finding the image of a triangle by a transformation

 The previous example consists in showing that a transformation is a one-to-one transformation. Now we are concerned about how to find the image of a triangle under a transformation. Let's use an example in this case.

Example

Solution

The triangle and its image are shown in the figure below. To understand how the sides of the triangle transform, call the side that joins (0,0) and (0,1) side A, the side that joins (0,0) and (1,1) side B and the side that joins (0,1) and (1,1) side C. 

Practice

   

Change of variables in multiple integrals. Planar Transformations

 Objective: Determine the image of a region under a given transformation of variables

Planar Transformation

A planar transformation is a function that transforms a region G in one plane into another region R into another plane by a change of variables. Both G and R are subsets of R². The figure bellow shows a region G in the (u,v) plane transformed into another region R in the (x,y) plane by the change of variables x = g(u,v) and y = h(u,v).

Definition

A transformation T: G → R defined as T(u,v) = (x,y) is said to be one- to- one transformation if no two points map the same image point.

Example

Solution

first quadrant of the xy plane. Hence R is a quarter circle bounded by x² + y² = 1 in the first quadrant.