VECTOR CALCULUS Grad Divergence and Curl In this

VECTOR CALCULUS Grad, Divergence and Curl In this section, we will learn about: The operations of Grad, divergence and curl and how they can be used to obtain vector forms of Green’s Theorem.

VECTOR CALCULUS • the gradient of a scalar field, • the divergence of a vector field, and • the curl of a vector field. Grad A gradient is a vector differential operator on a scalar field like temperature. Every point in space having a specific temperature. The gradient is a differential operator that gives you a vector field, which in every point shows you in what direction in 2 or 3 dimensional space the field of values is increasing the fastest. By moving in the opposite direction of the gradient you are seeing the fastest decline of the scalar value, say temperature. By moving perpendicularly to the gradient you are staying at the same scalar value. So gradient operates on a scalar field to give a vector field.

VECTOR CALCULUS Divergence is a differential operator that acts on a vector field to give a scalar field, so the opposite of gradient. If I have a vector field like a gravitational field or an electric field, by taking the divergence I get the scalar field that is proportional to the mass or charge density "causing" that field. Everywhere in space where we do have gravitational or electric fields, but where no mass or charge is located, the divergence is zero.

VECTOR CALCULUS curl The curl is a differential operator acting on a vector field to give another vector field. Curl alludes to something round or rotation, which is why it is also called rotation, abbreviated by rot. The curl measures the net boost that an element affected by the vector field that the curl operator acts upon would get when going in a small closed loop in a specific plane. This boost yields the curl vector component perpendicular to the specific plane that the local small loop, for which the boost is measured, lies in.

Introduction (Grad) The vector operator ∇ may also be allowed to act upon vector fields. Two different ways in which it may act, the subject of this package, are extremely important in mathematics, science and engineering. We will first briefly review some useful properties of vectors. Consider the (three dimensional) vector, a = a 1 i + a 2 j + a 3 k. We may also write this as a = (a 1, a 2, a 3). If we multiply it by a constant c, then every component of the vector is multiplied by c: ca = ac = (ca 1, ca 2, ca 3). If we introduce a second vector, b = (b 1, b 2, b 3), then we recall that there are two different ways of multiplying vectors together, the scalar and vector products. The scalar product (also called dot product) is defined by: a · b = a 1 b 1 + a 2 b 2 + a 3 b 3. It is a scalar (as the name scalar product implies).

vector product: We see that the cross product of the vectors a = a 1 i +a 2 j + a 3 k and b = b 1 i + b 2 j + b 3 k is:

Introduction (Grad) If U(r) = U(x, y, z) is a scalar field, ie a scalar function of position r= [x, y, z] in 3 dimensions, then its gradient at any point is defined in Cartesian co-ordinates by It is usual to define the vector operator which is called “del” or “nabla” Then

Introduction (Grad) The vector differential operator ∇ , called “del” or “nabla”, is defined in three dimensions to be: ∇ = ∂/∂x i + ∂/∂y j + ∂/∂z k Note that these are partial derivatives! If a scalar function, f(x, y, z), is defined and differentiable at all points in some region, then f is a differentiable scalar field. The del vector operator, ∇, may be applied to scalar fields and the result, ∇f, is a vector field. It is called the gradient of f (see the package on Gradients and Directional Derivatives). Quiz As a revision exercise, choose the gradient of the scalar field f(x, y, z) = xy 2 − yz. (a) i + (2 x − z)j − yk , (b) 2 xyi + 2 xyj + yk , (c) y 2 i − zj − yk , (d) y 2 i + (2 xy − z)j − yk.

Suppose: F = P i + Q j + R k is a vector field on . The partial derivatives of P, Q, and R all exist § We introduce the vector differential operator (“del”) as:

Del It has meaning when it operates on a scalar function to produce the gradient of f :

Grad “Vector operator acts on a scalar field to generate a vector field” Example:

Worked examples of gradient evaluation

The directional derivative: The rate of change of U wrt distance in direction

Example If we want to find the direction to move to increase our function the fastest, we plug in our current coordinates (such as 3, 4, 5) into the equation and get: So, this new vector (1, 8, 75) would be the direction we’d move in to increase the value of our function. In this case, our x-component doesn’t add much to the value of the function: the partial derivative is always 1.

Examples For the function z=f(x, y)=4 x^2+y^2. The gradient is For the function w=g(x, y, z)=exp(xyz)+sin(xy), the gradient is

The divergence of a vector field The divergence computes a scalar quantity from a vector field by differentiation. If a(x; y ; z) is a vector function of position in 3 dimensions, that is a = a 1^ı+a 2^+a 3^k, then its divergence at any point is defined in Cartesian coordinates by We can write this in a simplified notation using a scalar product with the r vector differential operator: Notice that the divergence of a vector field is a scalar field.

Divergence (Div) If F(x, y) is a vector field, then its divergence is written as div. F(x, y) = ∇ · F(r) which in two dimensions is: It is obtained by taking the scalar product of the vector operator r applied to the vector field F(x, y). The divergence of a vector field is a scalar field. Example 2 The divergence of

Examples of divergence evaluation We work through example (3). The x component of, and we The terms in y and z are similar, so that need to find of it.

Quiz Select the scalar product of a = (1, 2, 3) and b = (3, − 2, 1). (a) 2 , (b) 10 , (c) 3 x − 4 y + 3 z , (d) 4. The vector product (or cross product) is defined by: a × b = (a 2 b 3 − a 3 b 2)i − (a 1 b 3 − a 3 b 1)j + (a 1 b 2 − a 2 b 1)k It is a vector (as the name vector product implies). Note that the second line is a useful shorthand for the first. Quiz Choose the vector product of a = (1, 2, 3) and b = (3, − 2, 1). (a) 8 i − 8 j − 8 k , (b) − 4 i − 10 j + 4 k , (c) 8 i + 8 j − 8 k , (d) 8 i − 10 j − 8 k.

Example What is the directional derivative in the direction <1, 2> of the function z=f(x, y)=4 x^2+y^2 at the point x=1 and y=1. The gradient is <8 x, 2 y>, which is <8, 2> at the point x=1 and y=1. The direction u is <2, 1>. Converting this to a unit vector, we have <2, 1>/sqrt(5). Hence,

Conservative Fields In a Conservative Vector Field Which gives an easy way of evaluating line integrals: regardless of path, it is difference of potentials at points 1 and 2. Obvious provided potential is single-valued at the start and end point of the closed loop.

Div “Vector operator acts on a vector field to generate a scalar field” Example

Divergence Theorem S V Oover closed outer surface S enclosing V

Elemental volume for calculating divergence.

Divergence Theorem as aid to doing complicated surface integrals • Example z x • Evaluate Directly • Tedious integral over and (exercise for student!) gives y

Using the Divergence Theorem • So integral depends only on rim: so do easy integral over circle • as always beware of signs

Quiz Select the divergence of The definition of the divergence may be directly extended to vector fields defined in three dimensions, Exercise 1. Calculate the divergence of the vector fields F(x, y) and G(x, y, z) (click on the green letters for the solutions).

Example: Consider the example problem we had in the last section: Fx = ax 3 + bxy 2 + cz Fy = ay 3 + bx 2 y Fz = cx. We found that this force gave us a work that was independent of path (we tried two different paths and got the same result). Let’s look at the curl of this F. It should equal zero if the Work that this force does is independent of path. ΛXF = [x ¶ /¶x + y ¶ /¶y + z ¶ /¶z ] ´ [Fx x + Fy y + Fz z] = = (¶Fz/¶y - ¶Fy/¶z) x + (¶Fx/¶z - ¶Fz /¶x) y + (¶Fy /¶x - ¶Fx/¶y) z = (0 – 0) x + (c – c) y + (2 bxy – 2 bxy) z = 0.

DIVERGENCE Equation 9 If F = P i + Q j + R k is a vector field on and ∂P/∂x, ∂Q/∂y, and ∂R/∂z exist, the divergence of F is the function of three variables defined by:

DIVERGENCE Equation 10 In terms of the gradient operator the divergence of F can be written symbolically as the dot product of and F:

DIVERGENCE If Example 4 F(x, y, z) = xz i + xyz j – y 2 k find div F. § By the definition of divergence (Equation 9 or 10) we have:

The Divergence Theorem For the vector field in Figure 4, it appears that the vectors that end near P 1 are shorter than the vectors that start near P 1. Figure 4 The vector field F = x 2 i + y 2 j 37

Curl The curl of a vector field, F(x, y, z), in three dimensions may be written curl. F(x, y, z) = r× F(x, y, z), i. e. : It is obtained by taking the vector product of the vector operator r applied to the vector field F(x, y, z). The second line is again a formal shorthand. The curl of a vector field is a vector field. N. B. r×F is sometimes called the rotation of F and written rot. F. 38

Magnitude Curl Direction: normal to plane which maximises the line integral. Can evaluate 3 components by taking areas with normals in xyz directions “Vector operator acts on a vector field to generate a vector field” Example

Key Equations

Simple Examples Radial fields have zero curl Rotating fields have curl in direction of rotation

Example 3 The curl of F(x, y, z) = 3 x 2 i + 2 zj − xk is: Quiz Which of the following is the curl of F(x, y, z) = xi + yj + zk ? (a) 2 i − 2 j + 2 k , (b) xi + yj + zk , (c) 0 , (d) i + j + k.

Exercise 2. Calculate the curl of the following vector fields F(x, y, z) (click on the green letters for the solutions). (a) F = xi − yj + zk , (b) F = y 3 i + xyj − zk , Here is a review exercise before the final quiz. Exercise 3. Let f be a scalar field and F(x, y, z) and G(x, y, z) be vector fields. What, if anything, is wrong with each of the following expressions (click on the green letters for the solutions)? (a) ∇ f = x 3 − 4 y , (b) ∇ · F = i − x 2 yj − zk , (c) ∇ × G = r· F.

Exercise 2(a) The components of the vector field F = xi − yj + zk are F 1 = x , F 2 = −y , F 3 = z and its curl is: Therefore the vector field F = xi − yj + zk is an irrotational vector field.

Flux (Surface Integrals of Vectors Fields) Derivation of formula for Flux Suppose the velocity of a fluid in xyz space is described by the vector field F(x, y, z). Let S be a surface in xyz space. The flux across S is the volume of fluid crossing S per unit time. The figure below shows a surface S and the vector field F at various points on the surface.

Example Find the flux of the vector field <y, x, z> in the negative z direction through the part of the surface z=g(x, y)=16 -x^2 y^2 that lies above the xy plane (see the figure below). For this problem: It follows that the normal vector is <-2 x, -2 y, -1>. Computing Fo<-2 x, -2 y, -1>, we have

Example Here we use the fact that z=16 -x^2 -y^2. Hence, the integral becomes The region R is the disk of radius 4 centered at the origin shown below. It is convenient to convert to polar coordinates. The region of integration is 0<=r<=4 and 0<=theta<=2*pi. Substituting, we have

Example Here we have used the fact that x^2+y^2=r^2. (We can also integrate with respect to theta first, then r. ) The inner integral with respect to r yields (remember theta is held constant) which is equal to The inner integral is Hence, the flux through the surface in the downward z direction is -128*pi cubic units per unit time. This means a negative amount of fluid crosses the surface in the downward direction. In other words, there is a flux of 128*pi cubic units per unit time in the upward z direction.

Example 1 Find the flux of the vector field F(x, y, z) = z i + y j + x k over the unit sphere x 2 + y 2 + z 2 = 1. Solution: First we compute the divergence of F: The unit sphere S is the boundary of the unit ball B given by x 2 + y 2 + z 2 1.

Example 1 – Solution cont’d Thus the Divergence Theorem gives the flux as

CURL Equation 1 Then, the curl of F is the vector field on defined by:

CURL As a memory aid, let’s rewrite Equation 1 using operator notation. § We introduce the vector differential operator (“del”) as:

CURL It has meaning when it operates on a scalar function to produce the gradient of f :

CURL If we think of as a vector with components ∂/∂x, ∂/∂y, and ∂/∂z, we can also consider the formal cross product of with the vector field F as follows.

CURL

CURL Equation 2 Thus, the easiest way to remember Definition 1 is by means of the symbolic expression

CURL If Example 1 F(x, y, z) = xz i + xyz j – y 2 k find curl F. § Using Equation 2, we have the following result.

CURL Example 1

GRADIENT VECTOR FIELDS Theorem 3 If f is a function of three variables that has continuous second-order partial derivatives, then

GRADIENT VECTOR FIELDS By Clairaut’s Theorem, Proof

CONSERVATIVE VECTOR FIELDS A conservative vector field is one for which So, Theorem 3 can be rephrased as: If F is conservative, then curl F = 0. § This gives us a way of verifying that a vector field is not conservative.

CONSERVATIVE VECTOR FIELDS Example 2 Show that the vector field F(x, y, z) = xz i + xyz j – y 2 k is not conservative. § In Example 1, we showed that: curl F = –y(2 + x) i + x j + yz k § This shows that curl F ≠ 0. § So, by Theorem 3, F is not conservative.

CONSERVATIVE VECTOR FIELDS Example 3 a § As curl F = 0 and the domain of F is , F is a conservative vector field by Theorem 4.

Example • Irrotational and conservative are synonymous because by Stokes Theorem • So this is a conservative field, so we should be able to find a potential • All can be made consistent if

Stokes Theorem • Consider a surface S, embedded in a vector field • Assume it is bounded by a rim (not necessarily planar) � OUTER RIM SURFACE INTEGRAL OVER ANY SURFACE WHICH SPANS RIM

Example z (0, 0, 1) C 2 quarter circle C 1 x C 3 (0, 1, 0) y • Check via direct integration

2 nd-order Vector Operators Lecture 10 meaningless Laplace’s Equation is one of the most important in physics

CURL F VS. DIV F Observe that: § Curl F is a vector field. § Div F is a scalar field.

DIVERGENCE If F is a vector field on also a vector field on , then curl F is. As such, we can compute its divergence. § The next theorem shows that the result is 0.

DIVERGENCE Theorem 11 If F = P i + Q j + R k is a vector field on and P, Q, and R have continuous secondorder partial derivatives, then div curl F = 0

DIVERGENCE Proof By the definitions of divergence and curl, § The terms cancel in pairs by Clairaut’s Theorem.

DIVERGENCE Note the analogy with the scalar triple product: a. (a x b) = 0

DIVERGENCE Example 5 Show that the vector field F(x, y, z) = xz i + xyz j – y 2 k can’t be written as the curl of another vector field, that is, F ≠ curl G § In Example 4, we showed that div F = z + xz and therefore div F ≠ 0.

DIVERGENCE Example 5 § If it were true that F = curl G, then Theorem 11 would give: div F = div curl G = 0 § This contradicts div F ≠ 0. § Thus, F is not the curl of another vector field.

DIVERGENCE Again, the reason for the name divergence can be understood in the context of fluid flow. § If F(x, y, z) is the velocity of a fluid (or gas), div F(x, y, z) represents the net rate of change (with respect to time) of the mass of fluid (or gas) flowing from the point (x, y, z) per unit volume.

INCOMPRESSIBLE DIVERGENCE In other words, div F(x, y, z) measures the tendency of the fluid to diverge from the point (x, y, z). If div F = 0, F is said to be incompressible.

GRADIENT VECTOR FIELDS Another differential operator occurs when we compute the divergence of a gradient vector field . § If f is a function of three variables, we have:

Example: Find

Exercise: find the gradient of the scalar field f = x 2 y − z, we need the partial derivatives: Therefore, the gradient of f = x 2 y − z is

DEL Operator § DEL Operator in cylindrical coordinates: § DEL Operator in spherical coordinates: 1 -82

The Gradient in Polar Coordinates Suppose we have a function given to us as f(x, y) in two dimensions or as g(x, y, z) in three dimensions. We can take the partial derivatives with respect to the given variables and arrange them into a vector function of the variables called the gradient of f, namely which mean Suppose however, we are given f as a function of r and θ , that is, in polar coordinates, (or g in spherical coordinates, as a function of Ø, θ and ρ For example, suppose How do we find the gradient of f or g? One way to find the gradient of such a function is to convert r or ρ or θ into rectangular coordinates using the appropriate formulae for them, and perform the partial differentiation on the resulting expressions. Thus we can write and find, by ordinary partial differentiating It is a bit more convenient sometimes, to be able to express the gradient directly in polar coordinates or spherical coordinates, like it is expressed in rectangular coordinates as above. 1 -83

We now ask: what is f in polar coordinates? We know that if we make differential changes in r and θ the resulting change in f will be given by since this relation holds for any variables at all. But they must also obey 1 -84

Gradient of a scalar field § The gradient of a scalar field V is a vector that represents the magnitude and direction of the maximum space rate of increase of V. ü For Cartesian Coordinates ü For Cylindrical Coordinates ü For Spherical Coordinates 1 -85

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Divergence of a vector ü In Cartesian Coordinates: ü In Cylindrical Coordinates: ü In Spherical Coordinates: 1 -88

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1 -90

Curl of a vector ü In Cartesian Coordinates: ü In Cylindrical Coordinates: ü In Spherical Coordinates: 1 -91

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Stoke’s theorem 1 -96

Laplacian of a scalar 1 -97

Laplacian of a scalar 1 -98

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Example Find The Laplacian of F(x, y, z) if you know: Solution: The Laplacian is: Calculating the components in turn we find: So the Laplacian of F is: 1 -100

Exercise: 1 -

Exercise:

GREEN’S THEOREM Example 1 Evaluate where C is the triangular curve consisting of the line segments from (0, 0) to (1, 0) from (1, 0) to (0, 1) from (0, 1) to (0, 0)

GREEN’S THEOREM Example 1 The given line integral could be evaluated as usual by the methods of Section 16. 2. § However, that would involve setting up three separate integrals along the three sides of the triangle. § So, let’s use Green’s Theorem instead.

GREEN’S TH. (SIMPLE REGION) Example 1 Notice that the region D enclosed by C is simple and C has positive orientation.

GREEN’S TH. (SIMPLE REGION) Example 1 If we let P(x, y) = x 4 and Q(x, y) = xy, then

GREEN’S THEOREM Example 2 Evaluate where C is the circle x 2 + y 2 = 9. § The region D bounded by C is the disk x 2 + y 2 ≤ 9.

GREEN’S THEOREM Example 2 § So, let’s change to polar coordinates after applying Green’s Theorem:

UNION OF SIMPLE REGIONS Example 4 Notice that, though D is not simple, the y-axis divides it into two simple regions. § In polar coordinates, we can write: D = {(r, θ) | 1 ≤ r ≤ 2, 0 ≤ θ ≤π}

UNION OF SIMPLE REGIONS So, Green’s Theorem gives: Example 4