Vector Calculus

Pintu Samanta

Assistant ProfessorDepartment of Mathematics

Magadh University, Bodh Gaya

In real world, most of the things are more than one dimensional.Therefore calculus in one dimension is not enough to study real lifephenomena. Therefore, it is important to know calculus in manydimensions.We know that the equation of parabola is

y2 = 4ax , z = 0.

The parametric form is given by

x = at2, y = 2at, z = 0.

Now any vector in three dimension can be written asr = xi + yj + zk, where i, j, k are three mutually perpendicularunit vectors. Therefore if a particle moves along the parabola theposition vector is r(t) = at2i + 2atj + 0k.

Clearly, it is not possible to express the position of a movingparticle by a single valued function. We need a position vector r(t)having three components. Note that the first derivative of r(t)describes the velocity of the particle.

Definition (Scalar function)

A scalar function is a function f : Rn → R.

A scalar function always gives you a number.

Definition (Vector function)

A vector function is a function F : Rm → Rn.A vector function of a single scalar variable is a functionF : R→ Rn.

The function F always returns a vector.

Derivative of functions: Now the derivative of scalar continuousfunction of scalar variable, i.e., f : R→ R at point t = a is

f ′(a) = limh→0

f (a + h)− f (a)

h.

The derivative of vector function of scalar variable f : R→ R3,i.e., r(t) = f1(t)i + f2(t)j + f3(t)k is

dr

dt=

df1dt

i +df2dt

j +df3dt

k.

d

dt(φg) =

dφ

dtg + φ

dg

dt,

d

dt(g · h) =

dg

dt· h + g · dh

dt,

d

dt(g × h) =

dg

dt× h + g × dh

dt,

here φ is scalar function and g and h are vector functions.

Let g(t) = g1(t)i + g2j + g3k and h(t) = h1(t)i + h2j + h3k then

g × h =

∣∣∣∣∣∣i j kg1 g2 g3h1 h2 h3

∣∣∣∣∣∣Find

dg

dt,

dh

dt,

d

dt(g · h),

d

dt(g × h),

where g(t) = 5t2i + tj− t3k and h(t) = sin(t)i− cos(t)j

d(g · h)

dt=

d

dt

(5t2 sin(t)− t cos(t)

)=(5t2 − 1) cos(t) + 11t sin t

and

d(g × h)

dt=

d

dt

(−t3 cos ti− t3 sin tj− (5t2 sin t− 11t cos t− sin t)

)

Directional derivative:Directional derivative of a vector function f : R3 → R3 along theline with cosines l . m, n is

l∂f

∂x+ m

∂f

∂y+ n

∂f

∂z. (0.1)

Find directional derivative of f = f1i + f2j + f3k along x − axis,y − axis, and z − axis.

The Gradient (or slope) of a scalar point function:

Definition

The gradient of the scalar function f : Rn → R

∇f (x1, x2, · · · , xn) =n∑

i=1

∂f

∂xiei

Hence we can write the condition of differentiability as

δf =∂f

∂xiδxi + o(δx).

In differential notation, we write

df = ∇f · dr =∂f

∂xidxi, (df = f ′(x)dx for one dimension)

which is the chain rule for partial derivatives.

Example

Take f (x , y , z) = x + exy sin z . Then

∇f =∂f

∂xi +

∂f

∂yj +

∂f

∂zk

= (1 + yexy sin z)i + xexy sin zj + exy cos zk

At (x , y , z) = (0, 1, 0), ∇f = (1, 0, 1). So f increases/decreasesmost rapidly for n = ± 1√

2(1, 0, 1) with a rate of change of ±

√2.

There is no change in f if n is perpendicular to ± 1√2

(1, 0, 1).

Example

Let f (x , y , z) = 3x2y − y3z2. Find gradf at the point (1,−2,−1).

Example

Prove that gradient of a constant is zero.

Example

If φ and ψ be two scalar point functions, prove that

grad(φψ) = φgrad(ψ) + ψgrad(φ)

and

grad

(φ

ψ

)=ψgrad(φ)− φgrad(ψ)

ψ2

Example

Show that∇|r| =

r

|r|= r̂,

where r = xi + yj + zk.

Solution

Now |r| =√x2 + y2 + z2. Therefore

∂|r|∂x

=x√

x2 + y2 + z2=

x

|r|.

Hence

∇|r| =xi + yj + zk√x2 + y2 + z2

Example

Show that∇(log(|r|)) =

r

|r|2.

Hints

∂

∂x(log(|r|)) =

x

x2 + y2 + z2

and

∇(log(|r|)) =xi + yj + zk

x2 + y2 + z2=

r

|r|2.

Example

Prove that a.∇r = a and grad(a.r) = a, where a = a1i+ a2j+ a3k.

Definition (Divergence)

Let F : R3 → R3 be a vector valued function andF (x , y , z) = F1i + F2j + F3k. Then divergence or div of F is

∇ · F =∂F1∂x

+∂F2∂y

+∂F3∂z

.

Example

Let F = xez i + y2 sin xj + xyzk. Then

∇ · F =∂

∂xxez +

∂

∂yy2 sin x +

∂

∂zxyz = ez + 2y sin x + xy .

Definition (Curl)

Let F : R3 → R3 be a vector valued function andF (x , y , z) = F1i + F2j + F3k. Then curl of F is

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣

Example

Let F = xez i + y2 sin xj + xyzk. Then

∇× F = i

[∂

∂y(xyz)− ∂

∂z(y2 sin x)

]+ j

[∂

∂z(xez) +

∂

∂x(xyz)

]+ k

[∂

∂x(y2 sin x)− ∂

∂y(xez)

]= xz i + (xez − yz)j + y2 cos xk.

Problem

Show that ∇.r = 3 and ∇× r = 0, where r = xi + yj + xk.

Definition

A vector F is said to be irrotational, if curl F , i.e., ∇× F = 0.

Problem

Show that the vector F = (x2 − yz)i + (y2 − zx)j + (z2 − xy)k isirrotational.

Definition

A vector F is said to be solenoidal, if div F , i.e., ∇.F = 0.

Problem

Show that the vector F = (y − z)i + (z− x)j + (x− y)k issolenoidal.

Second order differential operators:The gradient of a scalar point function and the curl of a vectorpoint function being vector point function, they possess divergenceand curl, while the divergence of a vector point function beingscalar, it will have its gradient. Thus we have the following results

Proposition

Let f : R3 → R be a scalar point function. Then

div grad f = ∇.(∇f) = ∇2f =∂2f

∂x2+∂2f

∂y2+∂2f

∂z2.

Let F : R3 → R3 be vector valued function. Then

∇× (∇× F ) = ∇(∇.F )−∇2F

∇(∇.F ) = ∇× (∇× F ) +∇2F .

Proposition

Let f : R3 → R be a scalar function and F : R3 → R3 be a vectorfunction. Then

∇× (∇f ) = 0

∇ · (∇× F) = 0

Proof.

∇× (∇f ) =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

∂f∂x

∂f∂y

∂f∂z

∣∣∣∣∣∣∣

Problem

If f = x3 + 3yz2, then find ∇2f .

Problem

If f = 2x2y3z4, then find ∇.(∇f ).

Exercise

If the vector F = (ax + 3y + 4z)i+ (x− 2y + 3z)j+ (3x+ 2y− z)kis solenoidal, find a. Ans-a = 3.

Exercise

If F = x2z i− 2y3z2j + xy2zk. Find ∇.F and ∇× F at (1,−1, 1).Ans.-−3 and −6i.

Exercise

If F = ψgradψ = ψ∇ψ, then show that

F .(∇× F ) = 0.

Solution

Let G = ∇ψ. Then

∇× (ψG ) =(∇ψ)× G + ψ∇× G ,

= (∇ψ)× (∇ψ) + ψ(∇× (∇ψ)).

Now,

Solution

∇ψ × (∇ψ) =

∣∣∣∣∣∣∣i j k∂ψ∂x

∂ψ∂y

∂ψ∂z

∂ψ∂x

∂ψ∂y

∂ψ∂z

∣∣∣∣∣∣∣ = 0

and

∇× (∇ψ) =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

∂ψ∂x

∂ψ∂y

∂ψ∂z

∣∣∣∣∣∣∣ = 0.

Exercise

Sow that φ = x|r |3 satisfies the Laplace’s equation ∇2φ = 0.

Solution

Now

∇φ = ∇.(∇φ) =∂2φ

∂x2+∂2φ

∂y2+∂2φ

∂z2.

Note that∂|r |∂x

=x

|r |

∂φ

∂x=

1

|r |3− 3x

|r |4∂|r |∂x

=1

|r |3− 3x2

|r |5. (0.2)

Solution

∂2φ

∂x2= − 9x

|r |5+

15x3

|r |7,

∂2φ

∂y2= − 3x

|r |5+

15xy2

|r |7,

∂2φ

∂z2= − 3x

|r |5+

15xz2

|r |7.

Thus

∂2φ

∂x2+∂2φ

∂y2+∂2φ

∂z2= −15x

|r |5+

15x

|r |7(x2 + y2 + z2) = 0

Exercise

Prove that

∇2(x

|r |2) = − 2x

|r4|.

Exercise

Find the condition so that φ = ax2 + by2 + cz2 satisfies theLaplace’s equation ∇2φ = 0.

Exercise

If the vectors F and G be irrotational, then show that the vectorF × G is solenoidal.

Solution

Since F and G are irrotational, therefore

∇× F = ∇× G = 0.

Again we have

∇.(F × G ) = G .(∇× F )− F .(∇× G ) = 0.

Proposition

Let F : R3 → R3 be a vector valued function. Then

∇× F = 0 =⇒ F = ∇f

for some scalar valued function f : R3 → R.

Exercise

Show that the vector F = (4xy − z3)i + 2x2j− 3xz2k isirrotational. Show that F can be expressed as the gradient of somescalar function.

Solution

Let

F = ∇φ =∂φ

∂xi +

∂φ

∂yj +

∂φ

∂zk,

= (4xy − z3)i + 2x2j− 3xz2k.

Equating coefficients of like vectors, we have

∂φ

∂x= 4xy − z3,

∂φ

∂y= 2x2,

∂φ

∂z= −3xz2.

Solution

Integrating these equations with respect to x , y , z respectively, weget

φ = 2x2y − z3x + φ1(y , z),

φ = 2x2y + φ2(x , z),

φ = −xz3 + φ3(x , y).

These three equations agree, if we choose φ1 = 0, φ2 = −xz3, andφ3 = 2x2y , so that φ = 2x2y − xz3 to which we may add aconstant. Thus

φ = 2x2y − xz3 + a constant.

Exercise

Show that the vector

F = (2x − yz)i + (2y − zx)j + (2z− xy)k

is irrotational. For this F , find a scalar function φ so that F = ∇φ.

Ans:φ(x , y , z) = x2 + y2 + z2 − xyz + constant.

Vector Integration

Proposition

Let f : R2 → R be a continuous function on a compact set D.Then ∫

Df (x , y) dA =

∫Y

(∫xy

f (x , y) dx

)dy .

with xy ranging over {x : (x , y) ∈ D}.

Theorem (Fubini’s theorem)

If f is a continuous function on a compact set D, then∫∫f dx dy =

∫∫f dy dx .

Example

We integrate over the triangle bounded by (0, 0), (2, 0) and (0, 1).We want to integrate the function f (x , y) = x2y over the area. So∫

Df (xy) dA =

∫ 1

0

(∫ 2−2y

0x2y dx

)dy

=

∫ 1

0y

[x3

3

]2−2y0

dy

=8

3

∫ 1

0y(1− y)3 dy

=2

15

Example

We can integrate it the other way round:∫Dx2y dA =

∫ 2

0

∫ 1−x/2

0x2y dy dx

=

∫ 2

0x2[

1

2y2]1−x/20

dx

=

∫ 2

0

x2

2

(1− x

2

)2dx

=2

15

Proposition

Suppose we have a change of variables (x , y)↔ (u, v) that issmooth and invertible, with regions D,D ′ in one-to-onecorrespondence. Then∫

Df (x , y) dx dy =

∫D′

f (x(u, v), y(u, v))|J| du dv ,

where

J =∂(x , y)

∂(u, v)=

∣∣∣∣∣∣∣∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

∣∣∣∣∣∣∣is the Jacobian. In other words,

dx dy = |J| du dv .

Example

We transform from (x , y) to (ρ, ϕ) with

x = ρ cosϕ

y = ρ sinϕ

We have previously calculated that |J| = ρ. So

dA = ρ dρ dϕ.

Suppose we want to integrate a function over a quarter area D ofradius R.

Example

Let the function to be integrated bef = exp(−(x2 + y2)/2) = exp(−ρ2/2). Then∫

f dA =

∫f ρ dρ dϕ

=

∫ R

ρ=0

(∫ π/2

ϕ=0e−ρ

2/2ρ dϕ

)δρ

Note that in polar coordinates, we are integrating over a rectangleand the function is separable. So this is equal to

=[−e−ρ2/2

]R0

[ϕ]π/20

=π

2

(1− e−R

2/2).

Note that the integral exists as R →∞.

Example

Now we take the case of x , y →∞ and consider the originalintegral. ∫

Df dA =

∫ ∞x=0

∫ ∞y=0

e−(x2+y2)/2 dx dy

=

(∫ ∞0

e−x2/2 dx

)(∫ ∞0

e−y2/2 dy

)=π

2

So each of the two integrals must be√π/2, i.e.∫ ∞

0e−x

2/2 dx =

√π

2.

Volume Integral

Proposition ∫Vf dx dy dz =

∫Vf |J| du dv dw ,

with

J =∂(x , y , z)

∂(u, v ,w)=

∣∣∣∣∣∣∣∣∣∣∣∣

∂x

∂u

∂x

∂v

∂x

∂w∂y

∂u

∂y

∂v

∂y

∂w∂z

∂u

∂z

∂v

∂z

∂w

∣∣∣∣∣∣∣∣∣∣∣∣

Consider a volume within a sphere of radius a with a cylinder ofradius b (b < a) removed. The region is defined as

x2 + y2 + z2 ≤ a2

x2 + y2 ≥ b2.

We use cylindrical coordinates. The second criteria gives

b ≤ ρ ≤ a.

For the x2 + y2 + z2 ≤ a2 criterion, we have

−√a2 − ρ2 ≤ z ≤

√a2 − ρ2.

So the volume is∫V

dV =

∫ a

bdρ

∫ 2π

0dϕ

∫ √a2−ρ2

−√

a2−ρ2dz ρ

= 2π

∫ a

b2ρ√

a2 − ρ2 dρ

= 2π

[2

3(a2 − ρ2)3/2

]ab

=4

3π(a2 − b2)3/2.

Vector Integration

Proposition

Let F : R→ R3 be a vector valued function of a scalar variable,i.e., F (t) = F1(t)i + F2(t)j + F3(t)k. Then we define an indefiniteintegral of F (t) as∫

F (t)dt = i

∫F1(t)dt + j

∫F2(t)dt + k

∫F3(t)dt.

Example

Find∫ 21 F (t)dt, where F (t) = t2i + (t− 1)j− 4k.

Example

Find∫ 20 (R.S)dt and

∫ 20 (R × S)dt where R(t) = ti− t2j + (t− 1)k

and S(t) = 2t2i + 6tk.

Proposition

Let F : R3 → R3 be a vector valued function of a scalar variable,i.e., F (x , y , z) = F1(x , y , z)i + F2(x, y, z)j + F3(x, y, z)k. Then wedefine an indefinite integral of F (t) as∫

F (x , y , z).dr =

∫F1dx +

∫F2dy +

∫F3dz .

Example

If F = 3xy i− 5zj + 10xk, then evaluate∫F .dr along the curve C

given by x = t2 + 1, y = 2t2, z = t3 from t = 1 to t = 2.

Proposition

If F = F1i + F2j + F3k and r = x i + yj + zk, then∫cF×dr = i

∫c(F2dz−F3dy)+j

∫c(F3dx−F1dz)+k

∫c(F1dy−F2dx).

If φ be a continuous scalar point function, then∫cφdr = i

∫cφdx + j

∫cφdy + k

∫cφdz.

Theorem (Stokes’ theorem)

For a smooth vector field F(r),∫S∇× F · dS =

∫∂S

F · dr,

where S is a smooth, bounded surface and ∂S is a piecewisesmooth boundary of S .The direction of the line integral is as follows: If we walk along Cwith n facing up, then the surface is on your left.

Exercise

Evaluate by Stoke’s theorem∫c(exdx + 2ydy − dz), (0.3)

where c is the curve x2 + y2 = 4, z = 2.

Solution

In this case F = exi + 2yj− k.

Exercise

Verify Stoke’s theorem for the vector field defined byF = (x2 − y2)i + 2xyj in the rectangular region in the xy planebounded by lines x = 0, x = a, y = 0, y = b.

Solution

Now ∫cF .dr = 2ab2∫

s∇× F .ds = 2ab2.

Theorem (Divergence/Gauss theorem)

For a smooth vector field F(r),∫V∇ · F dV =

∫∂V

F · dS,

where V is a bounded volume with boundary ∂V , a piecewisesmooth, closed surface, with outward normal n.