Vector Calculus - (Part - 2) | Mathematics for Competitive Exams PDF Download

Surface Area And Surface Integrals

A smooth surface S is called Orientable or two sided if it is possible to define a field n of unit normal vectors on S that varies continuously with position. Spheres and their smooth closed surface in space are orientable. By convention, we choose n on a closed surface to point outward.

The surface together with its normal field is called oriented surface. Unit normal is at any point of it is called the positive direction at the point.
Let f(x, y, z) = c be any oriented surface S and R be its projection (Shadow) on coordinate plane (say xy-plane)
The angle between S and R is given by

{z-axis is normal to xy-plane}
area of the surface f(x, y, z) = c over a closed and bounded plane region R is

dS cos θ = dA
⇒

Surface Integral

If R be the projection (shadow) on xy-plane of surface S defined by equation f(x, y, z) = c and g is a continuous function defined at the points of S, then the integral of g over S is the integral.

Suppose that F is a continuous vector field defined over an oriented surface S and that  is unit normal to the surface S. The surface integral of the normal component of F, i.e. F. , is the flux of F across A in the positive direction and is given by

 

(∇ f.k ≠ 0)

Remark :

(a) To find the surface integral of a vector function or flux means the same thing.
(b) If S is a surface defined by a function z = f(x, y) that has continuous first order partial derivatives throughout a region R in the xy-plane then
Let φ(x, y, z) = f(x, y) - z be the level surface

(a) Unit normal to φ(x, y, z) is

(b) Area of surface S is

Example : Evaluate  where A = 18zi + 12j + 3yk and S is the part of the plane 2x + 3y + 6z = 12 which is located in the first octant.

We know that ds = dA/cosθ
where θ is angle between the normal to surface S and normal to its projection in xy-plane

∴

S : f(x, y, z) = 2x + 3y + 6z - 12

∴
Now




= 432 - 360 + 96
= 72 + 96 = 168

Example : Evaluate :  where F = yi + (x - 2xz)j - xyk and S is the surface of the sphere x² + y² + z² = a² above the xy-plane.

F = yi + (x - 2xz)j - xyk

= i (-x + 2x) - j (-y - 0) + k(1 - 2z - 1)
= xi + yj - 2zk ...(i)
Let S be the surface given by f(x, y, z) = x² + y² + z² - a² above xy-plane.
Thus, the projection on xy-plane is the circle x² + y² = a². Also
= unit normal to surface S

= (1/a)(xi + yj + zk) ...(ii)
and  ...(iii)
Now





= π/2(a³ - a³ + a³ - a³) = 0

Example : Evaluate :  over the entire surface S of the region bounded by the cylinder x² + z² = 9, x = 0, y = 0, z = 0 and y = 8 where A = 6zi + (2x + y)j - xk.

Here the entire surface S consists of 5 surfaces, namely. S₁ : lateral surface of the cylinder ABCD, S₂ : AOED, S₃ : OBCE, S₄ : OAB, S₅ : CDE.
Thus,

= l₁ + l₂ + l₃ + l₄ + l₅(say)
S₁ : ABCD : The curved surface S₁ is f = x² + z² = 9. The unit outward normal to S₁ is

∴
= 1/3(6xz - xz) = (5/3)xz
and

∴


=  (5 x 9 x 8)/2 = 180
S₂ : AOED : The surface S₂ is xy-plane i.e. z = 0. Unit outward normal to the surface is
∴

S₃ : OBCE : Surface S₃ is yz-plane i.e. x = 0. Unit outward normal to S₃ is
∴


S₄ : OAB : The section OAB is in xz-plane i.e. y = 0. The unit outward normal to S₄ is
∴


= (2/3)(-27) = -18
S₅ : CDE : The section S. is parallel to xz-plane, y = 8. The unit outward normal to S₅ is




= 18(1+π)
Thus, the required surface integral is

Example : Find the flux of the vector field A = (x - 2z)i + (x + 3y + z)j + (5x + y)k through the upper side of the triangle ABC with vertices at the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1).

Equation of the plane containing the given triangle ABC is f(x, y, z) ≡ x + y + z - 1
Unit normal h to ABC is

= (1/√3) (i + j + k)
Flux of A =

Example : Find the surface area of the plane 2z + x + 2y = 12 cut off by x = 0, y = 0, x = 1, Y = 1.

Let φ(x, y, z) = ((12 - x - 2y)/2)-z = 0 be the level surface.

∴

= (3/2)(y)¹₀
= 3/2

Volume Integral

Let V be a region is space enclosed by a closed surface S. Let F be a vector point function, and φ be a scalar point function, then the triple integrals are known as volume integral or space integrals.
In component form

Example : If V is the region of first octant bounded by y² + z² = 9 and the plane x = 2 and F = 2x² yi - y² j + 4xz²k. Then evaluate

= 4xy - 2y + 8xz

Example : If F = 2z i + xj + yk. Evaluate  where V is the region bounded by the surfaces x = 0, y = 0, x = 2, y = 4, z = x², z = 2.

= 32/15(3i + 5k)

Example : Evaluate  where A = (x + 2y)i - 3z j + xk and V is the closed region in the first octant bounded by the plane 2x + 2y + z = 4

= 3i - j - 2k

= (8/3)(3i - j - 2k)

Example : Find the volume enclosed between the two surfaces S₁ : z = 8 - x² - y² and S₂ : z  = x² + 3Y².

Eliminating z from the two surfaces S₁ and S₂, we get
8 - x² - y² = x² + 3y² 
⇒ x² + 2y² = 4
Thus, the two surface intersect on the elliptic cylinder.
So the solid region between S₁ and S₂ is covered when z varies from x² + 3y² to 8 - x² - y².
y varies from
x varies from - 2 to 2.
∴

Gauss's Divergence Theorem

Statement : The normal surface integral of a function F over the boundary of a closed region is equal to the volume integral of div F taking throughout the region.
If F is a continuously differentiable vector point function in a region V enclosed by the closed surface S, then :
 ...(1)
where  is the unit outward drawn normal vector to the surface S.
The Cartesian form of (1) is as follows :
 ...(2)
where F₁, F₂ and F₃ are scalar components of F along the axes.
Remark : The volume integral is reduced into surface integral with the help of this theorem. Conversely, the surface integral can also be reduced into volume integral.
Proof. Let V be the region wrt the axes such that if any straight line is drawn parallel to any axis, then that will meet the surface S in two points only.
In fig., A₃ is the projection of region V on the plane XOY. Let the coordinates of any point R in the plane A₃ be (x, y, 0).
Let any line through R meets the surfaces at the points P and Q.
Therefore z-co-ordinate of Q by ψ(x, y) and that of P be φ(x, y).
Since RP > RQ, therefore φ(x, y) > ψ(x, y)

Now

 ...(3)
Let S₁ and S₂ be the sub-parts of the surface S.
Let  be the outwards drawn unit normal vector at any point on the surface S.
Therefore dx dy = dS cos θ = .k dS,
where θ is the angle of the normal with x-axis.
∴  ...(4)
and  ...(5)
The negative sign of equation (5) shows that the normal drawn outwards make an obtuse angle with z-axis.
The positive sign in equation (4) is because the outwards normal on the surface S₁ make an acute angle with z-axis.
Substituting the value from the equations (4) and (5) in (3)

 ...(6)
Similarly,
 ...(7)
and  ...(8)
Adding the equations (6), (7) and (8),

This theorem is also true for the region V enclosed by two closed curves S₁ and S₂, where the surface S₂ is situated in side the surface S₁.
Remark : The volume integral is reduced into surface integral with the help of this theorem. Conversely, the surface integral can also be reduced into volume integral.

Example : Evaluate :  where F = 4xz i - y²j + yz k S is the surface of the cube bounded by the planes :

By Gauss’s theorem,


 [∵ dv = dx dy dz]

Example : Evaluate :  where S is the part of the sphere x² + y² + z² = 1 above the xy plane. 

F = y²z²i + z²x²j + z²y²k
∴
= 2zy²
By Gauss’s theorem,

This is the part of the sphere x² + y² + z² = 1 above the xy plane i.e., hemisphere above the xy plane and the limits for x, y, z are as follows :
First integrate wrt z from z = 0 to z =  then integrate wrt y, between the limit  and finally wrt x between the limits x = -1 to x = 1.
∴ Required integral =



put x = sin θ ⇒ dx = cosθ dθ

Example : Use Gauss’s divergence theorem to show that : 

 where the surface S is the sphere x² + y² + z² = a². 

We know that

where F₁ = x, F₂ = y, and F₃ = z
∴
∴ From equation (1)

  (1 + 1 + 1)dx dy dz, enclosed by surface S, where V is volume.
 dx dy dz = 3 (volume of the sphere)
= 3.(4/3)πa² = 4πa³

Example : If V is the volume enclosed by any closed surface S; show that:
(a)
(b)
(c)
(d)

(a) If F be any constant vector, then
  [∵ F is a constant vector]
By Gauss's theorem,
But

or,  [∵ F is a constant vector]
(b) If F be any constant vector, then

 (by interchanging the scalar and vector products)
 [by Gauss’s theorem] ...(1)
But ∇.(F x r) = r.(∇ x F) - F.(∇ x r) = 0 [∵ V x f = 0, ∵ F is constant and ∇ x r = 0]
By (1),
or,
(c) By Gauss’s theorem,

= 0 [∵ ∇.(Z∇ x F) = 0]
(d) By Gauss’s theorem,
 (∵ ∇.r = 3)
= 3V

Example : Evaluate : F = zi + xj + 3y²z k, where S is the surface of the cylinder x² + y² = 16 include in the first octant between z = 0 and z = 5.

The projection of the given surface on xz plane will be a rectangle whose sides will be x, z axes and lines parallel to these.
Therefore the given integral
 ...(1)
Normal vector on the given surface x² + y² = 16

∴
∴
= 1/4(xz + xy)
Now by (1),



= 5 x 8 - (25/2)(0 - 4) = 90

Example : If V is the volume enclosed by any closed surface S; show that :  where F = xi + 2yj + 3zk 

Given F = xi + 2yj + 3zk
∴

= 1 + 2 + 3 = 6
∴ By Gauss’s theorem,

Example : Verify Gauss’s Divergence theorem and show that : 

 where F = (x³ - yz)i - 2x²yj + 2k S is the surface of the cube bounded by the co-ordinate planes : x = y = z = 0; x = y = z = a 

Here
= 3x² - 2x² + 0 = x²
∴

Verification by direct Integration : In figure outward unit normal have been shown on each face of the cube. Now on the face ABCD = i, x = a, dS = dy dz

(∵ x = a)

= a⁵ - (1/4)a⁴
Taking  = - i on the face OEFC,


∴ On this face x = 0

Taking  = j on the face BEFC.

 [∵ on this face y = a]

Taking  = k on the face CDGF,

Taking  = -k on the face ABEO,

= -2x²
Adding the integrals on all the faces,
= a⁵ - (1/4)a⁴ + (1/4)a⁴ - (2/3)a⁵ + 0 + 2a² - 2a²
= (1/3)a⁵

Example : Evaluate : where F = xi + yj + z²k. where S is the closed surface bounded by the cone x² + y² = z² and the plane z = 1.

Here
= 1 + 1 + 2z = 2(1 + z)

 

where V is the volume of the given cone and  is the centre of gravity of the cone (situated on the axis of the one) which is at a distance y from the vertex O. The base of the given cone is a circle whose radius is 1 and height is also 1.
∴ OG = 3/4, V = (1 /3)π1² . 1 = π/3
∴ Required integral = 2V(1 + )
 

Stoke's Theorem

Statement : The line integral of a vector function F around any closed curve is equal to the surface integral of curl F taken over any surface of which of the curve is a boundary edge.
If F be any continuous differentiable vector function and S is the surface enclosed by a curve C, then :

where n is the unit normal vector at any point of S and drawn in the sense in which a right handed screw would move rotated in the sense of description of C.

Cartesian form :

 [∵ dz = 0]
Since  = k, therefore

Proof, (a) For surfaces in the planes :

Let the region S₁ be subdivided into sub-regions S_i such that if a straight line is drawn parallel to any axis then that will meet the curve C_i atmost in two points.
Let C_i be situated between the lines x = a and x = b.
Let any line the be drawn parallel to y-axis meets the curve at the point P and Q, where the ordinate of P is y = φ(x) and that of Q is y = ψ(x). In the figure limit of the curve has been divided by the line PQ in two parts C₁ and C₂
Now,


...(1)

Similarly,  ...(2)
From (1) and (2),

or,
(b) The Cartesian form of the surface in space,

  [*by Jacobian] ...(1)
where D is any region in u-v plane whose image is S.
Taking the terms of F₁ in the equation (1),
 (where the additional last term = 0)

Similarly, the values of the terms F₂ and F₃ in (1) are respectively

Therefore the RHS of (1)

Now by Stoke’s theorem,

Similarly,

and

Therefore

Now applying Stoke’s theorem for each subregion and adding the results,

Example : Verify Stoke’s theorem for the function F = zi + xj + yk, where the curve C is the unit circle in the xy-plane bounding the hemisphere z =

Given F = zi + xj - yk and r = xi + yj + zk
∴ F.dr = (zi + xj + yk).(idx + j dy + k dz)  
or, F.dr = z dx + x dy + y dz ...(1)
The unit circle C in xy plane x² + y² = 1, z = 0
∴ For the curve C, z = 0 and dz = 0
∴ by (1), F^.dr = x dy ...(2)
From the fig., x = cos φ, y = sin φ, where φ varies from 0 to 2π.
∴ From (2),

= 1/2[2π] = π ...(3)
From the fig, the direction cosines of the line OP are sin θ cos φ, sin θ sinφ, cos θ
Therefore  = (sin θ cos φ)i + (sin θ sin φ)j + (cos θ)k,
where  is outer normal at the point P.
Now

= i + j + k
∴  •(∇ x F) = [(sin θ cos φ)i + (sin θ sin φ)j + (cos θ)k]•[i + j + k]
= sin θ cos φ + sin θ sin φ + cos θ
∴  where dS = sin θ dφ dθ
 

 ...(4)
Therefore from (3) and (4),

Example : Using Stoke’s theorem, evaluate :  where C is the square in the xy plane with vertices respectively : (1, 0), (-1, 0), (0, 1); (0, 1)

Writing
 where dr = xi + yj
Here F = xy i + xy² j
∴
Taking  = k and dS = dx dy,
•(∇ x F) = k•(y² -x)k = y² - x
By Stokes’s theorem,

Example : Verify Stoke’s theorem for the function F = x²i + xy j integrated round the square in the plane z = 0, whose sides are along the lines x = y = 0 and x = y = 0.

In the plane z = 0, r = xi + yj
Therefore
dr = i dx + j dy
∴ F•dr = (x³i + xy j)•(i dx + j dy)
= x² dx + xy dy
Now

(a) On the line AB, y = 0, Therefore dy = 0
...(1)
(b) On the line BC, x = a, Therefore dx = 0
∴ Along the side BC of the square ...(2)
(c) On the line CD, y = a, Therefore dy = 0
∴ Along the side CD of the square  ...(3)
(d) On the line DA, x = 0, Therefore dx = 0
∴ Along the side DA of the square  ...(4)
Therefore from (1), (2), (3) and (4), along the square ABCD
 ...(5)
Check : curl F =
On square ABCD,  = k, Therefore curl F = yk•k = y
∴
 ...(6)
From (5) and (6),
Hence the Stoke’s theorem is verified.

Example : Evaluate by Stoke’s theorem  where C is the curve x² + y² = 4 and z = 2.

Writing
or
 where dr = i dx + j dy + k dz
Therefore taking F = e^xi + 2y j - k and by Stoke’s theorem
 ...(1)
where the surface S is the boundary C of the curves, x² + y² = 4 and z = 2.
Here C is a circle x² + y² = 4 and z = 2 whose centre is D(0, 0, 2) and radius is 2 and C is also the boundary of S₁, therefore  = k
∴  ...(2)
Now
= i(0) + j(0) + k(0) = 0

From (2), •(∇ x F) = 0
Therefore from (1), the given integral

Example : Prove by Stoke’s theorem :
(a)
(b)

(a) By Stoke’s theorem
 ...(1)
Let F = ∇(φψ), therefore ∇ x f = ∇ x (∇φψ)
or, ∇ x f = 0 [∵ ∇ x (∇φψ) = 0]
Therefore from (1),...(2)
and
From (2),
or,
(b)  (By Stoke's theorem) ...(1)
Let F = φ∇ψ
∴ ∇ x F = ∇ x (φ∇ψ)
= (∇φ) x ∇ψ + φ∇ x ∇ψ [∵ ∇ x fV = (∇φ) x V + φ(∇ x V)]
or, ∇ x F = ∇φ x ∇ψ [∵ ∇ x (∇ψ) = 0] ...(2)
Therefore from (1) and (2)

Green's Theorem

Statement: If φ and ψ are two continuously differentiable vector point functions such that ∇φ and ∇ψ are also continuously differentiable within V enclosed by a surface S, then

Proof. Using Gauss theorem for the vector point function φ∇ψ
 ...(1)
But  
 ...(2)
From (1) and (2),
...(3)
Interchanging φ and ψ,
 ...(4)
Subtracting (4) and (3),
 ...(5)
which is the Green’s theorem

Another Form :
Since
where ∂ψ/∂η is the derivative of ψ in the direction of outer normal at the point ψ.
Therefore,
From (5),

Cartesian form of Green’s Theorem :

If C is a regular closed curve in xy plane enclosing a region S, P(x, y) and Q(x, y) be two continuously differentiable functions in the region S, then :

Let F = Pi + Qj, Since n = k, therefore


∴
∴  ...(1)
and
 (P dx + Q dy) ...(2)
From (1) and (2),
 ...(3)
Cor. : When Q = x and P = -y, then
Therefore from (3),  (If A is the area of the enclosed region by the curve C)

∴

Example : Evaluate by Green’s theorem :  where C is the rectangle with vertices (π, 0), (0, 0), (π, π/2) and (0, π/2)

Here P = e^-x sin y; Q = e^-x cos y
∴ ∂P/∂Y = e-x cosy and ∂Q/∂x =  -e^-xcos y
∴   [- cosy - cosy] = -2e^-x cos y
Therefore by Green’s theorem,

Example : Use Green’s theorem to evaluate : 

 where C is the triangle enclosed by the lines y = 0, x = π/2 and y = 2x/π. 

Given P = y - sin x and Q = cos x
∴
By Green’s theorem,
 ...(1)
Substituting the values of P and Q in (1) and simplifying
 
 where S is the area of the triangle.

Example : Evaluate by Green’s theorem :  (x² - cosh y) dx + (y + sin x )dy,
where C is the rectangle with vertices (0, 0), (π, 0), (π, 1) and (0, 1).

Taking the given points as A, B, C and D respectively, then a rectangle ABCD is obtained as given in the fig.
Here P = x² - cosh y ⇒
and Q = y + sin x
Now by Green’s theorem,

= π cosh 1 - π = π[cosh 1 - 1]

Example : Evaluate :   [(3x² - 8y²)dx + (4y - 6xy)dy], where C is the region bounded by the parabolas y = √x and y = x². Also verify Green’s theorem.

Solving the equations of the parabola y =√x and y = x² the coordinates of the points of intersection are O(0, 0) and A(1, 1)
Here P = 3x² - 8y² and Q = 4y - 6xy
∴
By Green’s theorem,



= 3/2 ...(1)

Verification : The line integral towards C₁ (towards y = x²)

and line integral towards C₂ (towards y² = x)

Required integral = Line integral towards C₁ and C₂
 ...(2)
Therefore from (1) and (2), the Green’s theorem is verified.

Example : Show that the area bounded by a simple closed curve C is given by  Hence find the whole area of the ellipse.

Here P = -y and Q = x
Therefore
By Green’s theorem, we know that
 ...(1)
Substituting the values of  and Q in (1),

or  ...(2)
Therefore the enclosed area by the simple closed curve C = 1/2 (x dy-y dx)
The given curve is an ellipse whose parametric equations are
x = a cosθ and y = b sin θ.
Therefore x = a cosθ and y = b sinθ
∴ The required area of the ellipse