Vector Integration - Gate Study Material In Pdf 2g4r4a

Vector Integration - GATE Study Material in PDF In the previous article we have seen about the basics of vector calculus. And in these free GATE Study Notes we will learn about Vector Integration. A vector has both magnitude and direction whereas a scalar has only magnitude. Let us now see how to perform certain operations on vectors. These GATE study material are useful for GATE EE, GATE EC, GATE CS, GATE ME, GATE CE and all other branches. Also useful for exams such as DRDO, IES, BARC, BSNL, ISRO etc. You can have these notes ed as PDF so that your exam preparation is made easy and you ace your paper. Before you get started, go through the basics of Engineering Mathematics.

Recommended Reading – Types of Matrices Properties of Matrices Rank of a Matrix & Its Properties Solution of a System of Linear Equations Eigen Values & Eigen Vectors Linear Algebra Revision Test 1 Laplace Transforms Limits, Continuity & Differentiability Mean Value Theorems Differentiation Partial Differentiation Maxima and Minima Methods of Integration & Standard Integrals 1|Page

Vector Calculus

Rules of Vector Integration Let

d dt

⃗⃗(t) ⇒ ⃗f(t) = ∫ F ⃗⃗ (t)dt + c is called vector integration. [f⃗(t)] = F

b 1. ∫a ⃗f(t) dt = F(b) − F(a) b a 2. ∫a ⃗f(t) dt = − ∫b ⃗f(t) dt b c b 3. ∫a ⃗f(t) dt = ∫a ⃗f(t) dt + ∫c ⃗f(t) dt

Example 1: 1

Find the value of ∫0 (tî + t 2 ĵ + t 3 k̂) ⋅ dt

Solution: t2

1

∫0 (tî + t 2 ĵ + t 3 k̂) ⋅ dt = [ 2 î + 1

1

t3

ĵ + 3

t4

1

k̂] 4

0

1

= [2] î + 3 ĵ + 4 k̂ Generally, integrals are classified as line integrals, surface integrals and volume integrals.

Line Integrals An integral which is to evaluated along the curve is called line integral ⃗⃗ = F1 î + F2 ĵ + F3 k̂ and r⃗ = xî + yĵ + zk̂ Let F Consider r⃗ = xî + yĵ + zk̂ then dr⃗ = îdx + dyĵ + dzk̂ ∴ ∫c⃗F⃗ ⋅ dr⃗ = ∫c[F1 î + F2 ĵ + F3 k̂] ⋅ [îdx + ĵdy + k̂dz] = ∫c[F1 dx + F2 dy + F3 dz] where C is open curve If C is a close curve then ∮ ⃗F⃗ ⋅ dr⃗ is called the circulation then we need to Put x = r cos θ, y = r sin θ and θ = 0 to θ = 2π solve the given integration. 2|Page

Example 2: ⃗⃗ ⋅ dr⃗ if F = xî − yĵ and c is given by x 2 + y 2 = 4 Find the value of ∫cF

Solution: ⃗⃗ ⋅ dr⃗ = ∫ F1 dxî + F2 dyĵ ∫cF c = ∫c(xdx − ydy) Put, x = 2 sin θ , y = 2 cos θ dx = 2 cos θdθ and dy = −2 sin θdθ 2π

∫0 [2 sin θ(2 cos θ)dθ] − [2 cos θ(−2 sin θ)] dθ 2π

2π

∫0 8 cos θ sin θ dθ = 4 ∫0 sin 2θ dθ =4⋅

−1 2

[cos 2θ]2π 0 = −2[1 − 1] = 0

∴ ∫c⃗F⃗ ⋅ dr⃗ = 0

Note: If ⃗F⃗ is force then the total work done by a force is ∫c⃗F⃗ ⋅ dr⃗. If ∫c⃗F⃗ ⋅ dr⃗ = 0 ⇒ ⃗F⃗ is called conservative force.

Green’s Theorem If M(x,y) and N (x,y) are continuous and having continuous first order partial derivatives bounded by a closed curve ‘C’ then x

y

1

1

∂N

2 2 ∫c(Mdx + Ndy) = ∫x ∫y ( ∂x −

∂M ∂y

) dydx

Example 3: Find the value of ∫c(3x + 4y) dx + (2x − 2y)dy where C ∶ x 2 + y 2 = 4

Solution: 3|Page

Here M = 3x + 4y, N = 2x − 3y ∂M ∂y

∂N

=4

∂x

=2 ∂N

By Green’s theorem ∫R(Mdx + Ndy) = ∫ ∫ ( ∂x −

∂M ∂y

) dxdy

= −2 ∫ ∫ dxdy = (−2)(πr 2 ) = −8π

Multiple Integrals They are mainly classified as 1. Double Integrals 2. Triple Integrals

1. Double Integrals ∫R ∫ f(x,y) dx dy is called the double integral where R ϵ [(x1, x2)(y1, y2)]

Methods to Evaluate Double Integrals Method 1: If y1, y2 are functions of x only and x1, x2 are constants then the order of integration is first integral is with respect to “y” treating x- as a constant then the remaining expression integrate with respect to x. x2 =b y2 =f(x2 ) f(x, y) ∫ 1 =a y1 =f(x1 )

∫R ∫ f(x, y)dydx = ∫x

⋅ dy ⋅ dx

Example 4: √1+x2

1

Find the value of ∫x=0 ∫y=0

dydx 1+x2 +y2

Solution: √1+x2

1

∫x=0 ∫y=0 1

= ∫0

1

1+x2 +y2

[tan−1 √1+x2

4|Page

1

dydx

= ∫0

(√1+x2 ) √1+x2

1

y

⋅ [tan−1 (√1+x2)] √1+x2 ] ⋅ dx

√1+x2 0

⋅ dx

1

= ∫0

1

π

√1+x2

tan−1 (1) ⋅ dx = 4 ⋅ [sin h−1 (x)]10

π

= 4 ⋅ [sin h−1 (1) − sin h−1 (0)] π

π

= 4 ⋅ [log(1 + √1 + 1)] = 4 log(1 + √2)

Method 2: If x1, x2 are functions of y only, and y1, y2 are constant then the order of integration is first w.r.t x treating y as constant and then integrate remaining expression w.r.t. y y2 =d x2 =ϕ(y) f(x, y) ∫ 1 =c x1 =ϕ(y)

∫R ∫ f(x, y)dxdy = ∫y

⋅ dxdy

Example 5: 1

y

Find the value of ∫y=0 ∫x=√𝐲(x + y) dxdy

Solution: y

1 x2

∫0 [ 2 + xy] y3

= [6 + 1

y3 3

1

1 y2

⋅ dy = ∫0 (

√y

− 1

−

4

2y2 5

3

2

1

5

y2

2

y

+ y2) − ( + y2)

] 0

2

=6+3−4−5 =

10+20−15−24 60

=

30−39 60

=

−9 60

=

−3 20

Area Using Double Integral: x

y

1

1

Area = ∫ ∫ dA = ∫ ∫R dydx = ∫x 2 ∫y 2 dydx

Example 6: Find the area bounded between the curves y = x2, y = x

5|Page

Solution:

1

x

Area = ∫0 ∫x2⋅ dy. dx 1

x2

1

x3

1

1

1

= ∫0 [y]xx2 ⋅ dx = ∫0 (x − x 2 ) ⋅ dx = [ 2 − 3 ] = 2 − 3 = 6 0

2. Triple Integrals Triple integral is defined as ∫ ∫R ∫ f(x, y, z)dxdydz The order of integration:If z1, z2 are function of x and y and y1, y2 are function of x and x1, x2 are constant the order of integration is f(x, y, z). dz dy dx ↓ ] ↓ ] ↓ ] (1) (2) (3)

x2 =b y2 =f(x) z2 =ψ(x,y) ∫x =a ∫y =f(x) ∫z =ϕ(x,y) [[[ 1 1 1

Example 7: 1

x

y

Find the value of ∫0 ∫0 ∫0 xyz dz dy dx

Solution: 1

x

y

1

x

z2

y

1

1

x

∫0 ∫0 ∫0 xyz dz dy dx = ∫0 ∫0 xy [ 2 ] dy ⋅ dx = 2 ∫0 ∫0 xy 3 ⋅ dy ⋅ dx 0

6|Page

=

=

x

4 1 [y ]0 [ x ∫ 2 0 4

1

1 [∫ x 5 8 0 1

⋅ dx] 1

⋅ dx] =

[x6 ]0 8(6)

1

= 48

Surface Integrals An integral which is to be evaluated over a surface is called surface integral. ⃗⃗ ⋅ N ⃗⃗⃗ ds Mathematical formula for surface integral is = ∫s ∫ F where ⃗F⃗ = F1 î + F2 ĵ + F3 k̂ ⃗N ⃗⃗ = Outward unit normal vector and ds = projection of surface on to the planes Method To Evaluate Surface Integral: 1. If the surface S is on to XY (Z = 0) plane then ⃗⃗ ⋅ N ̂ ds = ∫x2 ∫y2 F ⃗⃗ ⋅ N ̂ dydx ∫ ∫S F ̂| ̂ ⋅k x y |N 1

1

⃗⃗ ̂ = ∇ϕ if ϕ(x, y, z) = c is given otherwise N ̂ = k̂. where N ⃗⃗ϕ| |∇

2. If the surface S is on to XZ (Y = 0) plane then ⃗⃗ ⋅ N ̂ ds = ∫x2 ∫z2 F ⃗⃗ ⋅ N ̂ dzdx ∫ ∫S F ̂ ⋅ĵ | x z |N 1

1

⃗⃗

̂ = ∇ϕ if ϕ(x, y, z) = c is given otherwise N ̂ = ĵ where N ⃗⃗ϕ| |∇ 3. If the surface S is on to YZ (X = 0) plane then ⃗⃗ ⋅ N ̂ ds = ∫x2 ∫y2 F ⃗⃗ ⋅ N ̂ ⋅ dydz ∫S ∫ F ̂ ⋅î| x y |N 1

⃗⃗

1

̂ = ∇ϕ if ϕ(x, y, z) = c is given otherwise N ̂ = î where N ⃗⃗ϕ| |∇

7|Page

Example 8: ⃗⃗ ⋅ N ̂ ds where F ⃗⃗ = zî + xĵ − 3y 2 k̂ and s is the surface of cylinder The value of ∫ ∫sF x 2 + y 2 = 16 included in the first octant between z = 0, z = 5 is.

Solution: ϕ = x 2 + y 2 − 16 and ⃗⃗

̂ = ∇ϕ = î(2x)+ĵ(2y) = N 2 2 ⃗⃗ϕ| |∇ 2√x +y

xî+yĵ √x2 +y2

=

xî+yĵ 4

⃗F⃗ ⋅ N ̂ = zx+xy 4 ⃗⃗ ⋅ N ̂ ds = ∫ ∫ ∫s ∫ F s

x(z+y) 4

⋅ ds

x 2 + y 2 = 16 put x = 0 ⇒ y = ±4 ̂ ⋅ î| = x Let the surface s is projected on to YZ plane, and |N 4 ̂ ds = ∫5 ∫4 x (z + y) ⋅ dzdy (4) ∫ ∫s⃗F⃗ ⋅ N z=0 0 4 x 5

∫z=0 (zy +

y2 2

5

) ⋅ dz = ∫0 (4z + 8)dz = [2z 2 + 8z]50

= 50 + 40 = 90

Stoke’s Theorem According to this theorem, let S be the two sided open surface bounded by a closed curve “C” and ⃗F⃗ be differentiable vector function then ⃗⃗ ⋅ dr⃗ = ∫ ∫ (∇ ⃗⃗ × F ⃗⃗) ⋅ N ̂ ds ∮C F S It gives relation between line integral and surface integral.

Example 9: By Stoke’s theorem the value of ∫ r ⋅ dr⃗ where c is x 2 + y 2 = 16, z = 0 is where 8|Page

r⃗ = xî + yĵ + zk̂

Solution: ⃗⃗ × r⃗) ⋅ N ̂ ⋅ ds ∫ r⃗ ⋅ dr⃗ = ∫ ∫(∇ î ĵ k̂ ⃗∇⃗ × r⃗ = |∂x ∂y ∂z| = î(0) + ĵ(0) + k̂(0) = 0 x y z ̂ (ds) = 0 ∫cr⃗ ⋅ dr = ∫ ∫(0) ⋅ N

Volume Integrals ⃗⃗dv is called volume integral and it is given as The integral ∫ ∫ ∫V F x y z = ∫x 2 ∫y 2 ∫z 2(F1 î + F2 ĵ + F3 k̂) ⋅ dzdydx 1

1

1

x y z x y z x y z = î ∫x 2 ∫y 2 ∫z 2 F1 dzdydx + ĵ ∫x 2 ∫y 2 ∫z 2 F2 dzdydx + k̂ ∫x 2 ∫y 2 ∫z 2 F3 dzdydx 1

1

1

1

1

1

1

1

1

Gauss Divergence Theorem It gives relation between surface integral to volume integral.

Definition: Let ‘V’ be the volume bounded by a closed surface S and ⃗F⃗ be a differentiable vector function then ⃗⃗ ⋅ N ̂ )ds = ∫ ∫ ∫(∇ ⃗⃗ ⋅ ⃗F⃗) dv ∫ ∫S(F V x

y

z

1

1

1

∂F

= ∫x 2 ∫y 2 ∫z 2 ( ∂x1 +

∂F2 ∂y

+

∂F3 ∂z

) dzdydx

Example 10: ̂ ds where r⃗ = xî + yĵ + zk̂ and S By Gauss Divergence theorem find the value of ∫ ∫sr⃗ ⋅ N is the surface of the sphere x 2 + y 2 + z 2 = 1 9|Page

Solution: ⃗∇⃗ ⋅ r⃗ = ∂ (x) + ∂ (y) + ∂ (z) = 3 ∂x ∂y ∂z ̂ ds = ∫ ∫ ∫(3) ⋅ dv ∫ ∫S r⃗ ⋅ N 4π

= 3 ( 3 r 3 ) = 4π

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