Test: Vector Calculus - Civil Engineering (CE) MCQ

Concept:

Let 

The curl of A is evaluated as:

Given

A_x = cos x cos y

A_y = cos x cos y

A_z = 0

∇ × A = (- sinx cosy + cosx siny) â_z

Gauss divergence theorem:

It states that the surface integral of the normal component of a vector function  taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function  taken over a volume enclosed by the closed surface ‘S’.

Given:

F = xi + yj + zk

Concept:

Let y = f(x) be the equation of a curve, then the slope of the tangent at any point say (x₁, y₁) is given by:

Calculation:

Given curve is xy³ - yx³ = 6

Now by partially differentiating the equation of curve with respect to x we get;

The slope(m) i.e. dy/dx of the tangent at (1, -1) is:

m =  -1

Revolution about x-axis: The volume of the solid generated by the revolution about the x-axis, of the area bounded by the curve
y = f(x), the x-axis and the ordinates
x = a and x = b is

similarly for revolution about y-axis:

V=∫πx²dy

Calculation:

Given:

V = 3π / 2

Hence the required volume will be
3π / 2.

Concept:

When two points (x₁, y₁. z₁) and (x₁, y₁. z₂) are mentioned find the relation in terms of the third variable in terms of x,y, and z:

Put the value of z,y, and z and use the end-points of one variable.

Given:

I = ∫(2xy²dx + 2x²ydy + dz), A (0, 0, 0) and B(1, 1, 1).

Equation of line i.e. path

∴ x = y = z = t and t : 0 → 1

∴ 

= 

∴ 

Given

Integral

∫∫∫v 8 xyz dv 

Limits for x, y and z is given as

[2, 3] × [1, 2] × [0, 1]

Volume of the integral

V = ∫∫∫v 8 xyz dv 

i.e. V = ∫ ∫ ∫_V 8 xyz dxdydz

V = 5 × 3 × 1

V = 15 

∴ Volume is 15 

If aî + bĵ + ck̂ and pî + qĵ + rk̂ are 2 different sides of the rhombus.

Suppose  = aî+bĵ+ck̂ and B = = pî+qĵ+rk̂

Then anyone diagonal of the rhombus is given by 

The other diagonal is given by 

The magnitude of the vector 

The magnitude of the vector diagonal D₁

D₁ = 7 

The other diagonal is given by 

D₂ = - 1î - 2ĵ + 8k̂

The magnitude of the vector diagonal D₂

∴ The length of diagonals of a rhombus is 7 and √69.

Concept:

For a vector F = F₁i + F₂j + F₃k

For irrotational (or) conservative field  (or) Null Vector.

Calculation:

Given that,

 is a conservative field.

k₃ – 3 = 0 ⇒ k₃ = 3

-4 + k₁ = 0 ⇒ k₁ = 4

k₂ – 5 = 0 ⇒ k₂ = 5

The required values are: k₁ = 4, k₂ = 5, k₃ = 3

Green's theorem

• It converts the line integral to a double integral. 
• It transforms the line integral in xy - plane to a surface integral on the same xy - plane.

If M and N are functions of (x, y) defined in an open region then from Green's theorem

The value of  From A(a, 0, 0) to B(a, 0, 2π b) along r̅ = (a cos t) î + (a sin t) ĵ + b t k̂ is: πa (bĵ + ak̂)