Test: Multiple Integrals - Civil Engineering (CE) MCQ

Concept:

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’.

Calculation:

Given:

F = xi + yj + zk

Stokes theorem:

It states that the line integral of a vector field  around any closed surface C is equal to the surface integral of the normal component of the curl of vector  over an unclosed surface ‘S’.

Green's theorem:

Calculation:

To cover shaded region 

x → 0 to 4 

y → -x to x 

I = 2× 4⁴ = 512

Concept:

Let’s change the Cartesian coordinate limits into cylindrical planar polar coordinate,

⇒ x = r cosθ; y = r sinθ

Replace dx.dy = rdθ .dr

Calculation:

Given:

θ → 0 to π; r → 0 to 5

Then the limit of the double integration will be,

⇒ I = - [cos π – cos 0] × 625

⇒ I = 1250

Concept:

The triple integral can be evaluated as repeated integral:

First, f(x, y, z) is integrated w.r.t z between limits z₁ and z₂ keeping x and y constant, then integrated w.r.t y between limits y₁ and y₂ keeping x constant.
The result is then integrated w.r.t x.

Calculation:

Given:

= 

Concept:

Triple integral can be evaluated as repeated integral:

First, f(x, y, z) is integrated w.r.t z between limits z₁ and z₂ keeping x and y constant, then integrated w.r.t y between limits y₁ and y₂ keeping x constant. The result is then integrated w.r.t x.

Calculation:

= 

= 

= 

Given,

and x² + y² ≤ 4

Putting x = r.cosθ, y = r.sinθ and dx.dy = r.dr.dθ

Calculation

Given equations are: y² = 9x, x – y + 2 = 0

By solving the above two equations,

The point of intersection of the two curves are: (1, 3) and (4, 6)

Now, the graph is shown below.

By considering the horizontal strip,

The limits of y are:  3 to 6

The limits of x are: (y – 2) to y²/9

Now, the required area is =∫∫dxdy

= 1/2

Concept:

Ellipse 

Length major axis of ellipse = 2a

Length of minor axis of ellipse = 2b

Area = 

Calculation:

For the first quadrant, take a vertical strip as shown. Here, y coordinate varies from 0 to

Also, the x-coordinate varies from 0 to a

∴ Area = 

= 

= 

= 

∴ The total area of ellipse = 4 × πab/4 = πab units

We have the integration given as,

Placing the limits we get,

Hence the required value of integration will be 26/105.

Concept:

In a double integral with variable limits, the change of order of integration can sometimes make the evaluation easy.

Calculation:

Since the limit of y is constant hence, we can say that a horizontal strip was taken for integration now for change of order we will take vertical strip as shown in the following diagram.

∴  is the correct solution.

y² = 4ax & x² = 4ay

We have to find shaded region area.

So area drawn by y= x² / 4a on x-axis = A₁ (say)

Then shaded area = |A₁ – A₂|

So, 

⇒ 

⇒ Shaded area = 

= 1

Concept:

The volume of solid generated by revolving around y-axis

V = ∫πx²dy         ----(1)

Volume generated by revolving around x-axis

V = ∫πy²dx         ----(2)

Calculations:

From the equation of ellipse

Subtitling in (1)

Given

a = 3

b = 2

V = 43π3²2 = 24π

= 75.39

Concept:

Since the inside limit is in terms of x, therefore we have to integrate first the 'y' terms and convert whole expression in terms of x.

Calculation:

Given:

= 1/3