Test: Multiple Integrals

# Test: Multiple Integrals

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## 15 Questions MCQ Test Engineering Mathematics | Test: Multiple Integrals

Test: Multiple Integrals for Civil Engineering (CE) 2023 is part of Engineering Mathematics preparation. The Test: Multiple Integrals questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Multiple Integrals MCQs are made for Civil Engineering (CE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Multiple Integrals below.
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Test: Multiple Integrals - Question 1

### The following surface integral is to be evaluated over a sphere for the given steady vector field, F = xi + yj + zk defined with respect to a Cartesian coordinate system having i, j, and k as unit base vectors.  , Where S is the sphere, x2 + y2 + z2 = 1 and n is the outward unit normal vector to the sphere. The value of the surface integral is

Detailed Solution for Test: Multiple Integrals - Question 1

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:

*Answer can only contain numeric values
Test: Multiple Integrals - Question 2

### The value of the integral , where D is the shaded triangular region shown in the diagram, is _____ (rounded off to the nearest integer).

Detailed Solution for Test: Multiple Integrals - Question 2

Calculation:

To cover shaded region

x → 0 to 4

y → -x to x

I = 2× 44 = 512

*Answer can only contain numeric values
Test: Multiple Integrals - Question 3

### A definite double integral is given below, then, evaluation of the double integral over the region R will be __ Where R is the region on X - Y plane for the function given as,   and r ∈ [0, 5]

Detailed Solution for Test: Multiple Integrals - Question 3

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

Test: Multiple Integrals - Question 4

Evaluate

Detailed Solution for Test: Multiple Integrals - Question 4

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:

Test: Multiple Integrals - Question 5

The value of integral  is

Detailed Solution for Test: Multiple Integrals - Question 5

Test: Multiple Integrals - Question 6

Evaluate the following integral  .

Detailed Solution for Test: Multiple Integrals - Question 6

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:

*Answer can only contain numeric values
Test: Multiple Integrals - Question 7

The integral , where D denotes the disc ��2 + ��2 ≤ 4, evaluates to__________.

Detailed Solution for Test: Multiple Integrals - Question 7

Given,

and x2 + y2 ≤ 4

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

Test: Multiple Integrals - Question 8

The area bounded by the curves y2 = 9x, x – y + 2 = 0 is given by

Detailed Solution for Test: Multiple Integrals - Question 8

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

Test: Multiple Integrals - Question 9

The area of an ellipse represented by an equation  is

Detailed Solution for Test: Multiple Integrals - Question 9

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

Test: Multiple Integrals - Question 10

The integral  equals to

Detailed Solution for Test: Multiple Integrals - Question 10

We have the integration given as,

Placing the limits we get,

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

Test: Multiple Integrals - Question 11

Change the order of integration in

Detailed Solution for Test: Multiple Integrals - Question 11

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.

Test: Multiple Integrals - Question 12

The area between the parabolas y2 = 4ax and x2 = 4ay is

Detailed Solution for Test: Multiple Integrals - Question 12

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 =

*Answer can only contain numeric values
Test: Multiple Integrals - Question 13

Let , where R is the region shown in the figure and c = 6 × 10-4. The value of I equals________. (Give the answer up to two decimal places.)

Detailed Solution for Test: Multiple Integrals - Question 13

= 1

*Answer can only contain numeric values
Test: Multiple Integrals - Question 14

An ellipse is revolved around the y-axis. The volume generated by the solid of revolution if a = 3 and b = 2 is______

Detailed Solution for Test: Multiple Integrals - Question 14

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

Test: Multiple Integrals - Question 15

The value of , where dA indicate small area in xy - plane, is

Detailed Solution for Test: Multiple Integrals - Question 15

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

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## Engineering Mathematics

47 videos|119 docs|75 tests