Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Test  >  Electromagnetic Fields Theory (EMFT)  >  Test: Volume Integral - Electrical Engineering (EE) MCQ

Volume Integral - Free MCQ Practice Test with solutions, GATE EE Electromagnetic


MCQ Practice Test & Solutions: Test: Volume Integral (10 Questions)

You can prepare effectively for Electrical Engineering (EE) Electromagnetic Fields Theory (EMFT) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Volume Integral". These 10 questions have been designed by the experts with the latest curriculum of Electrical Engineering (EE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

Sign up on EduRev for free to attempt this test and track your preparation progress.

Test: Volume Integral - Question 1

The divergence theorem converts

Detailed Solution: Question 1

Answer: b
Explanation: The divergence theorem is given by, ∫∫ D.ds = ∫∫∫ Div (D) dv. It is clear that it converts surface (double) integral to volume(triple) integral.

Test: Volume Integral - Question 2

The triple integral is used to compute volume. State True/False 

Detailed Solution: Question 2

Answer: a
Explanation: The triple integral, as the name suggests integrates the function/quantity three times. This gives volume which is the product of three independent quantities.

Test: Volume Integral - Question 3

The volume integral is three dimensional. State True/False

Detailed Solution: Question 3

Answer: a
Explanation: Volume integral integrates the independent quantities by three times. Thus it is said to be three dimensional integral or triple integral.

Test: Volume Integral - Question 4

Find the charged enclosed by a sphere of charge density ρ and radius a. 

Detailed Solution: Question 4

Answer: b
Explanation: The charge enclosed by the sphere is Q = ∫∫∫ ρ dv.
Where, dv = r2 sin θ dr dθ dφ and on integrating with r = 0->a, φ = 0->2π and θ = 0->π, we get Q = ρ(4πa3/3).

Test: Volume Integral - Question 5

Evaluate Gauss law for D = 5r2/2 i in spherical coordinates with r = 4m and θ = π/2 as volume integral.

Detailed Solution: Question 5

Answer: b
Explanation: ∫∫ D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r = 0->4, φ = 0->2π and θ = 0->π/4, we get Q = 588.9.

Test: Volume Integral - Question 6

Compute divergence theorem for D = 5r2/4 i in spherical coordinates between r = 1 and r = 2 in volume integral.

Detailed Solution: Question 6

Answer: c
Explanation: D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r = 1->2, φ = 0->2π and θ = 0->π, we get Q = 75 π.

Test: Volume Integral - Question 7

Compute the Gauss law for D = 10ρ3/4 i, in cylindrical coordinates with ρ = 4m, z = 0 and z = 5, hence find charge using volume integral.

Detailed Solution: Question 7

Answer: d
Explanation: Q = D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 10 ρ2 and dv = ρ dρ dφ dz. On integrating, ρ = 0->4, φ = 0->2π and z = 0->5, we get Q = 6400 π.

Test: Volume Integral - Question 8

Using volume integral, which quantity can be calculated?

Detailed Solution: Question 8

Answer: c
Explanation: The volume integral gives the volume of a vector in a region. Thus volume of a cube can be computed.

Test: Volume Integral - Question 9

Compute the charge enclosed by a cube of 2m each edge centered at the origin and with the edges parallel to the axes. Given D = 10y3/3 j.

Detailed Solution: Question 9

Answer: c
Explanation: Div(D) = 10y2
∫∫∫Div (D) dv = ∫∫∫ 10y2 dx dy dz. On integrating, x = -1->1, y = -1->1 and z = -1->1, we get Q = 80/3.

Test: Volume Integral - Question 10

Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.

Detailed Solution: Question 10

Answer: b
Explanation: Div (D) = 2y
∫∫∫Div (D) dv = ∫∫∫ 2y dx dy dz. On integrating, x = 0->1, y = 0->2 and z = 0->3, we get Q = 12.

10 videos|55 docs|56 tests
Information about Test: Volume Integral Page
In this test you can find the Exam questions for Test: Volume Integral solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Volume Integral, EduRev gives you an ample number of Online tests for practice
Download as PDF