1. The divergence theorem converts
a) Line to surface integral
b) Surface to volume integral
c) Volume to line integral
d) Surface to line integral
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.
2. The triple integral is used to compute volume.
a) True
b) False
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
3. The volume integral is three dimensional.
a) True
b) False
Explanation: Volume integral integrates the independent quantities by three times. Thus it is said to be three dimensional integral or triple integral
4. Find the charged enclosed by a sphere of charge density ρ and radius a.
a) ρ (4πa2)
b) ρ(4πa3/3)
c) ρ(2πa2)
d) ρ(2πa3/3)
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).
5. Evaluate Gauss law for D = 5r2/4 i in spherical coordinates with r = 4m and θ = π/2 as volume integral.
a) 600
b) 588.9
c) 577.8
d) 599.7
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
6. Compute divergence theorem for D = 5r2/4 i in spherical coordinates between r = 1 and r = 2 in volume integral.
a) 80 π
b) 5 π
c) 75 π
d) 85 π
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.
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.
a) 6100 π
b) 6200 π
c) 6300 π
d) 6400 π
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.
8. Using volume integral, which quantity can be calculated?
a) area of cube
b) area of cuboid
c) volume of cube
d) distance of vector
Explanation: The volume integral gives the volume of a vector in a region. Thus volume of a cube can be computed
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.
a) 20
b) 70/3
c) 80/3
d) 30
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.
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.
a) 10
b) 12
c) 14
d) 16
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.