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If a function is described by F = (3x + z, y2 − sin x2z, xz + yex5), then the divergence theorem value in the region 0<x<1, 0<y<3 and 0<z<2 will be
c Explanation: Div (F) = 3 + 2y + x. By divergence theorem, the triple integral of Div F in the region is ∫∫∫ (3 + 2y + x) dx dy dz. On integrating from x = 0->1, y = 0->3 and z = 0- >2, we get 39 units
c
See lessExplanation: Div (F) = 3 + 2y + x. By divergence theorem, the triple integral of Div F in
the region is ∫∫∫ (3 + 2y + x) dx dy dz. On integrating from x = 0->1, y = 0->3 and z = 0-
>2, we get 39 units
The divergence theorem value for the function x2 + y2 + z2 at a distance of one unit from the origin is
d Explanation: Div (F) = 2x + 2y + 2z. The triple integral of the divergence of the function is ∫∫∫(2x + 2y + 2z)dx dy dz, where x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3 units.
d
See lessExplanation: Div (F) = 2x + 2y + 2z. The triple integral of the divergence of the function is
∫∫∫(2x + 2y + 2z)dx dy dz, where x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3
units.
Find the Gauss value for a position vector in Cartesian system from the origin to one unit in three dimensions.
b Explanation: The position vector in Cartesian system is given by R = x i + y j + z k. Div(R) = 1 + 1 + 1 = 3. By divergence theorem, ∫∫∫3.dV, where V is a cube with x = 0->1, y = 0->1 and z = 0->1. On integrating, we get 3 units
b
See lessExplanation: The position vector in Cartesian system is given by R = x i + y j + z k.
Div(R) = 1 + 1 + 1 = 3. By divergence theorem, ∫∫∫3.dV, where V is a cube with x = 0->1,
y = 0->1 and z = 0->1. On integrating, we get 3 units
Evaluate the surface integral ∫∫ (3x i + 2y j). dS, where S is the sphere given by x2 + y2 + z2 = 9
b Explanation: We could parameterise surface and find surface integral, but it is wise to use divergence theorem to get faster results. The divergence theorem is given by ∫∫ F.dS = ∫∫∫ Div (F).dV Div (3x i + 2y j) = 3 + 2 = 5. Now the volume integral will be ∫∫∫ 5.dV, where dV is the volume of the sRead more
b
See lessExplanation: We could parameterise surface and find surface integral, but it is wise to
use divergence theorem to get faster results. The divergence theorem is given by ∫∫ F.dS
= ∫∫∫ Div (F).dV
Div (3x i + 2y j) = 3 + 2 = 5. Now the volume integral will be ∫∫∫ 5.dV, where dV is the
volume of the sphere 4πr3
/3 and r = 3units.Thus we get 180π
Find the area of a right angled triangle with sides of 90 degree unit and the functions described by L = cos y and M = sin x.
d Explanation: dM/dx = cos x and dL/dy = -sin y ∫∫(dM/dx – dL/dy)dx dy = ∫∫ (cos x + sin y)dx dy. On integrating with x = 0->90 and y = 0- >90, we get area of right angled triangle as -180 units (taken in clockwise direction). Since area cannot be negative, we take 180 units
d
See lessExplanation: dM/dx = cos x and dL/dy = -sin y
∫∫(dM/dx – dL/dy)dx dy = ∫∫ (cos x + sin y)dx dy. On integrating with x = 0->90 and y = 0-
>90, we get area of right angled triangle as -180 units (taken in clockwise direction).
Since area cannot be negative, we take 180 units
If two functions A and B are discrete, their Green’s value for a region of circle of radius a in the positive quadrant is
d Explanation: Green’s theorem is valid only for continuous functions. Since the given functions are discrete, the theorem is invalid or does not exist
d
See lessExplanation: Green’s theorem is valid only for continuous functions. Since the given
functions are discrete, the theorem is invalid or does not exist
If two functions A and B are discrete, their Green’s value for a region of circle of radius a in the positive quadrant is
d Explanation: Green’s theorem is valid only for continuous functions. Since the given functions are discrete, the theorem is invalid or does not exist
d
See lessExplanation: Green’s theorem is valid only for continuous functions. Since the given
functions are discrete, the theorem is invalid or does not exist
Calculate the Green’s value for the functions F = y2 and G = x2 for the region x = 1 and y = 2 from origin.
c Explanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(2x – 2y)dx dy. On integrating for x = 0->1 and y = 0->2, we get Green’s value as -2.
c
See lessExplanation: ∫∫(dG/dx – dF/dy)dx dy = ∫∫(2x – 2y)dx dy. On integrating for x = 0->1 and y
= 0->2, we get Green’s value as -2.
The resistivity of a material with resistance 200 ohm, length 10m and area twice that of the length is
c Explanation: Resistance calculated from Ohm’s law and Stoke’s theorem will be R = ρL/A. To get resistivity, ρ = RA/L = 200 X 20/10 = 400.
c
See lessExplanation: Resistance calculated from Ohm’s law and Stoke’s theorem will be R =
ρL/A. To get resistivity, ρ = RA/L = 200 X 20/10 = 400.
The conductivity of a material with current density 1 unit and electric field 200 μV is
d Explanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.
d
See lessExplanation: The current density is given by, J = σE. To find conductivity, σ = J/E =
1/200 X 10-6 = 5000.