JNTU B.Tech II MATHEMATICS-II Previous Question Paper (November/December - 2020 - R13)

The work done W by a force field F along a path C is given by the line integral W = ∫_C F · dr. Since the force field F = 3x²i + j + zk is conservative, we can express it as the gradient of a scalar potential function φ(x, y, z) such that F = ∇φ.

Integrating the components:

Combining these, the potential function is φ(x, y, z) = x³ + y + z²/2 + C. The work done is independent of the path and is equal to the difference in the potential function at the endpoints A(x₁, y₁, z₁) and B(x₂, y₂, z₂):

We evaluate the surface integral I = ∫∫_S F · n dS over the surface of the sphere x² + y² + z² = 1 in the first octant. Using the Divergence Theorem, ∫∫_S F · n dS = ∫∫∫_V (∇ · F) dV.

First, calculate the divergence of F = yzi + zxj + xyk:

Since the divergence is zero, the integral over the closed surface is 0. The surface in the first octant consists of the spherical part S₁ and three planar faces S₂, S₃, S₄ on the coordinate planes. Thus, ∫∫_S1 + ∫∫_S2 + ∫∫_S3 + ∫∫_S4 = 0.

On the coordinate planes, the normal vectors are -k, -j, -i respectively. For example, on the xy-plane (z=0), F = 0i + 0j + xyk, so F · n = (xyk) · (-k) = -xy. Integrating over the quarter circle x² + y² ≤ 1:

By symmetry, the integrals over the other two coordinate planes are also -1/8. Therefore, ∫∫_S1 F · n dS - 1/8 - 1/8 - 1/8 = 0, which gives ∫∫_S1 F · n dS = 3/8.