JNTU B.Tech II MATHEMATICS-II Previous Question Paper (October - 2020 - R13)

Stoke's Theorem states that the surface integral of the curl of a vector field F over a surface S is equal to the line integral of F around the boundary curve C: ∮_C F·dr = ∬_S (∇ × F)·n dS.

Given F = (x²+y²)i - 2xyj, we first compute the curl of F:

Since the rectangle lies in the xy-plane, the unit normal vector n is k. Thus, (∇ × F)·n = -4y. The surface integral over the rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b is:

Next, we evaluate the line integral ∮_C F·dr along the four sides of the rectangle:

1. Along C₁ (y=0, 0≤x≤a): r = xi, dr = dxi. F·dr = x²dx. ∫₀^a x²dx = a³/3.
2. Along C₂ (x=a, 0≤y≤b): r = ai + yj, dr = dyj. F·dr = -2ay dy. ∫₀^b -2ay dy = -ab².
3. Along C₃ (y=b, a≥x≥0): r = xi + bj, dr = dxi. F·dr = (x²+b²)dx. ∫_a⁰ (x²+b²)dx = -(a³/3 + ab²).
4. Along C₄ (x=0, b≥y≥0): r = yj, dr = dyj. F·dr = 0.

Summing these integrals: a³/3 - ab² - a³/3 - ab² = -2ab². Since both sides equal -2ab², Stoke's Theorem is verified.

Green's theorem states that if C is a positively oriented, piecewise smooth, simple closed curve in a plane and D is the region bounded by C, then for functions P(x, y) and Q(x, y) having continuous partial derivatives on an open region containing D:

To verify for P = x² - xy³ and Q = y² - 2xy, where C is bounded by y = x² and y = x, we first evaluate the line integral. The intersection points are (0,0) and (1,1). Along y = x² (from 0 to 1), dy = 2x dx:

Along y = x (from 1 to 0), dy = dx:

Summing these, the line integral is -11/120 + 24/120 = 13/120. Now evaluate the double integral where ∂Q/∂x = -2y and ∂P/∂y = -3xy²: