(a) Verify Gauss divergence theorem for F⃗ = 2x^2yî - y^2ĵ + 4xz^2k̂ taken over the region in the first octant bounded by y² +…

(a) Verification of the divergence theorem

F = 2x²y î − y² ĵ + 4xz² k̂; the region is 0 ≤ x ≤ 2, y, z ≥ 0, y² + z² ≤ 9.

Volume integral: div F = 4xy − 2y + 8xz.

∭ div F dV = ∫₀² ∫₀³ ∫₀^√(9−y²) (4xy − 2y + 8xz) dz dy dx = ∫₀² ∫₀³ [(4xy − 2y)√(9 − y²) + 4x(9 − y²)] dy dx.

With ∫₀³ y√(9 − y²) dy = 9 and ∫₀³ (9 − y²) dy = 18, the inner integral is 36x − 18 + 72x = 108x − 18. Then ∫₀² (108x − 18) dx = 216 − 36 = 180.

Surface integral:

Curved surface y² + z² = 9: take y = 3cos θ, z = 3sin θ, 0 ≤ θ ≤ π/2, n = (0, cos θ, sin θ), dS = 3 dθ dx. F·n = −y³/3 + 4xz³/3, so F·n dS = (−27cos³θ + 108x sin³θ) dθ dx. Since ∫₀^(π/2) cos³θ dθ = ∫₀^(π/2) sin³θ dθ = 2/3, the integral is ∫₀² (−18 + 72x) dx = −36 + 144 = 108.

Plane x = 2: n = î, F·n = 8y. Over the quarter disc of radius 3, ∬ 8y dA = 8 ∫₀^(π/2) ∫₀³ r² sin θ dr dθ = 8 × 9 = 72.

Planes x = 0, y = 0, z = 0: F·n equals −2x²y, y² and −4xz² respectively, each zero on its plane, so these contribute 0.

Total surface integral = 108 + 72 = 180, equal to the volume integral. The theorem is verified.

(b) y² log y = xy (dy/dx) + (dy/dx)²

Here y > 0. Put u = log y, so dy/dx = y u′; dividing by y² gives u = x u′ + u′², a Clairaut equation.

Differentiating: u′ = u′ + x u″ + 2u′u″, so (x + 2u′)u″ = 0.

Case u″ = 0: u′ = C, u = Cx + C², so y = e^(Cx + C²), with C arbitrary (the general solution).

Case x + 2u′ = 0: u′ = −x/2, and substituting in the equation gives u = −x²/4, so y = e^(−x²/4) (the singular solution, the envelope of the general family).

All solutions: y = e^(Cx + C²) and y = e^(−x²/4).

(c) Particle on a string

Let θ be the angle of the string from the downward vertical, so the height above the point of projection is a(1 − cos θ). Energy: v² = 2gh − 2ga(1 − cos θ).

Along the string: T − mg cos θ = mv²/a. The circular motion ceases when T = 0, that is v² = −ag cos θ.

Equating, 2g(h − a) + 2ga cos θ = −ag cos θ, so cos θ = −2(h − a)/(3a). The height is

a(1 − cos θ) = a + 2(h − a)/3 = (a + 2h)/3.

At this point v² = 2g(h − a)/3, and the velocity, perpendicular to the string, has vertical component v sin θ. The particle then moves as a projectile, rising a further

v² sin²θ/(2g) = ((h − a)/3)(1 − 4(h − a)²/(9a²)) = (h − a)/3 − 4(h − a)³/(27a²).

The greatest height above the point of projection is

(a + 2h)/3 + (h − a)/3 − 4(h − a)³/(27a²) = h − 4(h − a)³/(27a²) = (4a − h)(a + 2h)²/(27a²),

since 27a²h − 4(h − a)³ = (4a − h)(a + 2h)². The condition a < h < 5a/2 makes −1 < cos θ < 0, so the string goes slack above the horizontal level of the point of suspension.