(a) Show that the solution of the two-dimensional Laplace's equation ∂²φ(x,y)/∂x² + ∂²φ(x,y)/∂y² = 0, x ∈ (-∞, ∞), y ≥ 0 subject…

(a) Let Φ(k,y) = ∫ from −∞ to ∞ φ(x,y) exp(−ikx) dx, and let F(k) = ∫ from −∞ to ∞ f(x) exp(−ikx) dx. Applying the Fourier transform in x to Laplace's equation ∂²φ/∂x² + ∂²φ/∂y² = 0 gives Φ'' − k²Φ = 0, where the prime denotes the second derivative with respect to y. The solution that decays as y→∞ is Φ(k,y) = F(k) exp(−|k|y). By the inverse Fourier transform, φ(x,y) = (1/2π) ∫ from −∞ to ∞ F(k) exp(−|k|y) exp(ikx) dk = (1/2π) ∫ from −∞ to ∞ f(ξ) [ ∫ from −∞ to ∞ exp(−|k|y) exp(ik(x−ξ)) dk ] dξ. Let s = x−ξ. Since exp(−|k|y) is even, ∫ from −∞ to ∞ exp(−|k|y) exp(iks) dk = 2 ∫ from 0 to ∞ exp(−ky) cos(ks) dk = 2y/(y²+s²). Therefore, φ(x,y) = (1/2π) ∫ from −∞ to ∞ f(ξ) · 2y/(y²+(x−ξ)²) dξ = (y/π) ∫ from −∞ to ∞ f(ξ) dξ/[y²+(x−ξ)²]. As y→0⁺, the kernel y/[π(y²+(x−ξ)²)] tends to the Dirac delta δ(x−ξ), so φ(x,0⁺)=f(x). The decay condition φ→0 as |x|→∞ and y→∞ ensures uniqueness. This is the Poisson integral formula for the upper half-plane. Condition of validity: f continuous and bounded (or absolutely integrable) and vanishing at infinity as required.

(b) Simplify: Y = ABC̄ + BC̄ + ĀB = BC̄(A+1) + ĀB = BC̄ + ĀB = B(C̄+Ā) = B(AC)̄. Thus Y=1 exactly when B=1 and not(A=1 and C=1). The logical circuit is: feed A and C into a NAND gate to obtain (AC)̄; feed this output and B into an AND gate. Equivalently, invert A and C, OR the inverted outputs, then AND with B. Circuit: A and C enter a two-input NAND gate; its output and B enter a two-input AND gate, whose output is Y.

Now align the given bit sequences from left to right:

• Position 1: A=1, B=0, C=1 ⇒ Y=0
• Position 2: A=0, B=0, C=1 ⇒ Y=0
• Position 3: A=0, B=1, C=0 ⇒ Y=1
• Position 4: A=0, B=1, C=0 ⇒ Y=1
• Position 5: A=1, B=1, C=0 ⇒ Y=1
• Position 6: A=1, B=1, C=1 ⇒ Y=0
• Position 7: A=1, B=0, C=0 ⇒ Y=0
• Position 8: A=1, B=0, C=0 ⇒ Y=0 Final output: Y = 00111000.

(c) Consider the quadrant x≥0, y≥0, x²/a² + y²/b² ≤ 1. The density is ρ = kxy. Use elliptic polar coordinates: x = a r cos θ, y = b r sin θ, 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2. The Jacobian is J = ab r. The mass is M = ∫∫ ρ dx dy = ∫ from 0 to π/2 ∫ from 0 to 1 k (a r cos θ)(b r sin θ) (ab r) dr dθ = k a² b² ∫ from 0 to 1 r³ dr ∫ from 0 to π/2 sin θ cos θ dθ = k a² b² (1/4)(1/2) = k a² b² / 8. Hence k = 8M/(a² b²). The moment of inertia about the line through the centre perpendicular to the plane (the z-axis) is I = ∫∫ ρ (x²+y²) dx dy = ∫∫ kxy (x²+y²) dx dy. Substituting the elliptic polar coordinates, x²+y² = r² (a² cos² θ + b² sin² θ). Thus I = k a² b² ∫ from 0 to 1 r⁵ dr ∫ from 0 to π/2 sin θ cos θ (a² cos² θ + b² sin² θ) dθ. The r-integral is 1/6. For the θ-integral, put u = sin² θ, so du = 2 sin θ cos θ dθ. Then ∫ from 0 to π/2 sin θ cos θ (a² cos² θ + b² sin² θ) dθ = 1/2 ∫ from 0 to 1 [a²(1−u) + b²u] du = 1/2 [a² − a²/2 + b²/2] = (a² + b²)/4. Therefore I = k a² b² (1/6)(a² + b²)/4 = k a² b² (a² + b²)/24. Using k = 8M/(a² b²), I = (8M/(a² b²)) · a² b² (a² + b²)/24 = M (a² + b²)/3. Final answer: I = M(a² + b²)/3.