Question textThe volume of the solid of revolution formed by rotating the curve $y=\text{Arcsin}(\frac{x}{2}), 0\leq x\leq 2$ about the $y$-axis is Answer 1 Question 34[input]$\pi^2$.

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We start by restating the problem: find the volume of the solid obtained by rotating the curve y = arcsin(x/2), for 0 ≤ x ≤ 2, about the y-axis.

To form a volume around the y-axis, the shell method is natural: each vertical strip at position x, with height y = arcsin(x/2), generates a cylindrical shell of volume dV = 2π (radius) (height) dx = 2π x arcsin(x/2) dx. Integrating from x = 0 to x = 2 gives the total volume:
V = 2π ∫_{0}^{2} x · arcsin(x/2) dx.

We simplify the integral by substituting u = x/2, so x = 2u and dx = 2 du, with u ranging from 0 to 1:
V = 2π ∫_{0}^{2} x · arcsin(x/2) dx = 2π ∫_{0}^{1} (2u) · arcsin(u) · (2 du) = 8π ∫_{0}^{1} u · arcsin(u) du.

To evaluate I = ∫_{0}^{1} u · arcsin(u) du, use the substitution u = sin t, which implies t goes from 0 to π/2 and du = cos t dt, arcsin(u) = t, u = sin t:
I = ∫_{0}^{π/2} (sin t) · t · (cos t) dt = ∫_{0}^{π/2} t · sin t · cos t dt.

Using the identity sin t cos t = (1/2) sin(2t), we get I = (1/2) ∫_{0}^{π/2} t · sin(2t) dt. Integrate by parts with p = t and dq = sin(2t) dt, so q = -cos(2t)/2 and dp = dt:
∫ t · sin(2t) dt = - (t cos(2t))/2 + (1/2) ∫ cos(2t) dt = - (t cos(2t))/2 + (1/4) sin(2t).
Evaluating from 0 to π/2 yields J = ∫_{0}^{π/2} t · sin(2t) dt = π/4. Therefore I = (1/2) J = π/8.

Substituting back into the volume expression:
V = 8π · I = 8π · (π/8) = π^2.

Thus, the volume of the solid formed by rotating the given curve about the y-axis is π^2.

ENG1090 - MUM S1 2025 Mock Final Exam

Question textThe volume of the solid of revolution formed by rotating the curve \(y=\text{Arcsin}(\frac{x}{2}), 0\leq x\leq 2\) about the \(y\)-axis is Answer 1 Question 34[input]\(\pi^2\).

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