The area bounded by the curve $ y=2sin(2x) $, the y-axis and the line y = 2 is rotated about the x-axis. The volume formed is equal to:

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The area bounded by the curve $ y=2sin(2x) $, the y-axis and the line y = 2 is rotated about the x-axis. The volume formed is equal to:

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The problem asks for the volume generated when the region bounded by the curve y = 2 sin(2x), the y-axis (x = 0), and the horizontal line y = 2 is rotated about the x-axis.
First, determine the x-limits: the curve reaches y = 2 when 2 sin(2x) = 2, so sin(2x) = 1, which gives 2x = π/2, hence x = π/4. The region starts at x = 0 (the y-axis) and ends at x = π/4, with y ranging from 0 up to y = 2 sin(2x).
To find the volume about the x-axis using the disk method, each cross-section perpendicular to the x-axis is a disk of radius equal to the y-value, so the volume is V = π ∫_{0}^{π/4} [y(x)]^2 dx = π ∫_{0}^{π/4} (2 sin(2x))^2 dx.
This simplifies to V = 4π ∫_{0}^{π/4} sin^2(2x) dx. Using the identity sin^2(θ) = (1 − cos(2θ))/2 with θ = 2x gives sin^2(2x) = (1 − cos(4x))/2, so:
V = 4π ∫_{0}^{π/4} (1 − cos(4x))/2 dx = 2π ∫_{0}^{π/4} (1 − cos(4x)) dx.
Integrating, ∫ (1 − cos(4x)) dx = x − (1/4) sin(4x). Evaluating from 0 to π/4:
V = 2π [ x − (1/4) sin(4x) ]_{0}^{π/4} = 2π [ (π/4) − (1/4) sin(π) − (0 − 0) ].
Since sin(π) = 0, this becomes V = 2π (π/4) = π^2 / 2.
So the derived volume expression is π^2/2. If the answer choices included common numerical forms, this value corresponds to the correct one among typical options that represent π^2/2. The calculation relies on setting up the proper x-limits from the region, applying the disk method, and simplifying using the double-angle identity for sine squared.

MUF0102 Adv. Mathematics Unit 2 - Semester 2, 2025 3.8 Applications of calculus quiz

The area bounded by the curve \( y=2sin(2x) \), the y-axis and the line y = 2 is rotated about the x-axis. The volume formed is equal to:

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