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

Question

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

BlackTom AI · Accepted Answer

The region described is the area bounded by the curve y = 2 sin(2x), the y-axis (x = 0), and the horizontal line y = 2. First, find where the curve meets y = 2: 2 sin(2x) = 2 implies sin(2x) = 1, so 2x = pi/2 and x = pi/4. Thus the region of interest lies for x from 0 to pi/4, and for each x in this interval the vertical extent is between the curve y = 2 sin(2x) (bottom) and the line y = 2 (top).

Rotating this region about the x-axis forms washers. The outer radius is the distance from the x-axis to the top boundary y = 2, which is R = 2. The inner radius is the distance from the x-axis to the bottom boundary y = 2 sin(2x), which is r = 2 sin(2x).

Therefore, the volume is
V = π ∫_{0}^{π/4} [R^2 − r^2] dx
  = π ∫_{0}^{π/4} [2^2 − (2 sin(2x))^2] dx
  = π ∫_{0}^{π/4} [4 − 4 sin^2(2x)] dx
  = 4π ∫_{0}^{π/4} [1 − sin^2(2x)] dx
  = 4π ∫_{0}^{π/4} cos^2(2x) dx.

To integrate, use cos^2(2x) = (1 + cos(4x))/2. Hence
V = 4π ∫_{0}^{π/4} (1 + cos(4x))/2 dx
  = 2π ∫_{0}^{π/4} [1 + cos(4x)] dx
  = 2π [x + (1/4) sin(4x)]_{0}^{π/4}.

Compute the boundary values: sin(4x) evaluated at x = π/4 is sin(π) = 0, and at x = 0 is sin(0) = 0. Thus
V = 2π [ (π/4) + 0 − (0 + 0) ] = 2π · π/4 = π^2/2.

So the volume formed by revolving the described region about the x-axis equals π^2/2 (in the appropriate volume units).

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

The area bounded by the curve y=2sin(2x)[math] 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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