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:

Question

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:

BlackTom AI · Accepted Answer

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:

查看解析

登录即可查看完整答案

类似问题

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 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]\(\pi^2\).

Question textThe volume generated by rotating, about the \(X\) axis, the region enclosed by \(y=x^{\frac{3}{2}}\), \(x=1,x=2\), and the \(X\) axis, is Answer 1 Question 5[input] \(\pi \big/\) Answer 2 Question 5[input].

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 4[input]\(\pi^2\).

Question textThe volume generated by rotating, about the \(X\) axis, the region enclosed by \(y=x^2\), \(x=1,x=2\), and the \(X\) axis, is Answer 1 Question 2[input] \(\pi \big/\) Answer 2 Question 2[input].

A region is enclosed by the x-axis, the line 𝑥 = 1 , the line 𝑥 = 9 , and the curve 𝑦 = 𝑥 . What is the volume of the solid generated when this region is rotated around the 𝑥 -axis?

A solid of revolution is formed by rotating the region between the graph of g(y)= y3 3 and the y-axis over 1<y<2, around the y -axis. Which of the following integrals give the volume of the solid?

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单