EXAMPLE 4 The region R enclosed by the curves y = 2x and y = x is rotated about the x-axis. Find the volume of the resulting solid. SOLUTION The curves y = 2x and y = x intersect at the points (0, and (2. The region between them, the solid of y-2x rotation, and a cross section perpendicular to the x-axis are shown in the figures. A cross-section in the plane P, has the shape of a washer (an annular ring) with inner radius and an outer radius ,so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle: y-? A(x) = 4nx - m(x2)2 = Therefore, we have (4x - x) dx V = A(x) dx = A(X)

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.2: Properties Of Division
Problem 53E
icon
Related questions
Question

please box all answers in order of answers

EXAMPLE 4
The region R enclosed by the curves y = 2x and y = x2 is rotated about the x-axis. Find the volume of the resulting solid.
SOLUTION
The curves y = 2x and y = x2 intersect at the points (0,
and
The region between them, the solid of
y = 2 x
rotation, and a cross section perpendicular to the x-axis are shown in the figures. A cross-section in the plane P, has the shape of a washer (an annular ring)
with inner radius
and an outer radius
, so we find the cross-sectional area by subtracting the area of the inner circle from the
area of the outer circle:
1)
y = x²
A(x) = 4Tx2 – T(x²)² =
Therefore, we have
V =
A(x) dx =
T(4x2 - x*) dx
A(X)
2 x
Transcribed Image Text:EXAMPLE 4 The region R enclosed by the curves y = 2x and y = x2 is rotated about the x-axis. Find the volume of the resulting solid. SOLUTION The curves y = 2x and y = x2 intersect at the points (0, and The region between them, the solid of y = 2 x rotation, and a cross section perpendicular to the x-axis are shown in the figures. A cross-section in the plane P, has the shape of a washer (an annular ring) with inner radius and an outer radius , so we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle: 1) y = x² A(x) = 4Tx2 – T(x²)² = Therefore, we have V = A(x) dx = T(4x2 - x*) dx A(X) 2 x
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Basics of Inferential Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
Geometry
ISBN:
9781337614085
Author:
Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:
Cengage,