   Chapter 5.2, Problem 8E

Chapter
Section
Textbook Problem

# Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. y = 6 − x 2 ,   y = 2 ; about the x-axis

To determine

To find:

The volume of the solid obtained by rotating the region bounded by the given curves about the x axis and sketch the region, the solid, and a typical disk or washer.

Explanation

1) Concept:

i. If the cross section is a washer with the inner radius rin and the outer radius rout, then the area of the washer is obtained by subtracting the area of the inner disk from the area of the outer disk.

ii. The volume of a solid of revolution about the x-axis is

V= abA(x)dx

2) Given:

y=6-x2, y=2 ; about the x- axis

3) Calculation:

The region bounded by y=6-x2 and  y=2 and solid obtained by rotation about the x- axis are shown below.

Here, the region is rotated about the x – axis, so the cross-section is perpendicular to the x-axis.

A cross section of the solid is the washer with inner radius 2 and outer radius 6-x2.

So, its cross sectional area becomes

Ax=π6-x22-4=π36-12x2+x4-4

Ax=πx4-12x2+32

The region of integration is bounded by y=6-x2 and y=2,

At point of intersection of the curve and x-axis,

6-x2=2

x2=4

x=±2

The solid lies between  x=-2 and

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