√1 + (f′(x))2

√r2– x2

4

–

3

470 solution by radicals

For example, rotating a semicircle about its diameter

produces a

SPHERE

. Rotating a circle about an axis that

does not cut the circle produces a

TORUS

.

INTEGRAL CALCULUS

is used to calculate the volume

of a solid of revolution. For the case of a curve y= f(x)

above an interval [a,b] rotated about the x-axis, subdivide

[a,b] into small segments given by points a,x1,x2,…, xn,b.

The region under the curve is approximated as a collec-

tion of rectangles of heights f(xi) and widths (xi+1 – xi).

When revolved, each such rectangle produces a disc of

radius f(xi) and width (xi+1 – xi), and consequently vol-

ume π(f(xi))2. (xi+1 – xi). The volume of the solid in ques-

tion is then well approximated by the sum of these

individual volumes. Using finer and finer approxima-

tions shows that, in the limit, the true volume Vof the

solid of revolution is given by the integral:

V= ∫b

aπ(f(x))2dx

One can use this integration method, called the

disc method, to show, for example, that the volume of

a sphere of radius ris given by πr3(Use the function

y= over the interval [–r,r].) For solids that

arise from rotating curves about the y-axis, an analo-

gous integration technique, called the cylindrical shell

method, is employed to compute volumes.

By approximating a section of the curve y= f(x) as

a series of straight line segments, revolving each seg-

ment about the x-axis, and computing the surface area

of each small frustum of a cone that results, one can

show that surface area Sof a solid of revolution is

given by the integral:

S= ∫b

a2πf(x)dx

(This assumes that the derivative f′(x) is a continuous

function.) For example, one can use this formula to

show that the surface area of a sphere of radius ris 4πr2.

In the mid-fourth century

C

.

E

., without the aid of

formal integral calculus, Greek mathematician P

APPUS OF

A

LEXANDRIA

made two beautiful geometric discoveries.

If a figure in the plane is revolved around an

axis that does not cut through the figure, then

the volume of the solid of revolution produced

equals the product of the area of the region

and the distance traveled around the axis by

the figure’s centroid.

If a segment of a curve in a plane is

revolved around an axis that does not cut

through the segment, then the surface area of

the solid of revolution produced equals the

product of the

ARC LENGTH

of the segment and

the distance traveled around the axis by the

segment’s centroid.

The centroid of a plane figure or of a segment of a

curve is its

CENTER OF GRAVITY

or balance point if we

imagine the plane figure as made of uniformly dense

material or the curve segment of uniformly dense wire.

The centroid of a circle, regarded as a plane figure, for

example, is its center. The centroid of just its circumfer-

ence is also its center.

If a circle of radius rwith center Runits from the

x-axis is rotated about the x-axis, the distance traveled

by the centroid is 2πR. Consequently, by Pappus’s theo-

rem, the volume Vand the surface area Sof the torus

produced are:

V= (πr2) ×(2πR) = 2π2r2R

and

S= (2πr) ×(2πR) = 4π2rR

See also

HYPERBOLOID

; P

APPUS

’

S THEOREMS

;

PARABOLOID

.

solution by radicals If it is possible to express a

solution to a

POLYNOMIAL

equation in terms of the

COEFFICIENT

s that appear in the equation under the

application of a finite number of additions, subtrac-

tions, multiplications, divisions, and root extractions,

then we say we have a solution by radicals. For exam-

ple, the two solutions of a

QUADRATIC

equation ax2+

bx + c= 0 are given by . These are

solutions by radicals. C

ARDANO

’

S FORMULA

shows that

any solution to a

CUBIC EQUATION

is a solution by radi-

cals, and the work of L

UDOVICO

F

ERRARI

(1522–65)

showed that the same is true of any

QUARTIC EQUATION

.

Certainly some fifth-degree equations have solutions by

radicals (the solution of x5– a= 0, namely, x= 5

√

–

a, is a

solution by radicals), but the general question of

whether or not the solutions of all fifth (or higher)

degree equations are solutions by radicals remained an

important unsolved question for several centuries. This

question is equivalent to asking whether or not there