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,x 1 ,x 2 ,…, x n ,b.
The region under the curve is approximated as a collection of rectangles of heights f(x i ) and widths (x i+1 – x i ).
When revolved, each such rectangle produces a disc of
radius f(x i ) and width (x i+1 – x i ), and consequently volume π(f(x i ))
2
. (x i+1 – x i ). The volume of the solid in question is then well approximated by the sum of these
individual volumes. Using finer and finer approximations shows that, in the limit, the true volume V of the
solid of revolution is given by the integral:
V = ∫
b
a
π(f(x))
2 dx
One can use this integration method, called the
disc method, to show, for example, that the volume of
a sphere of radius r is given by πr
3 (Use the function
y =
over the interval [–r,r].) For solids that
arise from rotating curves about the y-axis, an analogous 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 segment about the x-axis, and computing the surface area
of each small frustum of a cone that results, one can
show that surface area S of a solid of revolution is
given by the integral:
S = ∫
b
a
2πf(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 r is 4πr
2
.
In the mid-fourth century C.E., without the aid of
formal integral calculus, Greek mathematician PAPPUS OF
ALEXANDRIA 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 circumference is also its center.
If a circle of radius r with center R units from the
x-axis is rotated about the x-axis, the distance traveled
by the centroid is 2πR. Consequently, by Pappus’s theorem, the volume V and the surface area S of the torus
produced are:
V = (πr
2
) × (2πR) = 2π
2
r
2
R
and
S = (2πr) × (2πR) = 4π
2
rR
See also HYPERBOLOID; PAPPUS’S THEOREMS;
PARABOLOID.
solution by radicals If it is possible to express a
solution to a POLYNOMIAL equation in terms of the
COEFFICIENTs that appear in the equation under the
application of a finite number of additions, subtractions, multiplications, divisions, and root extractions,
then we say we have a solution by radicals. For example, the two solutions of a QUADRATIC equation ax
2 +
bx + c = 0 are given by
. These are
solutions by radicals. CARDANO’S FORMULA shows that
any solution to a CUBIC EQUATION is a solution by radicals, and the work of LUDOVICO FERRARI (1522–65)
showed that the same is true of any QUARTIC EQUATION.
Certainly some fifth-degree equations have solutions by
radicals (the solution of x
5 – 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
x
b
b
ac
a
=
− ±
−
2
4
2
√1 + (f′(x))
2
√r
2 – x
2
4
–
3
470 solution by radicals
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,x 1 ,x 2 ,…, x n ,b.
The region under the curve is approximated as a collection of rectangles of heights f(x i ) and widths (x i+1 – x i ).
When revolved, each such rectangle produces a disc of
radius f(x i ) and width (x i+1 – x i ), and consequently volume π(f(x i ))
2
. (x i+1 – x i ). The volume of the solid in question is then well approximated by the sum of these
individual volumes. Using finer and finer approximations shows that, in the limit, the true volume V of the
solid of revolution is given by the integral:
V = ∫
b
a
π(f(x))
2 dx
One can use this integration method, called the
disc method, to show, for example, that the volume of
a sphere of radius r is given by πr
3 (Use the function
y =
over the interval [–r,r].) For solids that
arise from rotating curves about the y-axis, an analogous 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 segment about the x-axis, and computing the surface area
of each small frustum of a cone that results, one can
show that surface area S of a solid of revolution is
given by the integral:
S = ∫
b
a
2πf(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 r is 4πr
2
.
In the mid-fourth century C.E., without the aid of
formal integral calculus, Greek mathematician PAPPUS OF
ALEXANDRIA 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 circumference is also its center.
If a circle of radius r with center R units from the
x-axis is rotated about the x-axis, the distance traveled
by the centroid is 2πR. Consequently, by Pappus’s theorem, the volume V and the surface area S of the torus
produced are:
V = (πr
2
) × (2πR) = 2π
2
r
2
R
and
S = (2πr) × (2πR) = 4π
2
rR
See also HYPERBOLOID; PAPPUS’S THEOREMS;
PARABOLOID.
solution by radicals If it is possible to express a
solution to a POLYNOMIAL equation in terms of the
COEFFICIENTs that appear in the equation under the
application of a finite number of additions, subtractions, multiplications, divisions, and root extractions,
then we say we have a solution by radicals. For example, the two solutions of a QUADRATIC equation ax
2 +
bx + c = 0 are given by
. These are
solutions by radicals. CARDANO’S FORMULA shows that
any solution to a CUBIC EQUATION is a solution by radicals, and the work of LUDOVICO FERRARI (1522–65)
showed that the same is true of any QUARTIC EQUATION.
Certainly some fifth-degree equations have solutions by
radicals (the solution of x
5 – 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
x
b
b
ac
a
=
− ±
−
2
4
2
√1 + (f′(x))
2
√r
2 – x
2
4
–
3
470 solution by radicals