Calculus Challenge #10 2011-12
Solutions
Cylinders and Volumes of Revolution
One of the classic applications of calculus that
every student learns is that the volume of a right
circular cone created by revolving f  t   at for
0  t  x around the t-axis is one-third of the
corresponding right circular cylinder in which it
sits.
1)
Is something similar true for g  t   at 3 and
h  t   a t ? Find the ratio of the volume of the
solid of rotation on 0  t  x , V  x  , to the volume
of the appropriate cylinder, C  x  , for functions g
and h?
x
For g  t   at , we have the volume of the solid of revolution given by    at
3
 dt 
3 2
  a 2 x7 
7
0
and the volume of the associated cylinder is V   r 2 h    ax3  x . The ratio is
2
  a 2 x7 
7  ax

3 2

x
1
.
7
x
 
For h  t   a t , we have the volume of the solid of revolution given by   a t dt 
0

and the volume of the associated cylinder is V   a x
2)
Show that the ratio of volumes

2
x . The ratio is
2
  a2 x2 

2 a x

  a2 x2 
2

2
x
1
.
2
V  x
is a constant for any power function f  t   at k
C  x
where a  0 and k  0 .
In the general case, f  t   at k we have the volume of the solid of revolution given by
x
   at  dt 
0
k 2
  a 2 x 2 k 1 
2k  1
and the volume of the associated cylinder is V    ax k  x . The ratio
2
is
  a 2 x 2 k 1 
 2k  1  a  x 
2
2k

x
1
. The ratio is a constant for all power functions with positive
2k  1
exponents.
3)
Are there any other families of functions which exhibit this constant ratio of volumes?
The answer is no. Only power functions have this property. The next part of this challenge is to
prove this conjecture.
Let f be a positive, strictly increasing, twice differentiable function on the interval I = (0, b).
V  x
Suppose the ratio
defined above is a constant for all x  I . We must show that
C  x
f  t   at k for some positive values of a and k.
V  x
 Q for some constant Q. Differentiate both sides of this equation with
C  x
CV 
respect to x to show that V 
.
C
a)
Given
V  x
 Q , then by differentiating both sides of
C  x
CV   VC 
CV 
 0 , so CV   VC or V 
the equation, we find
.
2
C
C
It is helpful to suppress the function notation. If
x
Substitute C  x    x  f 2  x  and V  x     f 2  t  dt . Use the 2nd Fundamental
b)
0
Theorem of Calculus to find V . Simplify this result before proceeding. (Notice that C  x 
contains x  f 2  x  . It will get difficult to keep function notation separate from multiplications.
In the work that follows, I suggest you suppress the function notation and write x  f 2 or xy 2
instead of x  f 2  x  . Soon , we will have terms like x f  x  f 2  x  f   x  which is more easily
written as x  f  f 2  f  or x  y  y2  y .)
CV 
and C  x    x  f 2  x  , C  x    f 2  x   2 x  f  f  , V  x     f 2  t  dt , and
C
0
x
Since V 
V   x    f 2  x  by the 2nd Fundamental Theorem of Calculus.
x
So,   f
0
2
t 
 x  f  x    f  x   , and
dt 
2
2
 f 2  x   2 x  f  f 
x

f 2  t  dt 
0
xy 4
xy 3

y 2  2 xyy y  2 xy
by factoring out  and a common factor of y.
If V 
c)
CV 
, show by implicit differentiation that
C
 y  2 xy  3xy 2 y  y 3    xy 3   3 y  2 xy 
.
y2 
2
 y  2 xy
x
From c) above, we have
y 
2
 y  2 xy  3xy
2
0
2
y  y 3   xy 3  3 y  2 xy 
 y  2 xy
 y  2 xy
y

xy 3
f  t  dt 
. Differentiating both sides implicitly, we have
y  2 xy
2
2
2
. There is a factor of y 2 all around, so
  y  2 xy  3xy  y   xy  3 y   2 xy   and expanding,
 4 xyy  4 x 2 y2    3xyy  y 2  6 x 2 y2  2 xyy    3xyy  2 x 2 yy  .
Terms add out and there is a common factor of x, so  2 yy  2 xy2    3xy2  yy    xyy  and
finally, combining like terms we have
 yy  xy     xyy
2
d)
Being very careful with the algebra, show that the equation above simplifies to
x  y  y  x  y2   y  y .
This is just rewriting the expression above.
e)
The resulting equation is a second order differential equation, so we need to integrate
twice. To solve the differential equation above, divide both sides of the equation by x  y  y 
(note that all are positive) and integrate.
If x  y  y  x  y2   y  y , then
y y 1
x  y  y  x  y2  y  y
, so
  .

y y x
xyy
xyy
Integrating both sides with respect to x gives ln  y  ln  y    ln  x   C or
 y 
 y 
ln  x 1 
k
ln     ln  x   C . So,    ec e
 .
x
 y
 y
f)
Solve the resulting equation for
y
and integrate again.
y
ln  xk 
y k
 , we know that ln  y   k ln  x   d and y  ed e
 ax k . Written as a function of t,
y x
we see that only power functions f  t   at k have this property.
If
g)
Have you shown that the only solutions are function of the form f  t   at k for some
positive values of a and k?
Yes. We should that if a function is a power function, it has the property we desire. We also
showed that if a function has the property, then it must be a power function. So, the ratio of
V  x
volumes specified,
, is a constant if and only if the function defining the solid of rotation is
C  x
a power function with positive exponent.