Volume of the Paraboloid

## **Problem**

Derive the volume of a paraboloid of radius

r

{\displaystyle r}

 and height

h

{\displaystyle h}

. Compare the volume of the paraboloid to the volume of the cylinder with equal base and height.

## **Solution**

**Method 1: Using Integration**

Consider half a parabola where the interval of

x

{\displaystyle x}

 is

[
0
,
h
]

{\displaystyle [0, h]}

:

r
(
x
# )
a

x

{\displaystyle r(x)=a\sqrt{x} }

where

a

{\displaystyle a}

 is constant.

At

# x
0

{\displaystyle x = 0}

, we have

r
(
0
# )
a

0

=
0

{\displaystyle r(0) = a\sqrt{0} = 0 }

; at

# x
h

{\displaystyle x=h}

, we have

r
(
h
# )
a

h

=
r

{\displaystyle r(h) = a\sqrt{h} = r }

. Solving for the constant yields

# a

r

h

{\displaystyle a = \frac{r}{\sqrt{h}} }

; thus,

r
(
x
# )

r

h

x

.

{\displaystyle r(x) = \frac{r}{\sqrt{h}} \sqrt{x}.}

Use the disk method to calculate the volume:

V

=
π

∫
a
b

[
r
(
x
)

]
2

d
x

=
π

∫
0
h

r
2

h

x

d
x

=

π

r
2

h

∫
0
h

x

d
x

=

π

r
2

h

[

x
2

2

0

h

]

=

π

r
2

h

2

.

{\displaystyle \begin{align}
V &= \pi \int\_a^b [r(x)]^2 \; dx \\[5 pt]
&= \pi \int\_0^h \frac{r^2}{h} x \; dx \\[5 pt]
&= \frac{\pi r^2}{h} \int\_0^h x \; dx \\[5 pt]
&= \frac{\pi r^2}{h} \left[\frac{x^2}{2}\Biggr|\_{0}^{h}\right] \\[5 pt]
&= \frac{\pi r^2 h}{2}.
\end{align} }

When compared with the volume the cylinder

V

c
y
l
i
n
d
e
r

=
π

r
2

h

{\displaystyle {V}\_{cylinder} = \pi r^2 h }

, the volume of the paraboloid is half the volume of the cylinder.

**Method 2: Using Cavalieri's principle**

Let there be a cylinder of radius

r

{\displaystyle r}

 and height

h

{\displaystyle h}

. Inscribe an inverted paraboloid inside the cylinder. The paraboloid's apex is at the center of the lower base of the cylinder and whose base coincides with the upper base of the cylinder. The equation of this paraboloid's profile is

r
(
x
# )

r

h

x

.

{\displaystyle r(x) = \frac{r}{\sqrt{h}} \sqrt{x}.}

For every height

0
≤
x
≤
h

{\displaystyle 0\le x \le h}

, the disk-shaped cross-sectional areas of the inscribed paraboloid are

π

(

r

h

x

)

2

=
π

r
2

(

x
h

)

.

{\displaystyle \pi\left(\frac{r}{\sqrt{h}} \sqrt{x}\right)^2 = \pi r^2 \left(\frac{x}{h} \right).}

For every height

0
≤
x
≤
h

{\displaystyle 0\le x \le h}

, the washer-shaped cross-sectional area of the cylinder part outside the inscribed paraboloid.

π

r
2

−
π

(

r

h

x

)

2

=
π

r
2

(

1
−

x
h

)

.

{\displaystyle \pi r^2-\pi\left(\frac{r}{\sqrt{h}} \sqrt{x} \right)^2 = \pi r^2 \left(1-\frac{x}{h} \right).}

On the interval

0
≤
x
≤
h

{\displaystyle 0\le x \le h}

, the above two expressions are symmetric about the line

# x

h
2

{\displaystyle x = \frac{h}{2}}

. This implies that for every disk there exists a washer of equal area at the mirror opposite of the line of symmetry

# x

h
2

{\displaystyle x = \frac{h}{2}}

. For example, the area of the disk at

# x
0

{\displaystyle x = 0}

matches with the area of the washer at

# x
h

{\displaystyle x=h}

, the area of the disk at

# x

h
4

{\displaystyle x = \frac{h}{4}}

 matches with the area of the washer at

# x

3
h

4

{\displaystyle x = \frac{3h}{4}}

, the area of the disk at

# x

h
2

{\displaystyle x = \frac{h}{2}}

 matches with the area of the washer at

# x

3
h

2

{\displaystyle x = \frac{3h}{2}}

, the area of the disk at

# x
h

{\displaystyle x=h}

matches with the area of the washer at

# x
0

{\displaystyle x = 0}

, etc.

Therefore, the volume of the inscribed paraboloid must be equal to the volume of the cylinder part outside the inscribed paraboloid. This proves that the volume of the paraboloid is exactly half the volume of its circumscribing cylinder!

V

p
a
r
a
b
o
l
o
i
d

=

1
2

V

c
y
l
i
n
d
e
r

=

π

r
2

h

2

{\displaystyle V\_{paraboloid} = \frac{1}{2} V\_{cylinder} = \frac{\pi r^2h}{2}}