SME
Calculus II
Volume, arc length and work
Use this for “The region under y = 2x² from 0 to 1 is turned about the x-axis. Find the volume of the solid.”
Diagram
The region under y = √x on [0, 4], turned about the x-axis.
At x = 2 the slice is a disk of radius f(2) = 1.414, area π × 1.414² = 6.283.
V = ∫ πf² dx from 0 to 4 = 25.13.
This is an example. Type your own values to start a new problem.
∫
0
π
(
x
)
2
dx
=
V
x
Slice at
example
r
Radius of the slice
worked out (not typed)
Formulas
r
=
c
×
x
m
1.4142
=
1
×
2
0.5
V
=
π
×
c
2
2 ×
m
+ 1
×
b
2 ×
m
+ 1
25.1327
=
π
×
1
2
2 × 0.5 + 1
×
4
2 × 0.5 + 1
Assumptions
•
A slice at x, dx thick, is a disk whose radius is the curve’s height f(x), so its volume is π f(x)² dx.
•
Adding the disks from 0 to b gives V = ∫ from 0 to b of π f(x)² dx, here πc²x²ᵐ integrated.
•
With m = 1 the solid is a cone of radius cb and height b, so V = πr²h ÷ 3.
Step-by-step
WE KNOW
c = 1, m = 0.5, b = 4, x = 2
FIND
radius of the slice (r), volume of the solid (V)
Step 1 · Find radius of the slice (r)
Use r = c × xᵐ
The
slice
at
x
is
a
disk
whose
radius
is
the
curve’s
height
there,
f
(
x
)
=
c
·xᵐ.
r
=
1
×
2
0.5
r
=
1.4142
Step 2 · Find volume of the solid (V)
Use V = π × c² ÷ (2 × m + 1) × b2 × m + 1
Square
the
radius:
π
(
c
·xᵐ)
2
=
πc²
x
2
m
.
Raise
the
power
by
one
and
divide
by
the
new
power,
then
take
the
value
at
0
from
the
value
at
b
.
V
=
π
×
1
2
2 × 0.5 + 1
×
4
2 × 0.5 + 1
∫
from
0
to
4
of
π
×
x
dx
=
[π
×
x
2
2
]
from
0
to
4
=
25.1327
V
=
25.1327
Check
Put the numbers back in. Both sides should match.
1.4142
=
1
×
2
0.5
✓
25.1327
=
π
×
1
2
2 × 0.5 + 1
×
4
2 × 0.5 + 1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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