Find the volume of the solid that results when the enclosed region given by the functions
y=16-x^2 and y=0
Answers
Answer:
The volume of the solid generated by
y
=
2
x
,
y
=
x
2
revolved about the x-axis is
64
π
15
.
Explanation:
Graph of the two curves
Graph of the two curves
Revolving the area between these two curves about the x-axis, we end up with something that looks sort of like a cone... a hollow cone, with a curved inside. Now, imagine for a second taking a cross section parallel to the y-z plane, cutting the cone down the middle. The cross-section is going to look like a washer, with an inner radius equal to the height of
y
=
x
2
wherever we are in the solid, and the outer radius equal to
y
=
2
x
at that same
x
.
For reference, the area of a washer is equal to
π
⋅
(
r
outer
)
2
−
π
⋅
(
r
inner
)
2
, if
r
outer
is the outer radius and
r
inner
is the inner radius. This fact should be immediately obvious - we're just subtracting the area of a circle from the area of another circle.
Now, to find the volume of the solid, we need to sum (integrate) the area of each of cross-sectional washer in the solid. This procedure is commonly called the method of washers .
So, we'll use the method of washers to find the volume of this solid.
The general formula is
V
=
∫
b
a
π
(
f
(
x
)
)
2
d
x
−
∫
b
a
π
(
g
(
x
)
)
2
d
x
where
f
(
x
)
is a function giving the outer radius of the washer at any x, and
g
(
x
)
is a function giving the inner radius of the washer. Note that the integrands here represent the areas of circles - and together, they give the area of each infinitesimal washer.
In our case,
f
(
x
)
is equal to
2
x
and
g
(
x
)
is equal to
x
2
.
V
=
∫
b
a
π
(
2
x
)
2
d
x
−
∫
b
a
π
(
x
2
)
2
d
x
Next, we need to worry about the limits of the integrals. Since we're only concerned with rotating the region between
x
2
and
2
x
, the limits should involve the locations where
y
=
x
2
and
y
=
2
x
intersect. Intersection occurs at
x
=
0
and
x
=
2
.
V
=
∫
2
0
π
(
2
x
)
2
d
x
−
∫
2
0
π
(
x
2
)
2
d
x
The difficult part - setting up the appropriate integral with correct limits - is done, so I won't walk through evaluating the integral. It evaluates to
64
π
15
, but if you don't trust me you can evaluate it yourself as an exercise.