Section 6.3: Volume by Slicing
1. Use the general slicing √
method to find the volume of the solid with a semicircular
base defined by y = 5 cos x on the interval [−π/2, π/2]. The cross-sections of
this solid are squares perpendicular to the x-axis with bases running from the
x-axis to the curve.
√
A(x) = (5 cos x)2 = 25∗cosx
Z
V = 25
π/2
cos xdx
−π/2
π/2
25[sin x]−π/2
=
= 25[1 − −1] = 50
2. A tetrahedron (pyramid with four triangle faces), all of whose edges have length
4.
The p
relationship between the height
p of a tetrahedron and its edge length a is
h = 2/3 ∗ a. So, the height is 4 2/3
Let x be the distance from the top vertex down towrd the base perpendicularly. The cross sections perpendicular
to this axis are equilateral triangles with
√
side length
a
and
height(
3/2)a
.
The
area of an equilateral triangle
p is =
√
(1/2)(a)(( 3/2)a). But, using the relation above, we know that a = 3/2x,
Giving:
√ !
1
3
(a)
a
Area =
2
2
r ! √ r !
1
3
3 3
=
x
x
2
2
2
2
√
3 3 2
x
=
8
So...
Z 4√2/3 √
3 3 2
V =
x dx
8
0
√
√ 4 2/3
3 3 x3
=
8
3 0
√
16 2
=
3
3. Use the disk method to find the volume of the solid generated when the curve
y = cos x is revolved around the x-axis.
Z b
2
V =
πf (x) dx
a
Z
π/2
π cos2 xdx
V =
where cos2x = 1/2(1 + cos 2x)
0
Z
π/2
V = π/2
(1 + cos 2x)dx
o
π/2
= π/2[x + 1/2 sin 2x]0
= π 2/4
4. Use the washer method to find the volume of the solid generated when the region
bounded by y = ex/2 , y = e−x , x = ln 2, x = ln 8 is reveolved about the x-axis.
Z b
2
2
π(f (x) − g(x) )dx
V =
a
Z
ln 8
V =
π(ex/2∗2 − e−x∗2)dx
ln 2
Z
ln 8
(ex − e−2x)dx
ln 2
−2x ln 8
e
= π ex +
2 ln 2
8−2
2−2
= π 8+
−2−
2
2
753π
=
128
V = π
5. Use the washer method method to find the volume of the solid generated when
region bounded by y = x, y = 2x, y = 6 is revolved about the y-axis.
Z b
2
2
π f (y) − g(y) dy
V =
a
Z
V =
6
π y 2 − (y/2)2 dy
0
Z
6
2
2
y − y /4 dy
V = π
Z0 6
3/4y 2 dy
0 3 6
y
= π
4 0
= 54π
= π