Part II: The Answers
440
Answers
701–800
Find the point of intersection using y = x, x + y = 5, and substitution to get x + x = 5 so
that x = 5
2
. Notice that on the interval 0 5
2
,
( ) , the upper boundary of the region is x and
the lower boundary is y = 0; on 5
2
5
,
( ) the upper boundary of the region is 5 – x and the
lower boundary is y = 0. Because the line of rotation is x = –1, the shell radius is (x + 1).
Therefore, to find the volume using cylindrical shells, you use two integrals:
2
1
2
1 5
2
2
5 4
5 2
5
0
5 2
2
2
π
π
π
π
(
)( )
(
)(
)
x
x dx
x
x dx
x
x dx
x x
+
+
+
−
=
+
(
) +
+
−
(
)
∫
∫
d dx
x
x
x
x
x
5 2
5
0
5 2
3
2
0
5 2
2
3
5 2
5
2
3
2
2 5
2
3
2 1
3
5
∫
∫
=
+
+
+
−
=
π
π
π
2 2
1
2
5
2
2
5 5 2 5
5
3
5 5
2
2 5
2
1
3
3
2
2
3
2
( ) + ( )
+
( )+ ( ) −
− ( ) + ( ) −
π
5 5
2
175
4
3
( )
=
π
702.
2π
To find the volume of a solid obtained by rotating about the y-axis a region that’s under
the curve y = f
(x) and above the x-axis from x = a to x = b with cylindrical shells, use
the integral
V
xf x dx
a
b
= ∫ 2π ( )
More generically, you can use the formula
d
V
x
a
b
= ∫
2π (
) (
)
shell radius shell height
To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance
from x to the line of rotation. To find the shell height, let f
(x) be the function that
bounds the region above and let g(x) be the function that bounds the region below; the
shell height is given by f
(x) – g(x). Note that the limits of integration often correspond
to points of intersection.
In this example, the limits of integration are given as x = 0 and x = π , and the shell
height is sin(x
2
) – 0, so the integral becomes
2
2
0
π
π x
x dx
sin ( )
∫
To evaluate this integral, use the substitution u = x
2
so that du = 2
dx, or 1
2
du x dx
=
. You
can find the new limits of integration by noting that if x = π , then u = ( ) =
π
π
2
and
that if x = 0, then u = 0
2
= 0. Using these values gives you
2
1
2
0
2
0
0
π
π
π
π
π
π
π
sin
c os
(cos
cos )
u du
u
∫
= −
(
)
= −
−
=
440
Answers
701–800
Find the point of intersection using y = x, x + y = 5, and substitution to get x + x = 5 so
that x = 5
2
. Notice that on the interval 0 5
2
,
( ) , the upper boundary of the region is x and
the lower boundary is y = 0; on 5
2
5
,
( ) the upper boundary of the region is 5 – x and the
lower boundary is y = 0. Because the line of rotation is x = –1, the shell radius is (x + 1).
Therefore, to find the volume using cylindrical shells, you use two integrals:
2
1
2
1 5
2
2
5 4
5 2
5
0
5 2
2
2
π
π
π
π
(
)( )
(
)(
)
x
x dx
x
x dx
x
x dx
x x
+
+
+
−
=
+
(
) +
+
−
(
)
∫
∫
d dx
x
x
x
x
x
5 2
5
0
5 2
3
2
0
5 2
2
3
5 2
5
2
3
2
2 5
2
3
2 1
3
5
∫
∫
=
+
+
+
−
=
π
π
π
2 2
1
2
5
2
2
5 5 2 5
5
3
5 5
2
2 5
2
1
3
3
2
2
3
2
( ) + ( )
+
( )+ ( ) −
− ( ) + ( ) −
π
5 5
2
175
4
3
( )
=
π
702.
2π
To find the volume of a solid obtained by rotating about the y-axis a region that’s under
the curve y = f
(x) and above the x-axis from x = a to x = b with cylindrical shells, use
the integral
V
xf x dx
a
b
= ∫ 2π ( )
More generically, you can use the formula
d
V
x
a
b
= ∫
2π (
) (
)
shell radius shell height
To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance
from x to the line of rotation. To find the shell height, let f
(x) be the function that
bounds the region above and let g(x) be the function that bounds the region below; the
shell height is given by f
(x) – g(x). Note that the limits of integration often correspond
to points of intersection.
In this example, the limits of integration are given as x = 0 and x = π , and the shell
height is sin(x
2
) – 0, so the integral becomes
2
2
0
π
π x
x dx
sin ( )
∫
To evaluate this integral, use the substitution u = x
2
so that du = 2
dx, or 1
2
du x dx
=
. You
can find the new limits of integration by noting that if x = π , then u = ( ) =
π
π
2
and
that if x = 0, then u = 0
2
= 0. Using these values gives you
2
1
2
0
2
0
0
π
π
π
π
π
π
π
sin
c os
(cos
cos )
u du
u
∫
= −
(
)
= −
−
=