Part II: The Answers
422
Answers
601–700
In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional
slice is a washer, you can find A(x) by using
A
r
r
=
−
= ( ) − ( )
π
π
π
π
(
)
(
)
outer radius
inner radius
out
i n
2
2
2
2
where the outer radius, r out , is the distance of the function farther away from the line of
rotation and the inner radius, r in , is the distance of the function closer to the line of
rotation at a particular value of x.
Note that the lines x = 0 and x = 1 correspond to the limits of integration. Because
you’re rotating the region about the line y = –1, you have r out = e
–x
+ 1 and r in = 0 + 1 = 1.
Therefore, the integral to find the volume becomes
π
π
π
e
d x
e
e
dx
e
e
x
x
x
x
x
−
−
−
−
−
+
(
) − +
=
+
(
)
=
−
−
∫
∫
1
0 1
2
2
2
2
2
0
1
2
0
1
2
(
)
=
−
−
− − −
( )
=
−
−
−
−
0
1
2
1
2
2
2
1
2
2
5
2
1
2
2
π
e
e
e
e
π π
682.
1 024
3
,
Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x)
is the cross-sectional area of S in the plane P x that goes through x and is perpendicular
to the x-axis, where A is a continuous function, then the volume of S is
V
A x dx
a
b
= ∫ ( )
Here, you want to find an expression for A(x).
The base is a circle that has a radius of 4 and is centered at the origin, so the equation
of the circle is x
2
+ y
2
= 16.
422
Answers
601–700
In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional
slice is a washer, you can find A(x) by using
A
r
r
=
−
= ( ) − ( )
π
π
π
π
(
)
(
)
outer radius
inner radius
out
i n
2
2
2
2
where the outer radius, r out , is the distance of the function farther away from the line of
rotation and the inner radius, r in , is the distance of the function closer to the line of
rotation at a particular value of x.
Note that the lines x = 0 and x = 1 correspond to the limits of integration. Because
you’re rotating the region about the line y = –1, you have r out = e
–x
+ 1 and r in = 0 + 1 = 1.
Therefore, the integral to find the volume becomes
π
π
π
e
d x
e
e
dx
e
e
x
x
x
x
x
−
−
−
−
−
+
(
) − +
=
+
(
)
=
−
−
∫
∫
1
0 1
2
2
2
2
2
0
1
2
0
1
2
(
)
=
−
−
− − −
( )
=
−
−
−
−
0
1
2
1
2
2
2
1
2
2
5
2
1
2
2
π
e
e
e
e
π π
682.
1 024
3
,
Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x)
is the cross-sectional area of S in the plane P x that goes through x and is perpendicular
to the x-axis, where A is a continuous function, then the volume of S is
V
A x dx
a
b
= ∫ ( )
Here, you want to find an expression for A(x).
The base is a circle that has a radius of 4 and is centered at the origin, so the equation
of the circle is x
2
+ y
2
= 16.