Part II: The Answers
394
Answers
601–700
Begin by finding the points of intersection by setting the expressions equal to each
other and solving for y:
2
2
4 2
2
2
2
2
2
2
−
=
−
=
=
±
=
y
y
y
y
y
To determine which curve has the larger x values for y in the interval −
(
)
2 2
,
, take a point
inside the interval and substitute it into each function. So if y = 0, then x = 2 – 0
2
= 2 and
x = 0
2
– 2 = –2; therefore, 2 – y
2
> y
2
– 2 on (
, )
− 2 2 . That means the integral to find the area is
2
2
2
2
2
2
−
(
) − −
(
)
−
∫
y
y
dy
By symmetry, you can rewrite the integral as
2
4 2
2
0
2
−
(
)
∫
y dy
Now you can evaluate the integral:
2
4 2
2 4
2
3
2 4 2
2 2
3
0 0
2
0
2
3
0
2
3
−
(
) =
−
=
−
( ) − −
=
∫
y dy
y
y
(
)
8 8 2 8 2
3
16 2
3
−
=
The following figure shows the region bounded by the given curves:
394
Answers
601–700
Begin by finding the points of intersection by setting the expressions equal to each
other and solving for y:
2
2
4 2
2
2
2
2
2
2
−
=
−
=
=
±
=
y
y
y
y
y
To determine which curve has the larger x values for y in the interval −
(
)
2 2
,
, take a point
inside the interval and substitute it into each function. So if y = 0, then x = 2 – 0
2
= 2 and
x = 0
2
– 2 = –2; therefore, 2 – y
2
> y
2
– 2 on (
, )
− 2 2 . That means the integral to find the area is
2
2
2
2
2
2
−
(
) − −
(
)
−
∫
y
y
dy
By symmetry, you can rewrite the integral as
2
4 2
2
0
2
−
(
)
∫
y dy
Now you can evaluate the integral:
2
4 2
2 4
2
3
2 4 2
2 2
3
0 0
2
0
2
3
0
2
3
−
(
) =
−
=
−
( ) − −
=
∫
y dy
y
y
(
)
8 8 2 8 2
3
16 2
3
−
=
The following figure shows the region bounded by the given curves: