389
Answers
601–700
Answers and Explanations
Note that lines y = a and y = b may not be given, so the limits of integration often correspond to points of intersection of the curves.
In this case, you could integrate with respect to x, but you’d have to solve each equation for y by completing the square, which would needlessly complicate the problem.
Begin by finding the points of intersection by setting the expressions equal to each
other and solving for y:
y
y
y y
y
y
y y
y
2
2
2
3
2
4
0
2
2 0
0 2
− =
−
−
=
− =
=
(
)
,
To determine which function has larger x values for y in the interval (0, 2), pick a
point in the interval and substitute it into each function. So if y = 1, then x = 1
2
– 1 = 0
and x = 3(1) – 1
2
= 2. Therefore, the integral to find the area is
3
4
2
2
2
3
2 2
2
2
0
2
2
0
2
2
3
0
2
y y
y
y dy
y
y dy
y
y
−
(
) − −
(
)




=
−
(
)
=
−






= (
∫
∫
) ) − ( ) − −
(
)
= −
=
2
3
2 2
3
0 0
8 16
3
8
3
640.
4
3
In this case, you can easily solve the two equations for x, so integrating with respect to
y makes sense. If you instead solved the equations for y, you’d have to use more than
one integral when integrating with respect to x to set up the area.
Begin by solving the first equation for x to get x = y
2
– 1. Because y y
≥
−
2
1 on the
interval [0, 1], the integral to find the area is
y
y
dy
y
y
dy
y
y
y
−
−
(
)
(
) =
− +
(
)
=
−
+
(
)
= − +
=
∫
∫
2
0
1
1 2
2
0
1
3 2
3
0
1
1
1
2
3
1
3
2
3
1
3
1
4
3 3