6.1. Areas Between Curves
6.1. Areas Between Curves: 12, 20, 21, 28, 42, 49, 53, 55
12.
Solve 4x = 12 − y 2 = 4y to get y = −6, 2.
So the intersection points are (−6, −6) and (2, 2).
We write 4x + y 2 = 12 as x = 3 − y 2 /4.
In this, x = 3 − y 2 /4 is on the right side and x = y is on the left side.
Then the area is going to be
A=
Z
y=2
y=−6
"
= −
(xR − xL )dy =
y2
y3
−
12
2
#2
+ 3y Z
−6
2
−6
"
y2
3−
4
!
#
− y dy =
Z
2
−6
"
#
y2
3−
− y dy
4
2
64
= − − 2 + 6 − (18 − 18 − 18) =
3
3
1
20.
√
First of all, we write y = 2 − x as x = 2 − y 2 (y ≤ 0) .
In this, x = y 4 is on the left side and x = 2 − y 2 is on the right side.
They meet when x = y 4 = (2 − x)2 .
√
That is, x2 − 5x + 4 = 0 ⇔ x = 1 or 4 (not satisfying y = 2 − x).
So they must meet at (1,1).
The area is
A=
Z
0
1
(xR − xL )dy =
=
Z
0
1
[(2 − y 2 ) − y 4 ]dy
1
0
=
Z
(2 − y 2 − y 4 )dy
y3 y5
2y −
−
3
5
=2−
!1
1 1
22
− =
3 5
15
2
0
21.
In this, we note that y = tan x is below y =√
2 sin x while x > 0.
And it’s easy to check they meet at ±(π/3, 3).
The area is
A=
π
3
Z
− π3
|2 sin x − tan x|dx
By symmetry,
A=2
Z
0
π
3
(2 sin x − tan x)dx
π
= 2 (−2 cos(x) − ln | sec(x)|)|03
= 2 [(−1 − ln 2) − (−2 − 0)] = 2 − 2 ln 2
3
28.
We separate the area into two parts A and B.
A: Between y = x2 /4 to y = 2x2 , from x = 0 to x = 1. B: Between y = x2 /4
to y = 3 − x, from x = 1 to x = 2. We have
A=
Z
1
0
and
B=
!
x2
dx =
2x −
4
2
Z
1
2
"
Z
0
1
1
7
7x2
7x3 =
dx =
4
12 0 12
2
#
x2
x2 x3 3−x−
dx = 3x −
− 4
2
12 1
1
1
1
11
2
− 3− −
=1−
= .
= 6−2−
3
2 12
12
12
Hence, the total area is
A+B =
11
3
7
+
= .
12 12
2
4
42.
It’s routine to find the intersection points are (1/2, ±1/2). For given y, x is
between 2y 2 and 1 − |y|. i.e. 2y 2 ≤ x ≤ 1 = |y| from y = −1/2 to y = 1/2.
So the area is
A=
Z
1
2
− 12
[(1 − |y|) − (2y 2)]dy.
By symmetry,
A=2
Z
0
1
2
[(1 − y − 2y 2 ]dy
"
y 2 2y 3
=2 y−
−
2
3
# 1
2
0
7
1
1 1
= .
− −
=2
2 8 12
12
5
49.
We find that y 2 = x2 (x +√3) intersects itself at (0,0) and (-3,0).
Solve y, we have y√= ±x x + 3.
√
Note that y = −x x + 3 is above y = x x + 3.
So the area is
A=
Z
0
−3
h
√
i
√
−x x + 3 − x x + 3 dx
= −2
Let u =
√
Z
0
−3
√
x x + 3dx.
x + 3 so we have x = u2 − 3. Because dx = 2udu,
A = −2
Z
= −2
x=0
x=−3
Z
0
√
Z
√
x x + 3dx = −2
√
u= 3
u=0
(u2 − 3)u2udu
4u5
− 4u3
(2u4 − 6u2 )du = −
5
√
√
√
36 3
24 3
=−
− 12 3 =
.
5
5
3
6
!√3
0
53.
We want that
R |c|
−|c| [(c
2
− x2 ) − (x2 − c2 )] dx = 576.
2
Z
|c|
−|c|
c2 − x2 dx = 576
x3
c x−
3
2
|c|3
c |c| −
3
2
!
!|c|
= 288
−|c|
|c|3
− −c |c| +
3
2
4|c|3
= 288
3
3
|c|3 = 288 × = 216
4
So |c| = 6 and then c = ±6.
7
!
= 288
55.
Of course, y = mx and y = x/(x2 + 1) intersect at (0,0).
So we need to find another solution of mx = x/(x2 + 1) beside
q x = 0.
2
2
Solve mx = x/(x + 1) to find x + 1 = 1/m and then x = ± 1/m − 1.
So we need that 1/m − 1 > 0 then
1
1−m
−1>0⇔
> 0 ⇔ m(m − 1) > 0 ⇔ 0 < m < 1.
m
m
We want to find the area between two curves.
It equals
Z √1/m−1 Z √1/m−1 x
x
dx = 2
−
mx
−
mx
dx
√
x2 + 1
− 1/m−1 x2 + 1
0
Z √1/m−1
Z √1/m−1
Z √1/m−1
Z √1/m−1
x
1
=2
dx−2
dx2 −2
mxdx =
mxdx
x2 + 1
x2 + 1
0
0
0
0
√
1
1
2
2 1/m−1
= ln − m
− 1 = m − ln m − 1.
= ln |x + 1| − mx 0
m
m
8