# 12.11 Application of Integration

```Course: Accelerated Engineering Calculus I
12.11
Instructor: Michael Medvinsky
Application of Integration
12.11.1
Area between curves
Consider that we interesting to compute an area between two curves. How can we do
that? The idea is similar to the one used to find the area of
a
1
1
1
a+b
trapeze. For a example ha − hc = h ( b + c ) − hc = hb + hc = h
h
2
2
2
2
b
c
Let’s rewrite this example so the area will be between 2 curves
⎧x −b
h x≥b
⎪
, thus
y1 = h and y2 = ⎨ a − b
⎪⎩0
else
x−b
h ( x − b)
a
∫0 y1 − y2 dx = ∫0 h dx − ∫b a − b h dx = hx 0 − 2 a − b
a
a
a
2 a
= ha −
b
a−b
a+b
h=h
2
2
Thm: The area of region bounded by continuous curves y=f(x) and y=g(x) is given by
and the lines x=a and x=b, where f(x)&gt;g(x) for all x in [a,b] is given by
b
A = ∫ f ( x ) − g ( x ) dx
a
If f(x) and g(x) are velocities of moving objects and f(a)=g(a) the area is interpreted as
distance between the objects at time x=b.
Ex 1.
To find the area enclosed between parabolas 5 − x 2 and x 2 − 3 we first
need to find the points of intersection, which will be the limits of the integral.
Thus: 5 − x 2 = x 2 − 3 ⇒ 2x 2 = 5 + 3 = 8 ⇒ x 2 = 4 ⇒ x = &plusmn;2 and so
2
A=
2
2
2
2
∫ ( 5 − x ) − ( x − 3) dx = 2 ∫ 4 − x dx =
−2
−2
2
⎛
x ⎞
8⎞ ⎛
8⎞⎞
16 ⎞ 64
⎛⎛
⎛
2 ⎜ 4x − ⎟ = 2 ⎜ ⎜ 8 − ⎟ − ⎜ −8 + ⎟ ⎟ = 2 ⎜ 16 − ⎟ =
⎝
⎠
⎝
⎠
⎝
⎝
3 ⎠ −2
3
3 ⎠
3⎠ 3
⎝
3
Ex 2.
Draw curves y 2 = 5 − x and y 2 = 3 + x . How to find the areas enclosed
between them? It is possible to split it into 2 subdomains and sum their areas.
The other way is to see that this is the same graph as in previous example
(except that x and y were switched) and to integrate with respect to y.
Course: Accelerated Engineering Calculus I
Ex 3.
Consider we compute the area under a parametric curve given
by x = r cost,
y = r sint from x=-r to x=r:
r
∫ y ( x ) dx
−r
r2
=−
2
12.11.1.1
Instructor: Michael Medvinsky
0
0
0
∫ y (t ) x '(t ) dt = ∫ r sint ( −r sint ) dt = −r ∫ sin
=
2
x=r cost
dx=x '( t )dt=− r sintdt π
π
2
t dt =
π
1 ⎛ sin 2t ⎞
r 2π
1−
cos
2t
dt
=
−r
t
−
=
(
)
⎜
⎟
∫π
2⎝
2 ⎠π
2
0
0
2
Area in Polar coordinates (appendix H) Polar coordinates are given by x = r cosθ ,
y = r sin θ ⇒ r 2 = x 2 + y 2
tan θ =
y
.A
x
straight line (a ray) that passes through origin and making angle θ 0 with polar axis is
given by formula θ = θ 0 .
The rest of Appendix H.1 is for self-reading.
The area enclosed between θ1 and θ 2 and constant radius r is given
1
2
by A = r 2 (θ 2 − θ1 ) . The area enclosed between θ1 and θ 2 and a function r = f (θ ) in
θ2
θ
( )
2
n
1
1
1
polar coordinates is given by A = ∫ r 2 dθ = ∫ f 2 (θ ) dθ = ∑ f 2 θ *j Δθ
2
2
j=1 2
θ1
θ1
Ex 4. Find area enclosed by one loop of the four leaved rose r = cos2 2θ
π /4
π /4
π /4
1
1
1
1
A = ∫ r 2 dθ = ∫ cos 2 2θ dθ = ∫ (1+ cos 4θ ) dθ =
2
2 − π /4
2 − π /4 2
− π /4
=
1
2
π /4
∫
0
π /4
1
1
π
(1+ cos 4θ ) dθ = ⎛⎜⎝ θ + sin 4θ ⎞⎟⎠ =
2
4
8
0
```