Encontrando el área bajo una curva paramétrica
A
=
∫
a
b
y
(
t
)
x
′
(
t
)
d
t
A = \int\limits_a^b {y\left( t \right)\,} x'\left( t \right)dt
A
=
a
∫
b
y
(
t
)
x
′
(
t
)
d
t
A
=
∫
0
2
π
(
1
−
cos
t
)
(
1
−
cos
t
)
d
t
A = \int\limits_0^{2\pi } {\left( {1 - \cos t} \right)\,} \left( {1 - \cos t} \right)dt
A
=
0
∫
2
π
(
1
−
cos
t
)
(
1
−
cos
t
)
d
t
A
=
∫
0
2
π
(
1
−
2
cos
t
+
1
+
cos
2
t
2
)
d
t
A = \int\limits_0^{2\pi } {\left( {1 - 2\cos t + \frac{{1 + \cos 2t}}{2}} \right)\,} dt
A
=
0
∫
2
π
(
1
−
2
cos
t
+
2
1
+
cos
2
t
)
d
t
A
=
∫
0
2
π
(
3
2
−
2
cos
t
+
cos
2
t
2
)
d
t
A = \int\limits_0^{2\pi } {\left( {\frac{3}{2} - 2\cos t + \frac{{\cos 2t}}{2}} \right)\,} dt
A
=
0
∫
2
π
(
2
3
−
2
cos
t
+
2
cos
2
t
)
d
t
A
=
3
2
t
−
2
s
e
n
t
+
s
e
n
2
t
4
∣
t
=
0
2
π
=
3
π
A = \left. {\frac{3}{2}t - 2sent + \frac{{sen2t}}{4}} \right|_{t = 0}^{2\pi } = \boxed{3\pi }
A
=
2
3
t
−
2
se
n
t
+
4
se
n
2
t
∣
∣
t
=
0
2
π
=
3
π