evaluate the ellipse x= a cos x y = b sin x
Answers
Answer:
It is vital when dealing with parametric equations (or polar coordinates) to get a full understanding of the effect of the parameter on the curve (and sign) so that positive and negative areas can be determined and dealt with.
We have parametric equations:
x = a cosθ
y = b sinθ
Where
a, b > 0.
Let us first look at how
x = a cosθ
and
y = b sinθ
behave over the given domain
0 ≤ θ ≤ 2π
0 ≤ θ ≤ π/2
⇒x > 0, y > 0
π/2 ≤ θ ≤ π
⇒x < 0, y > 0
π ≤ θ ≤ 3π/2
⇒x < 0, y < 0
3π/2 ≤ θ ≤ 2/π
⇒x > 0, y< 0
So we can see that as
θ varies on the domain we move uniformly from Q1, Q2, Q3 then Q4.
Thus
0 ≤ θ ≤ π (Q1,Q2) will contribute positively and
π ≤ θ ≤ 2π (Q3,Q4) will contribute negatively.
If we now examine the parametric curve.
We can see that we can evaluate the area by symmetry as
4
times that of Q1
Thus we can represent the area of the ellipse by:
A
=
4
∫
x
=
a
x
=
0
y
d
x
Now we will change variable from
x
to
θ
to actually perform the integration:
x
=
a
cos
θ
⇒
d
x
d
θ
=
−
a
sin
θ
y
=
b
sin
θ
And for the limits of integration we have:
x
=
0
⇒
a
cos
θ
=
0
⇒
θ
=
π
2
x
=
a
⇒
a
cos
θ
=
a
⇒
θ
=
0
And so we can evaluate the integral as follows;
A
=
4
∫
0
π
2
(
b
sin
θ
)
(
−
a
sin
θ
)
d
θ
=
−
4
a
b
∫
0
π
2
sin
2
θ
d
θ
=
4
a
b
∫
π
2
0
sin
2
θ
d
θ
=
4
a
b
∫
π
2
0
sin
2
θ
d
θ
=
4
a
b
∫
π
2
0
1
−
cos
2
θ
2
d
θ
=
2
a
b
∫
π
2
0
1
−
cos
2
θ
d
θ
=
2
a
b
[
θ
−
sin
2
θ
2
]
π
2
0
=
2
a
b
{
(
π
2
−
0
)
−
(
0
−
0
)
}
=
π
a
b