The curve C is given by $x=a(\cos\theta-\cos2\theta)$ and $y=a(2\sin\theta-\sin2\theta)$.
Above is the curve when $a=1$.
I am supposed to find the area enclosed by the x-axis and the part of the curve above the x-axis.
The area is given by the definite integral $\int_{-3a}^{a} y dx$ and substitution was made to find the area $$\int_{\pi}^{0}(2\sin\theta-\sin2\theta)(2a(\sin2\theta-\sin\theta))d\theta$$
Question 1:
Is it correct to express the area as $\int_{-3a}^{a} y dx$? I thought this represents the area under the curve between $x=-3a$ and $x=a$?
Question 2:
Using the definite integral in terms of $\theta$, will the area near the "dent" part of the curve there counted twice as we trace from $\theta=\pi$ to $\theta=0$?
Question 3:
Am I right to say that for area enclosed by parametric curve I just need to integrate from end to end, even though the graph is one-to-many?
Appreciate if anyone can clarify. Thanks.