Consider the cardioid given by the equations:
$$x = a(2\cos{t} - \cos{2t})$$ $$y = a(2\sin{t} - \sin{2t})$$
I have to find the surface that the cardioid circumscribes, however, I don't know what limits for $t$ I have to take to integrate over. How can I know that, as I don't know how this shape looks like (or more precisely where it is located)?
In general, the area under a curve $ f(x) $ between $ x = a $ and $ x = b $ is given by the formula $ A = \int_{a}^{b} f(x) dx $.
When working with parametric equations (i.e. $ x(t) $ and $ y(t) $), this equation becomes $ A = \int_{t_1}^{t_2} f(x(t)) \frac{dx}{dt} dt = \int_{t_1}^{t_2} y(t) \frac{dx}{dt} dt $.
In this system of parametric equations, both $ x(t) $ and $ y(t) $ are periodic functions; one has period equal to $ 2\pi $ and the other has period equal to $ \pi $. Since the sum of two periodic functions is a periodic function (with period equal to the smallest common multiple of the two functions' periods), then we know that both $ x(t) $ and $ y(t) $ have period equal to $ 2\pi $. Therefore, we can use $ t_1 = 0 $ and $ t_2 = 2\pi $.
Since $ y = a(2sin(t) - sin(2t)) $, $ x = a(2cos(t) - cos(2t)) $, $ \frac{dx}{dt} = a(2sin(2t) - 2sin(t)) $, we get:
\begin{equation} A = \int_{0}^{2\pi} y(t) \frac{dx}{dt} = dt \int_{0}^{2\pi} a(2sin(t) - sin(2t)) \times a(2sin(2t) - 2sin(t)) dt \\ = 2a^2 \int_{0}^{2\pi} (2sin(t) - sin(2t)) \times (sin(2t) - sin(t))sin(t) dt = 2a^2 \times 3\pi = 6a^2\pi \end{equation}