I have this parametric equation:
$$
\text{C}: \qquad \vec r(t)=\left(\cos^3(t), \sin^3(t)\right), \qquad t \in [0, 2\pi]
$$
How to find the area inside of $\text{C}$?
I have this formula, but I don't know how to use it.
$$ -\int_C y \operatorname d\!x $$
If anyone could give me some hints as to what I should do, it would be appreciated!
$\endgroup$ 12 Answers
$\begingroup$You are given the parameterization $y = \sin^3 t$ and $x = \cos^3 t$ for $0 \le t \le 2\pi$.
Differentiating gives $dx = -3\cos^2 t \sin t \,dt$.
Now dump those into the formula to get $\text{Area} = -\displaystyle\int_{C} y\,dx = -\int_{0}^{2\pi}\underbrace{(\sin^3 t)}_{y}\underbrace{(-3\cos^2 t \sin t \,dt)}_{dx}$.
EDIT: Instead of using $\text{Area} = -\displaystyle\int_{C} y\,dx$ or $\text{Area} = \displaystyle\int_{C} x\,dy$, try using
$\text{Area} = \dfrac{1}{2}\displaystyle\int_{C} -y\,dx+x\,dy = \dfrac{1}{2}\int_{0}^{2\pi}\underbrace{(-\sin^3 t)}_{-y}\underbrace{(-3\cos^2 t \sin t \,dt)}_{dx}+\underbrace{(\cos^3 t)}_{x}\underbrace{(3\sin^2 t \cos t \,dt)}_{dy}$.
$= \displaystyle\dfrac{1}{2}\int_{0}^{2\pi}\left[3\sin^4 t\cos^2 t + 3\sin^2 t \cos^4 t\right]\,dt = \dfrac{3}{2}\int_{0}^{2\pi}(\sin^2 t + \cos^2 t)(\sin^2 t \cos^2 t)\,dt$
$= \displaystyle\dfrac{3}{2}\int_{0}^{2\pi}\sin^2 t \cos^2 t\,dt$, which should be easier to evaluate.
Remember the identities $\sin \theta \cos \theta = \dfrac{1}{2}\sin 2\theta$ and $\sin^2 \theta = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\theta$.
$\endgroup$ 2 $\begingroup$You can apply Green's Theorem, which states that $\oint_cP(x,y)dx+Q(x,y)dy=\int\int_R(\frac {\partial Q}{\partial x}-\frac {\partial P}{\partial y})dA$. In this case $Q(x,y)=0$ and $P(x,y)=y$. Thus, the line integral becomes: $\int\int_R(\frac {\partial 0}{\partial x}-\frac {\partial y}{\partial y})dA -\int\int_RdA$. Now, this just becomes a simple double integral.
J.B
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