Elementary calculus, looking for different approach We want to calculate the area of the shape $S$ bounded by the curve $\gamma$ defined by $\begin{cases}x = a\cos^3(\theta) \\ y = a\sin^3(\theta)\end{cases}$, where $a \in \mathbb R$ and $0 \leq \theta \leq 2\pi$.
I solved it. But I'm looking for a different approach, seeing as mine leads to a not very nice integral.
What I did:
$\gamma$ is a closed simple curve. Define the function $F(x,y) = (P(x,y), Q(x,y)) = (0, x)$. It's continuous everywhere with continuous derivatives and everything's nice,so we can use Green's theorem:
$\displaystyle \oint_{\gamma}\vec{F}d\vec{r} = \iint_S \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}dxdy = \iint_S1dxdy = Area(S)$
So we just need to calculate $\oint_{\gamma}\vec{F}d\vec{r}$ which by definition is $\int_{0}^{2\pi}(0,a\cos^3(\theta)) \cdot(-3a\cos^2(\theta)\sin(\theta), 3a\sin^2(\theta)\cos(\theta))d\theta = 3a^2\int_{0}^{2\pi}\sin^2(\theta)\cos^4(\theta)d\theta$.
This integral is doable, but hardly pleasant.
Is there a nicer way I'm not seeing?
 A: Generally, when trigonometric functions are involved, you should try using
$$A = \frac12\oint_C -y\,dx+x\,dy$$
rather than $\oint_C x\,dy$, as you did. This tends to make the calculations (far) simpler, as usually you can take advantage of the standard identities.
In this case, you will end up with $\displaystyle\int_0^{2\pi} 3\sin^2\theta\cos^2\theta\,d\theta$, which is easy with the double angle formula and the standard antiderivative $\int\sin^2u\,du$.
A: I think that your approach is good.
That integral is not pleasant, but if you write the $\sin\theta$ as $1-\cos^2\theta$ you get two integrals in $\cos^4\theta$ and $\cos^6\theta$.
Now you can use this reduction formula and you are done.
$$\int cos^n (\theta) d\theta = \frac{\sin(\theta)\cos^{n-1}(\theta)}{n} + \frac{n-1}{n}\int \cos^{n-2}(\theta)d\theta$$
A: your approach is good to me, one other thing that you could have thought about is using the Gauss's theorem with a function whose divergence is 1. Since it is not so difficult to find the normal vectors to the curve in $\mathbb{R}^2$ that you provided, you can compute it using
$$\int \int_{\Omega} \nabla \cdot F(x,y)dxdy=\int_{\partial\Omega} F \cdot \vec{n}ds$$
with $\vec{n}$ the vector field normal to the curve. I didn't do the calculations so I can't tell you wether it would simplify of not.
Otherwise, Using Green again as you did, and taking $F_1(x,y)=(-y,0)$, $F_2(x,y)=(0,x)$, you have curl($F_j$)=1 so
$$|\Omega|=\int -\gamma_2\dot{\gamma_1}dt = \int \gamma_1\dot{\gamma_2}dt = \frac{1}{2}\int\gamma_1\dot{\gamma_2}-\gamma_2\dot{\gamma_1}dt$$
I'm not so sure about it but in the case of your curve, you might obtain something as
$K\int\cos^2\sin^2dt$ with $K\propto a^2$ a constant. Tell me if I'm wrong I didn't check
A: Note $\sin^2 \theta \cos^4 \theta = \frac18\sin^22\theta(1+\cos2\theta)
$. Thus, the area is
$$4\cdot 3a^2\int_{0}^{\pi/2}\sin^2\theta\cos^4\theta d\theta
=\frac{3a^2}2 \int_{0}^{\pi/2}(\sin^22\theta d\theta +\frac12 \sin^22\theta d(\sin 2\theta ))=\frac{3\pi a^2}8
$$
