Find a parametrization of the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1$ and use it to compute the area of the interior. I have the following homework question that I know the answer to $(3\pi/8)$, however, I don't understand how to get this answer.
The question:
Find a parametrization of the curve $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1$ and use it to compute the area of the interior.
I found the answer at the following link example #5.
http://www2.math.umd.edu/~jmr/241/lineint2.htm
I don't understand though. I understand how they make $u = x^{\frac{1}{2}}$ and $y = y^{\frac{1}{2}}$ so they could get $u^2 + v^2 = 1$, which is a nice trick. I can see how they parameterized that to $u = \cos t$, $v = \sin t$ from $t = 0$ to $t = 2\pi$. I just don't get what they do next.
 A: They are using "Green's Theorem" to compute areas using line integrals.
Roughly Green's theorem tells you how to turn a double integral into a line integral or vice-versa. 
A little more detail: Let $C$ be the boundary of some 2D region $R$ and let $C$ be oriented counter-clockwise. In addition, suppose that $P(x,y)$ and $Q(x,y)$ have continuous first partials. Then 
$$ \iint_R \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\,dA = \int_C P\,dx+Q\,dy $$
Next, $\iint_R 1\,dA$ computes the area of $R$ just as $\int_a^b 1\,dt$ computes the length of $[a,b]$, $\int_C 1\,ds$ computes the arc length of $C$, $\iiint_E 1\,dV$ computes the volume of $E$, etc.
If you can pick out $P$ and $Q$ such that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1$, then by Green's theorem, the corresponding line integral will compute the area of $R$. Note that, for example, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1$ when $Q=x$ and $P=0$ or $Q=0$ and $P=-y$ or $Q=x/2$ and $P=-y/2$.
Therefore, $$\iint_R 1\,dA = \int_C x\,dy = \int_C -y\,dx = \int_C -y/2\,dx+x/2\,dy$$
So if $R$ is the region bounded by $C$ where $C$ is the curve $x^{2/3}+y^{2/3}=1$. Then $\int_C x\,dy$ will compute the area of $R$ (given $C$ is oriented counter-clockwise). So we parametrize $C$ using ${\bf r}(t) = \langle x(t),y(t) \rangle = \langle \cos^3(t), \sin^3(t) \rangle$ and $0 \leq t \leq 2\pi$. [Note: $x^{2/3}+y^{2/3} = (\cos^3(t))^{2/3}+(\sin^3(t))^{2/3} = \cos^2(t)+\sin^2(t)=1$.]
Thus $x = \cos^3(t)$ and $dy = y'(t)\,dt = 3\sin^2(t)\cos(t)\,dt$. Therefore,
$$ \mathrm{Area}(R) = \iint_R 1\,dA = \int_C x\,dy = \int_0^{2\pi} \cos^3(t) \cdot 3\sin^2(t)\cos(t)\,dt = \frac{3\pi}{8} $$
See Wolfram Alpha - indefinite and Wolfram Alpha - definite for the details regarding the evaluation of this integral.
