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This is the last part of an exercise in Apostol Vol. II. (p.385, 1 (e), to be precise.) No doubt there's a trick I'm missing, because evaluating the double integral over the region involved seems unduly complicated.

We are supposed to use Green's Theorem to evaluate the line integral $\oint_CPdx + Qdy$ (where $C$ is traversed counterclockwise), with $P = y^2$, $Q = x$, and $C$ described by the parametric equations $x = 2\cos^3t$, $y = 2\sin^3t$ where $t$ ranges from $0$ to $2\pi$.

We have $\frac{\partial Q}{\partial x} = 1$, $\frac{\partial P}{\partial y} = 2y$. So what we want to evaluate is

\begin{align} \iint\limits_R (1 - 2y)dydx \end{align} where $R$ is the interior of the the region bounded by $C$. Trying to evaluate by iterated integration gives us \begin{align} \int_{-2}^2\left[\int_{-2\cos^3\left[\arcsin\left(\sqrt[3]{\frac{x}{2}}\right)\right]}^{2\cos^3\left[\arcsin\left(\sqrt[3]{\frac{x}{2}}\right)\right]}(1-2y)dy\right]dx & = \int_{-2}^24\cos^3\left[\arcsin\left(\sqrt[3]{\frac{x}{2}}\right)\right]dx \end{align} and, I mean, give me a break. There's got to be a better way, right?

solitaireartist
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    Seems it would be helpful to use a change of variables to $(r,t)$, where $x=r\cos^3t$ and $y=r\sin^3t$, so that your region becomes defined by $r\in(0,2)$ and $t\in(0,2\pi)$. – symplectomorphic Jan 16 '15 at 21:51

2 Answers2

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The curve $C$ is known as astroid. This is what it looks like

enter image description here

The identity $\cos^2t+\sin^2t=1$ gives us after a bit of tinkering that in $xy$-coordinates it has the equation $$ x^{2/3}+y^{2/3}=2^{2/3}. $$ This means that the region is bounded by $-2\le x\le2$, $-(2^{2/3}-x^{2/3})^{3/2}\le y\le(2^{2/3}-x^{2/3})^{3/2}$.

Thus the inner integral is $$ \int_{y=-(2^{2/3}-x^{2/3})^{3/2}}^{(2^{2/3}-x^{2/3})^{3/2}}(1-2y)\,dy =\mathop{\Bigg/}\nolimits_{\hspace{-2mm}y=-(2^{2/3}-x^{2/3})^{3/2}}^{\hspace{1mm}(2^{2/3}-x^{2/3})^{3/2}}(y-y^2)=2(2^{2/3}-x^{2/3})^{3/2}, $$ as the substitutions into $y^2$ cancel.

The function and this region are both symmetric w.r.t. the $y$-axis, so the answer is $$ \iint_R(1-2y)\,dy\,dx=4\int_{x=0}^2(2^{2/3}-x^{2/3})^{3/2}\,dx. $$ Here the substitution $x=2\cos^3t$ stands out. Then you get $$ (2^{2/3}-x^{2/3})^{3/2}=(2^{2/3}(1-\cos^2t))^{3/2}=2\sin^3t $$ and $$ dx=-6\cos^2t\sin t. $$ This gives a standard trig integral that I'm sure you can manage. I got $3\pi/2$.

Jyrki Lahtonen
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    I believe you meant astroid, not to be confused with asteroids which are chunks of rocks flying around in space. ;) – David H Jan 17 '15 at 06:15
  • Thanks @DavidH! This was my mistranslation. In Finnish the curve is called (confusingly enough!) asteroidi. And the exact same word is used for those chunks of rock :-) – Jyrki Lahtonen Jan 17 '15 at 06:58
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    How odd! The etymological history of math and science related terminology kind of a hobby of mine. Apparently the 'astroid' derives from the ancient Greek 'ástron+eîdos', whereas 'asteroid' derives from the ancient Greek 'astḗr+eîdos'. I guess this is the words are spelled differently in English. But apparently these two Greek roots were actually synonyms in the ancient Greek language, both meaning "star". Maybe this accounts for the lack of spelling difference in other languages such as Finnish. Cheers =) – David H Jan 17 '15 at 07:45
  • Thanks so much. I suspect that what Apostol's looking for is a combination of this reasoning with the observation about the $y$-coordinate of the centroid in the answer below. In any case, I'm very grateful. – solitaireartist Jan 17 '15 at 14:50
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Hint:

$$\begin{align} \iint_{R}\left(1-2y\right)\,\mathrm{d}A &=\iint_{R}\mathrm{d}A-2\iint_{R}y\,\mathrm{d}A\\ &=\operatorname{area}{\left(R\right)}-2\,S_{y}{\left(R\right)}\\ &=\operatorname{area}{\left(R\right)}-2\,\bar{y}\cdot\operatorname{area}{\left(R\right)},\\ \end{align}$$

where $S_{y}{\left(R\right)}$ is the first moment of area of the region $R$ in the $y$-direction, and where $\bar{y}$ is the $y$-coordinate of the centroid of $R$. By symmetry, the coordinates of the centroid of $R$ are simply $(\bar{x},\bar{y})=(0,0)$, which should be immediately obvious from a plot of the region. Thus the value of the sought integral reduces to just the area of the region.

David H
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