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I have the following equation

$$(x+y)^{4} = ax^{2}y$$

I need to find the area limited by the equation above. I know I have to transform x and y in polar coordinates:

$$\begin{align*} &x = r\cos^2\theta\\ &y = r\sin^2\theta \end{align*}$$

I also know the double integral formula. But what should I do after is unknown to me... Thanks!

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You mean $x=r\cos\theta$ and $y=r\sin\theta$, without the square, don't you? – Berci Feb 4 '13 at 21:55
I put square, because the x + y sum's exponent is 4 ( i.e 2^2). Are the polar coordinates x=rcosθ and y=rsinθ, no matter what? – SpaceNecron Feb 4 '13 at 21:58
Draw a circle with radius r, and find out! – Bob Feb 4 '13 at 22:01
@SpaceNecron: actually, you are wise to use these alternatives to polar coordinates, as I outline below. Just keep in mind that the familiar period, plot, etc., will be different. – Ron Gordon Feb 5 '13 at 2:40

1 Answer 1

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The main thing is that you need to compute the Jacobian $J$ of your transformation and use it in the area integral. For $x=r \cos^2{\theta}$, $y=r \sin^2{\theta}$ you get

$$J(r,\theta) = 2 r \cos{\theta} \sin{\theta} $$

The curve you are given is $r = a \cos^4{\theta} \sin^2{\theta}$. The area integral is then

$$\begin{align}A &= \int_0^{\pi} d \theta \: \int_0^{a \cos^4{\theta} \sin^2{\theta}} dr \: J(r,\theta) \\ &= a^2 \int_0^{\pi/2} d \theta \: \cos^9{\theta} \sin^5{\theta} \\ &= a^2 \int_0^1 du \: u^9 (1-u^2)^2 \end{align}$$

And I'll leave it for you to fill in the gaps and finish this off.

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Edited to fix a mistake in the Jacobian and to take advantage of symmetry to compute area. – Ron Gordon Feb 4 '13 at 22:34
As you've written the double integral, we have $$A=\pi\cdot a\cos^4\theta\sin^2\theta\cdot2r\cos\theta\sin\theta=2\pi a r\cos^5\theta\sin^3\theta,$$ which doesn't make sense. It should instead be $$\int_0^\pi\int_0^{a\cos^4\theta\sin^2\theta}J(r,\theta)\,dr\,d\theta.$$ – Cameron Buie Feb 4 '13 at 23:26
@CameronBuie: I wrote the integral in a way that makes what I am doing very clear. Perhaps, though, you are not used to the $dr$ term being next to the integral. This is a convention commonly used in physics to make reading double integral s easier. The inner integral is on the right, and you work right to left. – Ron Gordon Feb 4 '13 at 23:41
Ah, I see! You're treating them as successive operators being applied to the function $J(r,\theta)$, rather than two integrals (one a scalar, one a function in $\theta$) multiplied by $J(r,\theta)$. I am, indeed, unfamiliar with that convention. – Cameron Buie Feb 5 '13 at 0:21
@Cameron Buie: bingo. – Ron Gordon Feb 5 '13 at 0:26

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