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I need to find the area enclosed by the function $x^4+y^4=2xy$. I know that putting x=r cos theta and y= r sin theta might help but I don’t know how to integrate in polar coordinates. I think I should integrate $r(r d theta)/2$ but I am not sure about the limits of theta.

Any help would be greatly appreciated. Thanks in advance!

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  • $\begingroup$ The basic idea is that $dA = dx dy = r dr d\theta$. Graphing can be a big help in seeing the limits of $\theta$. $\endgroup$ – RobertTheTutor Apr 12 at 17:12
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We have, letting $x = r\cos(\theta)$ and $y = r\sin(\theta)$:

$$(r\cos(\theta))^{4} + (r\sin(\theta))^{4} = 2(r\cos(\theta))(r\sin(\theta))$$

$$r^{4}(\sin^{4}(\theta) + \cos^{4}(\theta)) = r^{2}\sin(2\theta)$$

$$r^{2}((\sin^{2}(\theta) + \cos^{2}(\theta))^{2} - 2\sin^{2}(\theta)\cos^{2}(\theta)) = \sin(2\theta)$$

$$r^{2} = \frac{\sin(2\theta)}{1 - \frac{\sin^{2}(2\theta)}{2}}$$

$$r = \sqrt{\frac{2\sin(2\theta)}{2-\sin^{2}(2\theta)}} = \sqrt{\frac{2\sin(2\theta)}{1 + \cos^{2}(2\theta)}}$$

First, note that the area "under" a curve $r(\theta)$ in polar coordinates is given by:

$$A = \int_{\theta_{1}}^{\theta_{2}}\frac{r(\theta)^{2}d\theta}{2}$$

This is derived by approximating the curve as a bunch of little circular arcs and then finding the area under each arc. Also note that the given curve can be split into $4$ congruent sections, with one from $\theta = 0$ to $\frac{\pi}{4}$. Thus, the area is:

$$A = 4\int_{0}^{\frac{\pi}{4}}\frac{\sqrt{\frac{2\sin(2\theta)}{1 + \cos^{2}(2\theta)}}^{2}d\theta}{2}=4\int_{0}^{\frac{\pi}{4}}\frac{\sin(2\theta)}{1 + \cos^{2}(2\theta)}d\theta$$

Now, we let $u = 1 + \cos^{2}(2\theta)$, so $d\theta = -\frac{du}{4\sin(2\theta)\cos(2\theta)}$. Then:

$$A = 4\int_{2}^{1}-\frac{du}{4u\cos(2\theta)}$$

Simplifying using integral properties:

$$A = \int_{1}^{2}\frac{du}{u\cos(2\theta)}$$

Because $\cos(2\theta) = \sqrt{u-1}$:

$$A = \int_{1}^{2}\frac{du}{u\sqrt{u-1}}$$

Now, let $t = \sqrt{u-1}$, so $du =2t\ dt$:

$$A = \int_{0}^{1}\frac{2\ dt}{u}$$

Because $u = t^{2} + 1$:

$$A = 2\int_{0}^{1}\frac{dt}{1 + t^{2}} = 2(\arctan(t))\bigg\vert_{0}^{1} = 2\bigg(\frac{\pi}{4} - 0\bigg) = \boxed{\frac{\pi}{2}}$$

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  • $\begingroup$ Thank you so much ! $\endgroup$ – Nil Apr 12 at 19:25
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The conversion from Cartesian-coordinates to polar coordinates is given by $x= r cos(\theta)$ and $y= r sin(\theta)$ so that $x^4+ y^4= r^4(cos^4(\theta)+ sin^4(\theta))$ and $2xy= 2r^2(sin(\theta)cos(\theta)$.

The equation becomes $r^2(cos^4(\theta)+ sin^4(\theta))= 2r^2 sin(\theta)cos(\theta)$.

You should immediately see that the $r^2$ on each side, as long as r is not 0, will cancel leaving $cos^4(\theta)+ sin^4(\theta)= 2 sin(\theta)cos(\theta)$

The graph is one or more straight lines through the origin at angle $\theta$ satisfying that equation.

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  • $\begingroup$ The $r$'s don't completely cancel out, you have a fourth power on the LHS and a square on the RHS. $\endgroup$ – Joshua Wang Apr 12 at 17:47

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