# Integration using polar coordinates and finding limits of theta

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!

• The basic idea is that $dA = dx dy = r dr d\theta$. Graphing can be a big help in seeing the limits of $\theta$. – RobertTheTutor Apr 12 at 17:12

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

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

• Thank you so much ! – Nil Apr 12 at 19:25

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.

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