areaintegrationpolar coordinates
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!
Finding the area by polar coordinates without any further transformations is not the easy way to do it. That said, your integral has several errors, the most important being that you need to split the defined region into three parts.
The graph shows the region and its limits, though you can find this purely analytically. Region 1's angle $\theta$ goes from $-\frac{\pi}2$ to $\operatorname{atan}\left(-\frac 23\right)$ and its $r$ goes from $0$ to $\frac{4\cos\theta}{\sin^2\theta}$. You almost had these limit right. Region 3's angle goes from $\operatorname{atan}(2)$ to $\frac{\pi}2$ and has the same limits on $r$. Region 2's angle goes from $\operatorname{atan}\left(-\frac 23\right)$ to $\operatorname{atan}(2)$, and $r$ starts at zero with its upper limit coming from
$$y-3 = x$$ $$r\sin\theta = 3-r\cos\theta$$ $$r = \frac{3}{\sin\theta+\cos\theta}$$
The final expression, the sum of three double integrals, should be clear. If you really need just one double integral, you could make $-\theta$ from $\frac{\pi}2$ to $\frac{\pi}2$ and $r$ from $0$ to the minimum of the expressions for the parabola and for the line. But this would be an improper integral, since the max value of $r$ would not be properly defined for the parabola at $\theta=0$.
The problem is much easier in Cartesian coordinates. I think it would also be easier in polar coordinates if you translated the region $3$ units to the left, so the line would set the limits on $\theta$ and the parabola would set the limits on $r$. If you are allowed to go that route, you should. I'll leave the details to you. (I have not worked out the details myself.)
Area of a curve given in polar coordinate is given by $$\rm{A} = \frac 12\int_{\theta_1}^{\theta_2} r^2(\theta) \:\rm{d} \theta$$
We can obtain this formula by dividing the curve into infinitely many thin sectors. (each subtending angle $\rm d \theta$ at the origin)
Then use the formula of area of sector to obtain $$dA=\frac{1}{2} r^2 d \theta$$
Now integrate both the sides,
$$\int \rm dA=A= \frac 12\int_{\theta_1}^{\theta_2} r^2(\theta) \:\rm{d} \theta$$
Best Answer
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 $u = t^{2} + 1$:
$$A = 2\int_{0}^{1}\frac{dt}{1 + t^{2}} = 2(\arctan(t))\bigg\vert_{0}^{1} = 2\bigg(\frac{\pi}{4} - 0\bigg) = \boxed{\frac{\pi}{2}}$$