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Find the area bounded by $ \ x^4+y^4=4(x^2+y^2) \ $ .

Answer:

The graph is above : enter image description here

Since the region is symmetrical ,

$ Area =4 \times \int_{0}^{2} \int_{0}^{\sqrt{2+\sqrt{4x^2-x^4+4}}} dxdy $

Am I right ? Is there any help ?

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  • $\begingroup$ That does not look completely correct. How did you obtain those boundaries? $\endgroup$ – mickep Jul 31 '17 at 7:36
  • $\begingroup$ this is correct the boundaries terms are well defined. but it could be easy to employee polar coordinates . $\endgroup$ – Guy Fsone Jul 31 '17 at 7:42
  • $\begingroup$ Since $ x^4+y^4=4(x^2+y^2) \ \ or, y=\sqrt{2+\sqrt{4x^2-x^4+4}} $ $\endgroup$ – M. A. SARKAR Jul 31 '17 at 7:43
  • $\begingroup$ But for some values of $x$ (typically larger than $2$) $y$ you will have two $y$-values. What you calculate is the area in the part of the domain (of the first quadrant) where $x$ is less than $2$. On the other hand, that should suffice. $\endgroup$ – mickep Jul 31 '17 at 7:51
  • $\begingroup$ Any help setting up the integral $\endgroup$ – M. A. SARKAR Jul 31 '17 at 7:53
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Convert to polar form $r^2=\dfrac{16}{\cos 4 t+3}$

The area is $$A=\frac{1}{2} \int_0^{2 \pi } \frac{16}{\cos 4 t+3} \, dt=32 \int_0^{\frac{\pi }{2}} \frac{1}{\cos 4 t+3} \, dt$$

$\cos 4t=\cos^2 2t-\sin^2 2t=2\cos^2 2t -1$

so we have to integrate

$\int \dfrac{dt}{2(\cos^2 2 t+1)}=\dfrac{1}{2}\int \dfrac{dt}{1+\cos^2 2t}$

Remember the identity $\cos 2t =\dfrac{1-\tan^2 t}{1+\tan^2 t}$

with the substitution $\tan t=u;\;t=\tan^{-1}u;\;dt=\dfrac{du}{1+u^2};\;\cos 2t = \dfrac{1-u^2}{1+u^2}$

Limits of integration become $t=0\to u=0;\;t=\pi/2\to u=\infty$

$$\dfrac{1}{2}\int \dfrac{dt}{1+\cos^2 2t}=\dfrac{1}{2}\int \dfrac{\dfrac{du}{1+u^2}}{1+\left(\dfrac{1-u^2}{1+u^2}\right)^2}$$ Which simplifies to $$\dfrac{1}{2}\int \frac{u^2+1}{ u^4+1} \, du=\frac{\tan ^{-1}\left(\sqrt{2} u+1\right)-\tan ^{-1}\left(1-\sqrt{2} u\right)}{4 \sqrt{2}}$$ Area is $$A=32\left[\frac{\tan ^{-1}\left(\sqrt{2} u+1\right)-\tan ^{-1}\left(1-\sqrt{2} u\right)}{4 \sqrt{2}}\right]_0^{\infty}=4 \sqrt{2} \pi$$

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Let us consider the part of the domain in the first quadrant, and use Green's theorem. The only difficult part is to find a good parametrization of the curved part of the boundary.

We look for a parametrization $(x(t),y(t))$ where $y(t)=tx(t)$. Inserting this into the equation, we find that $$ x^4+t^4x^4=4(x^2+t^2x^2). $$ Solving for $x$, we find that (in this part of the domain) $$ x=2\frac{\sqrt{1+t^2}}{\sqrt{1+t^4}} $$ will do. The range of $t$ becomes $[0,+\infty)$.

Green's theorem says that (here $\Omega$ is the planar domain (in the first quadrant) and $\Gamma$ its boundary, positively oriented) $$ A(\Omega)=\iint_\Omega 1\,dx\,dy=\frac{1}{2}\int_\Gamma x\,dy-y\,dx. $$ A simple, but tedious, caluclation gives $$ x(t)y'(t)-y(t)x'(t)=4\frac{1+t^2}{1+t^4}. $$ Thus, (the path integrals along the vertical and horizontal part do not contribute) $$ A(\Omega)=2\int_0^{+\infty}\frac{1+t^2}{1+t^4}\,dt=\sqrt{2}\pi. $$ I leave it to you to fill in details (and to multiply by $4$ to get the area of the full planar domain).

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According to your own sketch, the integral you wrote would not cover the area $|y| > 2$, never mind that the order of integration does not match the limits of integration. So the setup is problematic in multiple ways.

The suggestion to use polar coordinates is worth exploring. Let $(x,y) = (r \cos \theta, r \sin \theta)$ so that we have $$x^2 + y^2 = r^2,$$ and $$x^4 + y^4 = (x^2 + y^2)^2 - 2 x^2 y^2 = r^4 - 2 r^4 \sin^2 \theta \cos^2 \theta = r^4 (1 - \tfrac{1}{2} \sin^2 2\theta).$$ Thus we have $$0 = (x^4 + y^4) - 4(x^2 + y^2) = r^4 (1 - \tfrac{1}{2} \sin^2 2\theta) - 4r^2,$$ and eliminating the trivial solution $r = 0$, we get $$r^2 = 4(1 - \tfrac{1}{2} \sin^2 2\theta)^{-1}.$$ In the first quadrant, $0 \le \theta \le \pi/2$ and we can write $$\frac{A}{4} = \int_{\theta=0}^{\pi/2} \frac{1}{2}r^2 \, d\theta = 2 \int_{\theta=0}^{\pi/2} \frac{1}{1-\frac{1}{2}\sin^2 2\theta} \, d\theta = 8\int_{\theta=0}^{\pi/2} \frac{d\theta}{3+\cos 4\theta},$$ and the rest is quite straightforward with the standard trigonometric substitutions.

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