2
$\begingroup$

Find the area of surface with given equation $$\big(x^2 + y^2\big)^2 = a(x^3+y^3)$$

I tried to use polar coordinates $x=r\cos\alpha$ and $y=r\sin\alpha$ and so $$r = a(\cos\alpha + \sin\alpha)\bigg(1-{\sin2\alpha\over{2}}\bigg)$$ Maybe anyone can suggest better way?

$\endgroup$
  • 1
    $\begingroup$ do you mean the area of the region enclosed by the given curve? $\endgroup$ – daulomb Dec 23 '17 at 10:32
  • $\begingroup$ oh yes, you are right $\endgroup$ – Spike Bughdaryan Dec 23 '17 at 10:38
  • 1
    $\begingroup$ usage of polar coordinates seems okey you need to determine the upper and lover limits of $\theta$. $\endgroup$ – daulomb Dec 23 '17 at 10:45
  • $\begingroup$ I think it's a best way here. $\endgroup$ – Michael Rozenberg Dec 23 '17 at 10:47
  • $\begingroup$ @MichaelRozenberg Where "here"? $\endgroup$ – DonAntonio Dec 23 '17 at 11:07
3
$\begingroup$

Okay i solved, with $\alpha \in [-\pi/4, 3\pi/4]$

and $${a^2\over2}\int_{-\pi\over4}^{3\pi\over4}(\sin\alpha + \cos\alpha)^2 (1-{\sin2\alpha \over 2})^2 d\alpha = \bigg(\dfrac{\cos\left(6x\right)+9\sin\left(4x\right)-9\cos\left(2x\right)-48\sin^4\left(x\right)+48\cos^4\left(x\right)-96\cos^2\left(x\right)+60x}{96}\bigg)_{-\pi\over4}^{3\pi\over4} = {5\pi a^2\over 16}$$

$\endgroup$
3
$\begingroup$

Why not directly the double integral (in polar coordinates)?

$$\int_0^{2\pi}\int_0^{a(\cos t + \sin t)\bigg(1-{\sin2t\over{2}}\bigg)}r\,dr\,dt=\frac{a^2}2\int_0^{2\pi}(\cos t+\sin t)^2\left(1-\frac{\sin2t}2\right)^2dt=$$

$$=\frac{a^2}2\int_0^{2\pi}\left(1+\sin2t\right)\left(1-\sin2t+\frac{\sin^22t}4\right)dt=\frac{a^2}2\int_0^{2\pi}\left[1-\frac34\sin^22t+\frac{\sin^32t}4\right]dt=$$

$$=\frac{a^2}4\int_0^{4\pi}\left[1-\frac34\sin^2u+\frac{\sin^3u}4\right]du=$$

$$=\frac{a^2}4\left(4\pi-\frac384\pi+\frac14\overbrace{\int_0^{4\pi}\left(\sin u-\sin u\cos^2u\right)du}^{=0}\right)=\frac{5\pi a^2}8$$

Seeing your solution, at least one of us two is wrong. I wouldn't be surprised at all if it is me...yet in your expression you didn't divide by two $\;\sin2\alpha\;$ ... Check this.

Added. Since it must be $\;t\in\left[-\frac\pi4,\,\frac{3\pi}4\right]\;$ for $\;r\;$ to be positive, the above integral must be changed accordingly to these limits. The outcome indeed is $\;\cfrac{5\pi a^2}{16}\;$

$\endgroup$
  • $\begingroup$ oh yes, forgot about it, i'm not sure if my solution is all the way right, but the answer is $5\pi a^2 \over 16$ and also we need to chose the $\alpha$ where $sin\alpha + cos\alpha > 0$ and so $[−π/4,3π/4]$ $\endgroup$ – Spike Bughdaryan Dec 23 '17 at 11:14
  • $\begingroup$ Didn't mean the restriction, but from $r=a(cosα+sinα)(1−{\sin2α\over2})$ we need to chose the angle $\alpha$ , where $r>0$ $\endgroup$ – Spike Bughdaryan Dec 23 '17 at 11:40
  • $\begingroup$ @SpikeBughdaryan You are completely right, of course... $\endgroup$ – DonAntonio Dec 23 '17 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.