# Find area of surface enclosed with given curve

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?

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

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

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

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 $\;\cfrac{5\pi a^2}{16}\;$

• 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 $[−π/4,3π/4]$ – Spike Bughdaryan Dec 23 '17 at 11:14
• Didn't mean the restriction, but from $r=a(cosα+sinα)(1−{\sin2α\over2})$ we need to chose the angle $\alpha$ , where $r>0$ – Spike Bughdaryan Dec 23 '17 at 11:40
• @SpikeBughdaryan You are completely right, of course... – DonAntonio Dec 23 '17 at 11:42