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I've been solving one following problem, but my answer was different to the answer on Wolfram Mathematica. But I couldn't find any mistakes in my actions, so I've tried to hand over it to my lecturer, but he, expectedly, said that my answer is incorrect, but did not say where I'm making a mistake. Can you please point me to my mistakes? Thank you!

Here is what I have:

Find the area of the shape bounded by a curve $(x+y)^2=a(y-x),x\ge0,y\ge0,$ in polar coordinates

Here is how I solved it (wrong answer):

1. Rewrote the equation of the curve in polar coordinates and expressed $r$:

$$ \left\{ \begin{array}{c} x=r\cos{\phi}; \\ y=r\sin{\phi}; \end{array} \right. $$ $ \left\{ \begin{array}{c} x\ge0 \\ y\ge0 \end{array} \right. $ $\Rightarrow$ $ \left\{ \begin{array}{c} r\cos{\phi}\ge0 \\ r\sin{\phi}\ge0 \end{array} \right. $ $\Rightarrow$ $ \left\{ \begin{array}{c} \cos{\phi}\ge0 \\ \sin{\phi}\ge0 \end{array} \right. $ $\Rightarrow$ $ \left\{ \begin{array}{c} -\frac{\pi}{2}\le\phi\le\frac{\pi}{2} \\ 0\le\phi\le\pi \end{array} \right. $ $\Rightarrow$ $ 0\le\phi\le\frac{\pi}{2} $ $$(x+y)^2=a(y-x)$$ $$\Leftrightarrow x^2+2xy+y^2=ay-ax$$ $$\Leftrightarrow r^2\cos^2{\phi}+2r\cos{\phi}r\sin{\phi}+r^2\sin^2{\phi}=ar\sin{\phi}-ar\cos{\phi}$$ $$\Leftrightarrow r^2(\cos^2{\phi}+2\cos{\phi}\sin{\phi}+\sin^2{\phi})=ar(\sin{\phi}-\cos{\phi})|:r$$ $$\Rightarrow r(\sin{\phi}+\cos{\phi})^2=ar(\sin{\phi}-\cos{\phi})$$ $$\Rightarrow r=\frac{a(\sin{\phi}-\cos{\phi})}{(\sin{\phi}+\cos{\phi})^2}$$

2. Found area using $S=\frac 12\int\limits_{\phi_1}^{\phi_2} r \, d\phi$

$$\frac 12\int\limits_{0}^{\frac \pi 2} a\frac{(\sin\phi-\cos\phi)}{(\sin\phi+\cos\phi)^2} \, d\phi=\Biggl[\begin{matrix} t=\sin{\phi}+\cos{\phi} \\ dt=(\cos{\phi}-\sin{\phi})d\phi \\ \end{matrix}\Biggr]=-\frac 1 2 a\int\limits_{0}^{\frac \pi 2} \frac{1}{t^2} \, dt=\frac 1 2 a\dot(\frac{1}{1+0}-\frac{1}{0+1})=\frac a 2 0 = 0$$ So, the answer $0$ is wrong. Please, help me see my mistakes. Thank you!

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  • $\begingroup$ Integrate from $\frac{\pi }{4}$ to $\frac{\pi }{2}$ $\endgroup$ – Lozenges May 16 '18 at 20:34
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The polar expression $r=\frac{a(\sin{\phi}-\cos{\phi})}{(\sin{\phi}+\cos{\phi})^2}$ is correct but recall that

$$S=\frac 12\int\limits_{\phi_1}^{\phi_2} \color{red}{r^2} \, d\phi=\frac 12\int\limits_{\frac \pi 4}^{\frac \pi 2} a^2\frac{(\sin\phi-\cos\phi)^2}{(\sin\phi+\cos\phi)^4} \, d\phi=$$

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  • $\begingroup$ integral should go from $\frac{\pi }{4}$ to $\frac{\pi }{2}$. answer: $\frac{1}{12}a^2$ $\endgroup$ – Lozenges May 16 '18 at 21:00
  • $\begingroup$ @Lozenges Yes of course! Thanks I fix $\endgroup$ – user May 16 '18 at 21:10
  • $\begingroup$ Could you explain please, why should I integrate from $\frac{\pi}{4}$ to $\frac{\pi}{2}$? I've got my integration boundaries from translation to polar coordinates. Were there any mistakes then? $\endgroup$ – Alex May 16 '18 at 21:16
  • $\begingroup$ @Alex One simple check is that for $(x,y)=(0,0)$ we need to have $r=0$ and since $r=\frac{a(\sin{\phi}-\cos{\phi})}{(\sin{\phi}+\cos{\phi})^2}$ we need $\sin{\phi}-\cos{\phi}=0$. That's not a proof of course but a good check to do. To prove that we can calculate the slope of the tangent of the curve at $(x,y)=(0,0)$ that is indeed 45°. We can also ask to Lozenges for that point. $\endgroup$ – user May 16 '18 at 21:23
  • $\begingroup$ @Alex assume $a>0$. From the equation we get $y\geq x$ therefore $\theta \geq \frac{\pi }{4}$ $\endgroup$ – Lozenges May 17 '18 at 7:36

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