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!
The polar expression $r=\frac{a(\sin{\phi}-\cos{\phi})}{(\sin{\phi}+\cos{\phi})^2}$ is correct but recall that
$$S=\frac 12\int\limits_{\phi_1}^{\phi_2} \color{red}{r^2} \, d\phi=\frac 12\int\limits_{\frac \pi 4}^{\frac \pi 2} a^2\frac{(\sin\phi-\cos\phi)^2}{(\sin\phi+\cos\phi)^4} \, d\phi=$$