Question

Find the area of region:
$(x,y):x^2+y^2\le 1,\le x+\frac{y}{2}$

Answer 1

$\begin{array}{c}Given,\\ x^2+y^2\le 1,andx+\frac{y}{2}=2x+y\ge 2,\therefore y=2-2x\\ then,x^2+{\left[2(1-x)\right]}^2=1\\ x^2+4(1+x^2-2x)=1\\ x^2+4+4x^2-8x-1=0\\ 5x^2-8x+3=0\\ 5x^2-5x-3x+3=0\\ (5x-3)(x-1)=0\Rightarrow x=\frac{3}{5},x=1\\ andy=2-2\left(\frac{3}{5}\right)=\frac{4}{5}\\ Now\\ Areaofregion-{\int }_{\frac{3}{5}}^{1}y.dy-\frac{1}{2}(1-\frac{3}{5})\times \frac{4}{5}\\ {\int }_{\frac{3}{5}}^1\sqrt{1-x^2}.dx-\frac{4}{5}[\int \sqrt{a^2-x^2}dx=\frac{a^2}{2}{\tan }^{-1}\frac{x}{\sqrt{a^2-x^2}}+\frac{x}{2}\sqrt{a^2-x^2}\\ {{\left[\frac{1}{2}{\tan }^{-1}\frac{x}{\sqrt{1-x^2}}+\frac{x}{2}\sqrt{1-x}\right]}_{\frac{3}{5}}}^1-\frac{4}{5}\\ \left[\frac{1}{2}\times \frac{\pi }{2}-\frac{1}{2}{\tan }^{-1}\frac{\frac{3}{5}}{\sqrt{1-\frac{9}{25}}}-\frac{\frac{3}{5}}{2}\sqrt{1-\frac{9}{25}}\right]-\frac{4}{5}\\ \left[\frac{\pi }{4}-\frac{1}{2}{\tan }^{-1}\frac{3}{4}-\frac{3}{5\times 2}\times \frac{4}{5}\right]-\frac{4}{25}\\ \left[\frac{\pi }{4}-\frac{1}{2}{\tan }^{-1}\frac{3}{4}-\frac{6}{25}\right]-\frac{4}{25}\\ \therefore \frac{\pi }{4}-\frac{1}{2}{\tan }^{-1}\frac{3}{4}-\frac{2}{5}\end{array}$

So, that the Area of region is $\frac{\pi }{4}-\frac{1}{2}{\tan }^{-1}\frac{3}{4}-\frac{2}{5}$.