Let y = p(x) be the parabola passing through the points ( - 1,0),(0,1) and (1,0).

Question

Let $y = p(x)$ be the parabola passing through the points $( - 1,0),(0,1)$ and $(1,0)$. If the area of the region $$\{ (x,y):{(x + 1)^2} + {(y - 1)^2} \le 1,y \le p(x)\} $$ is A, then $$12(\pi  - 4A)$$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 16

Accepted Answer

Let, $y=p(x)$ be the parabola passing through the points $(-1,0)(0,1)(1,0)$.

Now, to find the area of the region

${(x,y);(x+1{)}^{2}+(y-1{)}^{2}\le 1,y\le p(x)}$

Thus, area of shaded region is

$\begin{gathered} A=\int _{-1}^{0}{(1-{x}^{2})-(1-\sqrt{1-(x+1{)}^{2}}}dx \ =\int _{-1}^{0}{-{x}^{2}+\sqrt{1-(x+1{)}^{2}}}dx \ =\int _{-1}^{0}-{x}^{2}dx+{\int }_{0}^{-1}\sqrt{1-(x+1{)}^{2}}dx \ =-{(\frac{{x}^{3}}{3})}_{0}^{-1}+{[\frac{\sqrt{1-(x+1{)}^{2}}}{2}+\frac{1}{2}{\sin }^{-1}(x+1)]}_{0}^{-1} \ =-\frac{1}{3}+[\frac{1}{2}+\frac{\pi }{4}-\frac{1}{2}] \ A=\frac{\pi }{4}-\frac{1}{3} \end{gathered}$

Now, $12(\pi -4A)$

$\begin{gathered} =12[\pi -4(\frac{\pi }{4}-\frac{1}{3})] \ =12(\pi -\pi +\frac{4}{3})=16 \end{gathered}$

Solutions · 1