Question

Let $A_1,A_2,$ and $A_3$ be the regions on $R^2$ defined by
$A_1=\left\{\right(x,y):x\ge 0,y\ge 0,2x+2y-x^2-y^2>1>x+y\},A_2=\left\{\right(x,y):x\ge 0,y\ge 0,x+y>1>x^2+y^2\},A_3=\left\{\right(x,y):x\ge 0,y\ge 0,x+y>1>x^3+y^3\}.$
Denote by $|A_1|,|A_2|,$ and $|A_3|$ the areas of the regions $A_1,A_2$ and $A_3$ respectively. Then

• A. $|A_1|>|A_2|>|A_3|$
• B. $|A_1|>|A_3|>|A_2|$
• C. $|A_1|=|A_2|<|A_3|$
• D. $|A_1|=|A_3|<|A_2|$

Answer 1

The correct option is C $|A_1|=|A_2|<|A_3|$

The area of the $A_1$
$2x+2y-x^2-y^2>1\Rightarrow x^2+y^2-2x-2y+1<0\Rightarrow (x-1{)}^2+(y-1{)}^2<1$
and
$x+y<1$

$|A_1|=\frac{1}{4}\times \pi \times 1^2-\frac{1}{2}\times 1\times 1=\frac{\pi -2}{4}$
Now the region bounded by $A_2$
$x^2+y^2<1$ and $x+y>1$


$|A_2|=\frac{1}{4}\times \pi \times 1^2-\frac{1}{2}\times 1\times 1=\frac{\pi -2}{4}$
So $|A_1|=|A_2|$ which only matches with option c.