What is the area bound by relation lxl + lyl = 2?

Question

please help

Answer 1

Given equations are
lxl + lyl = 2 —(1)
x² + y² =4 —(2)

For eq(1),
For x>0, y>0 (Ist quadrant),
x+y=2
For x<0, y>0 (2nd quadrant),
-x+y=2
For x<0, y<0 (3rd quadrant),
-x-y=2
For x>0, y<o (4th quadrant),
x-y=2

This forms a square. The other equation is of a circle with radius 2. So, the square is inside the circle except that the vertices of the square touches the circle. Each side of the square is now √(2²+2²)=2√2
Area=side²
=(2√2)²
=8.

Else, if you wanted to find the area using integration. Find the area of curve in first quadrant and multiply by 4.
Area of region in Ist quadrant=(0 to 2)∫(2-x)dx
=[2x-x²/2](0 to 2)
=4-2
=2
Required area=4*2
=8.

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