Question

Find the area bounded by curves $(x - 1)^2 + y^2 = 1$ and $x^2 + y^2 = 1$.

Answer 1

Answer:

Area = 2 π / 3 - √3 / 2

Step-by-step explanation:

The second is a circle with radius 1 centred at the origin (0, 0).

The first is a circle with radius 1 centres at (1, 0), which is on the second circle.

Let's work out the area of the half of the region above the x-axis.  The whole region is then just double this (must remember to double at the end then!).

Let's write A = (0, 0), B = (1, 0), C = the point above the x-axis where the circles meet.

Then ABC is an equilateral triangle with base AB.  Its height (use Pythagoras' Theorem, if necessary) is then √3 / 2.  Its area is

area of triangle ABC = base × height / 2 = √3 / 4.

Now look at the sector ABC of the second circle (radius AB, radius AC and arc BC).  This is just one sixth of the whole circle.  The area of the circle is π × radius² = π × 1² = π.  So the area of this sector is π / 6.

The sector BAC of the first circle (radius BA, radius BC and arc AC) is congruent to the other sector, so its area is also π / 6.

Adding the areas of the two sectors then gives the area we're working out except that we've counted the area of the triangle twice.  So the area of the region above the x-axis is

area of sector ABC of second circle + area of sector BAC of first circle - area of triangle ABC

= π / 6 + π / 6 - √3 / 4

= π / 3 - √3 / 4.

The whole enclosed area is double this, so the answer we're looking for is

2 π / 3 - √3 / 2