Question

Set-up the integral for the area A(R) of the region R bounded by the circles y

2 + x

2 = 1, and y

2 + x

2 = 4,

below the x-axis and to the left by the y-axis using the vertical strip.

Answer 1

$$A=\displaystyle\int_{-2}^{-1}(0-(-\sqrt{4-x^2}))dx$$

$$+\displaystyle\int_{-1}^{0}(-\sqrt{1-x^2}-(-\sqrt{4-x^2}))dx$$

$$=\displaystyle\int_{-2}^{0}\sqrt{4-x^2}dx-\displaystyle\int_{-1}^{0}\sqrt{1-x^2}dx$$

$$=\dfrac{1}{4}\pi(2)^2-\dfrac{a}{4}\pi(1)^2=\dfrac{3\pi}{4} ({units}^2)$$