Question

Using integration find the area of the region bounded by the curve y = ✓(4-x²), x²+y²-4x=0 and the x-axis.

Answer 1

So the area required is
$\frac{8\pi}{3} - 2 \sqrt{3}$
.
Hope it helped u..

Answer 2

$\int_{0}^{\sqrt 3}(\sqrt {4x-x^2})dx + \int_{\sqrt 3}^{2}(\sqrt {4-x^2})$ is required Area

Step-by-step explanation:

y=$\sqrt {4-x^2}$....(i) is a semicircle with centre (0,0) and radius 2 units.

$x^2+y^2-4x=0$ .....(ii) is a circle with centre (2,0) and radius 2 units.

Using (i) and (ii)

$x^2+4-x^2-4x=0$

⇒4−4x=0

⇒x=1

y= $\sqrt {4-x^2}$

Hence we have  the diagram as shown in attachment.

By (ii)

$y^2=4x-x^2$

⇒y = $\sqrt {4x-x^2}$

Hence required area is  

A = $\int_{0}^{\sqrt 3}(\sqrt {4x-x^2})dx + \int_{\sqrt 3}^{2}(\sqrt {4-x^2})$