Question

Find the area of ​​the region covered by the circle x^2+y^2=8x, parabola y^2 = 4x and the first phase from the X-axis with help of integration.
And the answer is 4/3(8+3pi)

Answer 1

Solution :

x² + y² = 8x

x² - 8x + y² = 0

x² - 8x + y² + 16 - 16 = 0

(x-4)² + y² = 16

(x-4)² + y² = 4²

Equation of circle

r = 4

Coordinate = (4,0)

x = 4 unit, y = 0 from origin

Equate

x² + y² = 8x & y² = 4x

x² + 4x = 8x

x² = 4x

(x= 0), (x=4)

Area of rquired region

$\displaystyle\sf\int_{0}^{4} y dx + \int_{4}^{8} y dx$

$\displaystyle\sf\int_{0}^{4} \sqrt{4x} dx + \int_{4}^{8} \sqrt{8x - x^2} \: dx$

$\displaystyle\sf 2 \int_{0}^{4} \sqrt{x} dx + \int_{4}^{8} \sqrt{4^2 - (x -4)^2} \: dx$

$\displaystyle\sf 2 \Bigg[ \dfrac{2}{3} x^\frac{3}{2} \Bigg]_{0}^{4} + \Bigg[\dfrac{(x -4)}{2} \sqrt{4^2 - (x - 4)^2} + \dfrac{16}{2} Sin^{-1} \Bigg[\dfrac{(x -4)}{4}\Bigg] \Bigg]_{4}^{8}$

$\displaystyle\sf 2 \Bigg[ \dfrac{2}{3} . 8 \Bigg] + \Bigg[2 \times \sqrt{16 - 16} + 8 \: sin^{-1} (1) - 0 - 8 sin^{-1} (0) \Bigg]$

$\displaystyle\sf\dfrac{32}{3} + 8 \times \dfrac{\pi}{2}$

$\displaystyle\sf 4\pi + \dfrac{32}{3}$