Question

Using integration, find the area of the region given by x^2+y^2<8x and y^2<4x

Answer 1

Answer:To find area {(x, y): y2≤ 8x, x2 + y2≤9}y2 = 8x ...(i)x2 + y2 = 9 ...(ii)On solving the equation (i) and (ii),Or, x2 + 8x = 9Or, x2 + 8x – 9 = 0Or, (x + 9)(x – 1) = 0Or, x = – 9 or x = 1And when x = 1 then y = ±2√2Equation (i) represents a parabola with vertex (0,0) and axis as x – axis, equation (ii) represents a circle with centre (0,0) and radius 3 units, so it meets area at (±3, 0), (0,±3).Point of intersection of parabola and circle is (1,2√2) and (1, – 2√2).I hope this will help uh