bounded by the curve, x -axis and the ordinates x = and x = a ) [as the circle is symmetrical about both x -axis and y -axis] a ydx (taking vertical strips) x dx Since x + y = a gives y = ± As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get the whole area enclosed by the given circle =π Fig . Fig . APPLICATION OF INTEGRALS Alternatively , considering horizontal strips as shown in Fig .
, the whole area of the region enclosed by circle a xdy π = π Example Find the area enclosed by the ellipse Solution From Fig . , the area of the region ABA ′ B ′ A bounded by the ellipse in , , area of theregion AOBA the first quadrantbounded bythecurve x axisand theordinates x (as the ellipse is symmetrical about both x -axis and y -axis) (taking verticalstrips) a ydx Now = gives =± , but as the region AOBA lies in the first quadrant, y is taken as positive. So, the required area is a b x dx b x × + b a ab π =π Fig . Fig .
Alternatively , considering horizontal strips as shown in the Fig . , the area of the ellipse is = xdy