Q1: Smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2 is.
Ans: Given,
Equation of circle is x2+ y2 = 4……….(i)
x2+ y2 = 22
y = √(22 – x2) …………(ii)
Equation of a lines is x + y = 2 ………(iii)
y = 2 – x
Therefore, the graph of equation (iii) is the straight line joining the points (0, 2) and (2, 0).
From the graph of a circle (i) and straight-line (iii), it is clear that points of intersections of circle
(i) and the straight line (iii) is A (2, 0) and B (0.2).
Area of OACB, bounded by the circle and the coordinate axes is
= [ 1 × √0 + 2 sin-1(1) – 0√4 – 2 × 0]
= 2 sin-1(1)
= 2 × π/2
= π sq. units
Area of triangle OAB, bounded by the straight line and the coordinate axes is
= 4 – 2 – 0 + 0
= 2 sq.units
Hence, the required area = Area of OACB – Area of triangle OAB
= (π – 2) sq.units
Q2: Find the area of the curve y = sin x between 0 and π.
Ans: Given,
y = sin x
Area of OAB
= – [cos π – cos 0]
= -(-1 -1)
= 2 sq. units
Q3: Find the area enclosed by the ellipse x2/a2 + y2/b2 =1.
Ans: Given,
We know that,
Ellipse is symmetrical about both x-axis and y-axis.
Area of ellipse = 4 × Area of AOB
Substituting the positive value of y in the above expression since OAB lies in the first quadrant.
= 2ab × sin-1(1)
= 2ab × π/2
= πab
Hence, the required area is πab sq.units.
Q4: Find the area of the region bounded by the two parabolas y = x2 and y2 = x.
Ans: Given two parabolas are y = x2 and y2 = x.
The point of intersection of these two parabolas is O (0, 0) and A (1, 1) as shown in the below figure.
Now,
y2 = x
y = √x = f(x)
y = x2 = g(x), where, f (x) ≥ g (x) in [0, 1].
Area of the shaded region
= (⅔) – (⅓)
= ⅓
Hence, the required area is ⅓ sq. units.
Q3: Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Ans: We can draw the figure of y2 = 9x; x = 2, x = 4 and the x-axis in the first curve as below.
y2 = 9x
y = ±√(9x)
y = ±3√x
We can consider the positive value of y since the required area is in the first quadrant.
The required area is the shaded region enclosed by ABCD.
= 2 [(2)3 – (√2)3]
= 2[8 – 2√2]
= 16 – 4√2
Hence, the required area is 16 – 4√2 sq.units.
