Exercise 8.1
Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1.
Answer
Here a = 4, b = 3. The ellipse is symmetric about both axes, so Area = 4 × (area in the first quadrant).
In the first quadrant, y = (3/4)√(16 − x²), 0 ≤ x ≤ 4.
Area = 4 ∫₀⁴ (3/4)√(16 − x²) dx = 3 [ (x/2)√(16 − x²) + (16/2) sin⁻¹(x/4) ]₀⁴
⇒ = 3 [0 + 8 × (π/2) − 0]
⇒ = 12π square units.
Find the area of the region bounded by the ellipse x²/4 + y²/9 = 1.
Answer
Here a = 2, b = 3. In the first quadrant, y = (3/2)√(4 − x²), 0 ≤ x ≤ 2.
Area = 4 ∫₀² (3/2)√(4 − x²) dx = 6 [ (x/2)√(4 − x²) + 2 sin⁻¹(x/2) ]₀²
⇒ = 6 [0 + 2 × (π/2)]
⇒ = 6π square units.
Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is (A) π (B) π/2 (C) π/3 (D) π/4
Answer
Area = ∫₀² √(4 − x²) dx = [ (x/2)√(4 − x²) + 2 sin⁻¹(x/2) ]₀² = 2 × (π/2) = π.
Correct answer: (A) π.
Area of the region bounded by the curve y² = 4x, y-axis and the line y = 3 is (A) 2 (B) 9/4 (C) 9/3 (D) 9/2
Answer
The region lies between the y-axis and the parabola x = y²/4, from y = 0 to y = 3.
Area = ∫₀³ x dy = ∫₀³ (y²/4) dy = [y³/12]₀³ = 27/12 = 9/4.
Correct answer: (B) 9/4.
Miscellaneous Exercise on Chapter 8
Q1. Find the area under the given curves and given lines
(i) y = x², x = 1, x = 2 and x-axis
Ans. Area = ∫₁² x² dx = [x³/3]₁² = 8/3 − 1/3 = 7/3 square units.
(ii) y = x⁴, x = 1, x = 5 and x-axis
Ans. Area = ∫₁⁵ x⁴ dx = [x⁵/5]₁⁵ = 3125/5 − 1/5 = 3124/5 = 624.8 square units.
Sketch the graph of y = |x + 3| and evaluate ∫−60 |x + 3| dx.
Answer
y = |x + 3| = x + 3 for x ≥ −3 and −(x + 3) for x < −3. Its graph is a V-shape with vertex at (−3, 0), passing through (−6, 3) and (0, 3); it is symmetric about the line x = −3.
∫−60 |x + 3| dx = ∫−6−3 −(x + 3) dx + ∫−30 (x + 3) dx
= −[x²/2 + 3x]−6−3 + [x²/2 + 3x]−30
⇒ = −[(9/2 − 9) − (18 − 18)] + [0 − (9/2 − 9)]
⇒ = 9/2 + 9/2 = 9.
(Geometrically, two triangles each of base 3 and height 3: 2 × 9/2 = 9.)
Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
Answer
sin x ≥ 0 on [0, π] and sin x ≤ 0 on [π, 2π], so take absolute values of the two parts.
Area = ∫0π sin x dx + |∫π2π sin x dx|
⇒ = [−cos x]0π + |[−cos x]π2π|
⇒ = (1 + 1) + |−1 − 1| = 2 + 2 = 4 square units.
Area bounded by the curve y = x³, the x-axis and the ordinates x = −2 and x = 1 is (A) −9 (B) −15/4 (C) 15/4 (D) 17/4
Answer
y = x³ is negative on [−2, 0] and positive on [0, 1].
Area = |∫−20 x³ dx| + ∫₀¹ x³ dx = |[x⁴/4]−20| + [x⁴/4]₀¹
⇒ = |0 − 4| + 1/4 = 17/4.
Correct answer: (D) 17/4.
The area bounded by the curve y = x|x|, x-axis and the ordinates x = −1 and x = 1 is given by (A) 0 (B) 1/3 (C) 2/3 (D) 4/3 [Hint: y = x² if x > 0 and y = −x² if x < 0].
Answer
Area = |∫−10 (−x²) dx| + ∫₀¹ x² dx = 1/3 + 1/3 = 2/3.
Correct answer: (C) 2/3.
Important Extra Questions
Find the area enclosed by the circle x² + y² = a².
Answer
Area = 4 ∫0a √(a² − x²) dx = 4 [ (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) ]0a = 4 × (πa²/4) = πa² square units.
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Answer
Area = ∫₂⁴ 3√x dx = 3 × (2/3)[x3/2]₂⁴ = 2(8 − 2√2) = 16 − 4√2 square units.
Find the area of the region bounded by x² = 4y, y = 2, y = 4 and the y-axis in the first quadrant.
Answer
Area = ∫₂⁴ x dy = ∫₂⁴ 2√y dy = (4/3)[y3/2]₂⁴ = (4/3)(8 − 2√2) = (32 − 8√2)/3 square units.
Find the area of the region bounded by the line y = 3x + 2, the x-axis and the ordinates x = −1 and x = 1.
Answer
The line meets the x-axis at x = −2/3. Area = |∫−1−2/3 (3x + 2) dx| + ∫−2/31 (3x + 2) dx = 1/6 + 25/6 = 13/3 square units.
Find the area bounded by y = cos x, the x-axis and the lines x = 0 and x = 2π.
Answer
Area = ∫0π/2 cos x dx + |∫π/23π/2 cos x dx| + ∫3π/22π cos x dx = 1 + 2 + 1 = 4 square units.