CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Application of Integrals

+ y² ≤ 16}.

Solution :

o yourself. [Ans. = (32π/3 – (4/3)√3) sq. unit.]

Q.5. Find the area of the region : {(x, y) : y² ≤ 4x, 4x² + 4y² ≤ 9 }.

Solution :

Do yourself. [Ans. = √2/6 + 9/8π – 9/4 sin –1(1/3)]

Q.6. Using integration, find the area lying above x – axis and included between the circle x² + y² = 8x and the parabola y² = 4x.

Solution :

Do yourself. [Ans. = 4/3(8 + 3π) sq. unit.]

Q.7. Find the area of the region bounded by the parabola y² = 4ax and x² = 4ay.

Solution :

Fig.

We have, y² = 4ax --------------------------- (1)
x² = 4ay ---------------------------- (2)
(1) and (2) intersects hence
x = y2/4a (a > 0)
=> (y²/4a)² = 4ay
=> y⁴ = 64a³y
=> y⁴ – 64a³y = 0
=> y[y³ – (4a)³] = 0
=> y = 0, 4a
When y = 0, x = 0 and when y = 4a, x = 4a.
The points of intersection of (1) and (2) are O(0, 0) and A(4a, 4a).
The area of the region between the two curves
= Area of the shaded region
= ₀∫^4a(y1 – y²)dx
= ₀∫^4a[√(4ax) – x2/4a]dx
= [2√a.(x3/2)/(3/2) – (1/4a)(x³/3)]0^4a
= 4/3√a(4a)3/2 – (1/12a)(4a)³ – 0
= 32/3a² – 16/3a²
= 16/3a² sq. units. [Ans.]

Paper By Mr. M.P.Keshari
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