## Rajasthan Board RBSE Class 12 Maths Chapter 11 Application of Integral:Quadrature Ex 11.2

Question 1.

Find the area between the region of parabola. y^{2} = 2x and x^{2} + y^{2} = 8.

Solution:

Required area is shaded in following figure.

Question 2.

Find the area of the region enclosed by parabola 4y = 3x^{2} and line 3x – 2y + 12 = 0.

Solution:

Parabola 4y = 3x^{2} and line 3x – 2y + 12 = 0 intersect each other at point, A(-2, 3) and B(4, 12). Required area is shaded in the follows figure.

Question 3.

Find the area of the region enclosed by curve y = , x = y and x-axis.

Solution:

Curve y = is circle whose vertex is origin and radius is 2.

Question 4.

Find the area of the region in the first quadrant enclosed by circle x^{2} + y^{2} = 16 and line y = x.

Solution:

Centre of circle x^{2} + y^{2} = 16 is origin and radius is 4 line y = x passes through origin and cuts the circle at A.

Question 5.

Find the area of common region between parabola y^{2} = 4x and x^{2} = 4y.

Solution:

Equation of given parabolas are

y^{2} = 4x …..(i)

x^{2} = 4y ……(ii)

Solving these, we get (0, 0) and (4, 4) their intersecting points as

Question 6.

Find the area of the region in the first quadrant enclosed by curve x^{2} + y^{2} – 1 and x + y = 1 equation.

Solution:

Given circle x^{2} + y^{2} = 1 its centre passes through origin and radius is 1. x + y = 1 is equation of line which passes through points (1,0) and (0, 1).

Question 7.

Find the area of the region enclosed by curve y^{2} = 4ax, and line y = 2a and y – axis.

Solution:

Shaded part of following figure shows area enclosed by curve y^{2} = 4ax, line y = 2a and y axis.

Question 8.

Find the area of that portion of circle x^{2} + y^{2} = which lies outside the parabola y^{2} = 6x.

Solution:

Radius of given circle x^{2} + y^{2} = 16 is 4 unit and it passes through origin. Let parabola y^{2} = 6x is interested by circle at points P and Q then solving two equations.

x^{2} + 6x = 16 (∵y^{2} – 6x)

⇒ x^{2} + 6x – 16 = 0

⇒ x + 8x – 2x – 16 = 0

(x + 8) (x – 2) = 0

Thus x = – 8, + 2

Here we take positive value of x.

Thus we will take limits 0 and 2, 2 and 4

Area POQSP = 2 × area PORSP

= 2 [area PORP + area PRSP]

Question 9.

By using integrating method, find the area of ∆ABC whose coordinates of vertices are A(2, 0), 5(4, 5), C(6, 3).

Solution:

In figure, ∆ABC is shaded.

Question 10.

By using intersection method, find the area of triangular prism equations of whose sides are 3x – 2y + 3 = 0, x + 2y – 7 = 0 and x – 2y + 1 = 0. Solution:

Given lines

3x – 27 + 3 = 0 …..(i)

x + 27 – 7 = 0 …..(ii)

and x – 27 + 1 = 0 …..(3)

Solving equation (1) and (2)

x = 1, y = 3

Solving equation (2) and (3)

x = 3, y = 2

Solving equation (3) and (1)

x = -1, y = 0

Now draw the graph of threee lines

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