Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 8 Application of Integrals Ex 8.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 8 Application of Integrals Ex 8.1

Question 1.

Find the area of the region bounded by the curve y^{2} = x and the lines x = 1, x = 4 and the x-axis.

Solution:

The curve y^{2} = x is a parabola with vertex at origin.

Axis of x is the line of symmetry, which is the axis of parabola.

The area of the region bounded by the curve, x = 1, x – 4 and the x-axis.

Question 2.

Find the area of the region bounded by y^{2} = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Solution:

The given curve is y^{2} = 9x,

which is a parabola with vertex at (0, 0) and axis along x-axis.

It is symmetrical about x-axis as it contains only even powers of y,

x – 2 and x = 4 are straight lines parallel to y-axis at a positive distances of 2 and 4 units from it respectively.

∴ Required area = area ABCD

Question 3.

Find the area of the region bounded by x^{2} = 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Solution:

The given curve x^{2} = 4y is a parabola with vertex at (0, 0).

Also, since it contains only even powers of x, it is symmetrical about y-axis.

y = 2 and y = 4 are straight lines parallel to x-axis at a positive distances of 2 and 4 from it respectively.

∴ Required area = area ABCD

Question 4.

Find the area of the region bounded by the ellipse \(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{9}\) = 1.

Solution:

The equation of the ellipse if \(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{9}\) = 1.

The given ellipse is symmetrical about both axes as it contains only even powers of y and x.

Now, area bounded by the ellipse

= 4 × (area of ellipse in first quadrant)

= 4(area OAC) = 4\(\int_{0}^{4}\)y dx

= \(\int_{0}^{4}\)\(\frac{3}{4}\)\(\sqrt{16-x^{2}}\)dx

[∵ y ≥ 0 in first quadrant]

Put x = 4 sin θ so that dx = 4 cos θ dθ.

Now, when x = 0, θ = 0 and when x – 4, θ = \(\frac{π}{2}\).

Question 5.

Find the area of the region bounded of the ellipse \(\frac{x^{2}}{4}\) + \(\frac{y^{2}}{9}\) = 1.

Solution:

It is an ellipse centre (0, 0), the length of semi-major axis = 3 and that of semi-minor axis = 2.

Its rough sketch is shown in the figure.

∴ Area bounded of the ellipse

= 4 × area of the region AOB

= 6[(0 + 2 sin^{-1}) – (0 + sin^{-1}(0))]

= 6[2.\(\frac{π}{2}\)] = 6π sq.units.

Question 6.

Find the area of the region in the first quadrant enclosed by the x-axis,

the line x = \(\sqrt{3}\)y and the circle x^{2} + y^{2} = 4.

Solution:

Consider the two equations

x^{2} + y^{2} = 4 ………………. (1)

and x = \(\sqrt{3}\)y, i.e; y = \(\frac{1}{\sqrt{3}}\)x ………………. (2)

(1) x^{2} + y^{2} = 4 is a circle with centre O (0, 0) and radius = 4.

(2) y = \(\frac{1}{\sqrt{3}}\)x is a straight line passing through (0, 0) and intersecting the circle at B(\(\sqrt{3}\), 1).

The required area = shaded region

= area OBL + area LBA

Question 7.

Find the area of the smaller part of the circle

x^{2} + y^{2} = a^{2} cut off by the line x = \(\frac{a}{\sqrt{2}}\).

Solution:

The equations of the given curves are

x^{2} + y^{2} = a^{2} …………………. (1)

and x = \(\frac{a}{\sqrt{2}}\) ………………. (2)

Clearly, (1) represents a circle and (2) is the equation of a straight line parallel to y-axis at a distance

\(\frac{a}{\sqrt{2}}\) units to the right of y-axis.

Solving (1) and (2), we get

\(\frac{a^{2}}{2}\) + y^{2} = a^{2} ⇒ y^{2} = \(\frac{a^{2}}{2}\)

⇒ y = \(\frac{a}{\sqrt{2}}\).

∴ P(\(\frac{a}{\sqrt{2}}\), \(\frac{a}{\sqrt{2}}\)) is the point of intersection of curves (1) and (2) in the first quadrant.

The smaller region bounded by these two curves is the shaded portion shown in the figure.

∴ Required area = shaded region

=

Question 8.

The area between x = y^{2} and x = 4 is divided into two equal parts by the line x_{1} = a. Find the value of a.

Solution:

Graph of the curve x = y^{2} is a parabola as shown in the figure.

Its vertex is O and axis is x-axis.

QR is the ordinate along x = 4

∴ Area of the region

Area of the region bounded by the curve and ordinate x = a is

Now PS [x = a] divides the area of the region ORQ into two equal parts.

∴ Area of region ORQ = 2 Area of the region OSP

∴ A_{1} = 2A_{2}

∴ \(\frac{32}{3}\) = 2.\(\frac{4}{3}\)a^{3/2}

a^{3/2} = 4

∴ a = 4^{2/3} units.

Question 9.

Find the area of the region bounded by the parabola y = x^{2} and lines y = |x|.

Solution:

Clearly, x^{2} = y represents a parabola with vertex at (0, 0),

positive direction of y-axis as its axis and it opens upwards.

y = |x|, i.e., y = x and y = – x represent two lines passing through the origin and making an angle of 45° and 135° with the positive direction of the x-axis.

The required region is the shaded region as shown in the figure. Since both the curves are symmetrical about y-axis. So,

Required area = 2(Shaded area in the first quadrant)

Question 10.

Find the area bounded by the curve x^{2} = 4y and the straight line x = 4y – 2.

Solution:

Given curve is x^{2} = 4y, ……………… (1)

which is an upward parabola with vertex at (0, 0) and is symmetrical about y-axis.

Equation of the line is x = 4y – 2 …………….. (2)

Solving (1) and (2) simultaneously, we get

(4y – 2)^{2} = 4y

⇒ 16y^{2} – 16y + 4 = 4y

⇒ 16y^{2} – 20y + 4 = 0

⇒ 4y^{2} – 5y + 1 = 0

⇒ (4y – 1)(y – 1) = 0

⇒ y = \(\frac{1}{4}\), 1

From (2), when y = \(\frac{1}{4}\), x = 1 – 2 = – 1.

When y = 1, x = 4 – 2 = 2.

∴ A(- 1, \(\frac{1}{4}\)) and (2, 1) are the points of intersection of (1) and (2).

Clearly, from the figure the shaded portion OBDAO is the region of the required area.

Required area = area LOMBDA – area LMBOAL ……………. (3)

Now, area LMBDA = \(\int_{-1}^{2}\) ydx

(where x = 4y – 2 ⇒ y = \(\frac{1}{4}\)(x + 2))

From (3), (4) and (5), the required area

= \(\frac{15}{8}\) – \(\frac{3}{4}\) = \(\frac{9}{8}\) sq.units.

Question 11.

Find the area of the region bounded by the curve y^{2} = 4x and the line x = 3.

Solution:

The curve y^{2} = 4x is a parabola as shown in the figure.

Axis of the parabola is .r-axis.

The area of the region bounded by the curve y^{2} = 4x and the line x = 3 is A.

= Area of region PQO

= 2 area of the region OLQ

[Note : Areas below and above x-axis are equal.]

Choose the correct answers in the following questions 12 and 13:

Question 12.

Area lying in the first quadrant and bounded by the circle x^{2} + y^{2} = 4 and the lines x = 0 and x = 2 is

(A) π

(B) \(\frac{π}{2}\)

(C) \(\frac{π}{3}\)

(D) \(\frac{π}{4}\)

Solution:

A circle of radius 2 is drawn.

The line OP is x = 0 and AQ is x = 2.

∴ Area A of the required region OAP

= \(\int_{0}^{2}\) y dx …………… (1)

Equation of the circle is x^{2} + y^{2} = 4

∴ Part (A) is the correct answer.

Question 13.

Area of the region bounded by the curve y^{2} = 4x, y-axis and the line y = 3 is

(A) π

(B) \(\frac{9}{4}\)

(C) \(\frac{9}{3}\)

(D) \(\frac{9}{2}\)

Solution:

The curve y^{2} = 4x is the parabola and MP is the line y = 3.

The area of the region bounded by y^{2} = 4x, y-axis and y = 3 is

∴ Part (B) is the correct answer.