Odisha State Board Elements of Mathematics Class 12 CHSE Odisha Solutions Chapter 10 Area Under Plane Curves Ex 10 Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 10 Area Under Plane Curves Exercise 10

Question 1.

Find the area bounded by

(i) y = e^{x}, y = 0, x = 4, x = 2

Solution:

Area = \(\int_2^4\)e^{x} dx

= \(\left[e^x\right]_2^4\)

= e^{4} – e^{2}

(ii) y = x^{2}, y = 0, x = 1

Solution:

Area = \(\int_0^1\)x^{2} dx

= \(\left[\frac{x^3}{3}\right]_0^1\)

= \(\frac{1}{3}\)

(iii) xy = a^{2}, y = 0, x = α, x = β (β > α > 0)

Solution:

Area = \(\int_\alpha^\beta y\)y dx

= \(\int_\alpha^\beta \frac{a^2}{x}\) dx

= a^{2}\([\ln x]_\alpha^\beta\)

= a^{2} ln (β/α)

(iv) y = sin x, y = 0, x = \(\frac{\pi}{2}\)

Solution:

Area = \(\int_0^{\frac{\pi}{2}}\)sin x dx

= \([-\cos x]_0^{\frac{\pi}{2}}\)

= -cos\(\frac{\pi}{2}\) + cos θ = 1

Question 2.

Find the area enclosed by

(i) y = e^{x}, x = 0, y = 2, y = 3

Solution:

(ii) y^{2} = x, x = 0, y = 1

Solution:

Area = \(\int_0^1\)x dy

= \(\int_0^1\)y^{2} dy

= \(\left[\frac{y^3}{3}\right]_0^1\)

= \(\frac{1}{3}\)

(iii) xy = a^{2}, x = 0, y = α, y = β (β > α > 0)

Solution:

Area = \(\int_\alpha^\beta\)x dy

= \(\int_\alpha^\beta \frac{a^2}{y}\)dy

= a^{2}\([\ln y]_\alpha^\beta\)

= a^{2} ln (β/α)

(iv) y^{2} = x^{3}, x = 0, y = 1

Solution:

Given curve is y^{2} = x^{3}

⇒ x = y^{2/3}

It passes through the origin. So the required area

= \(\int_0^1\)x dy

= \(\int_0^1 y^{\frac{2}{3}}\) dy

= \(\left[\frac{y^{\frac{5}{3}}}{5 / 3}\right]_0^1\)

= \(\frac{3}{5}\)

Question 3.

(i) Determineellipse the area with in the ellipse

\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1

Solution:

The ellipse is symmetrical about x-axis and y-axis.It is divided into 4 equal parts by the coordinate axes. So required area

(ii) Find the area of the circle x^{2} + y^{2} = 2ax.

Solution:

Given circle is x^{2} + y^{2} = 2ax

⇒ x^{2} – 2ax + a^{2} + y^{2} = a^{2}

⇒ (x – a)^{2} + y^{2} = a^{2} … (1)

The centre of the circle is (a, 0) and radius is a.

(iii) Find the area of the portion of the parabola y^{2} = 4x bounded by the double ordinate through (3, 0).

Solution:

Given parabola is y^{2} = 4x

(iv) Determine the area of the region bounded by y^{2} = x^{3} and the double ordinate through (2, 0)

Solution:

Given curve is y^{2} = x^{3}

⇒ y = ±x^{3/2} … (1)

The curve passes through the origin and symmetrical about x-axis because the power of y is even.

Question 4.

(i) Find the area of the regions into which the circle x^{2} + y^{2} = 4 is divided by the line x + √3y = 2.

Solution:

Given circle and the straight line are x^{2}+ y^{2} = 4 and x+ √3y = 2

The circle has the centre at (0, 0) and radius ‘2’.

The eqn. (2) can be written as

y = –\(\frac{1}{\sqrt{3}}\)x + \(\frac{2}{\sqrt{3}}\)

Slope of the strainght line = –\(\frac{1}{\sqrt{3}}\)

The line makes and angle of 150° with x-axis making intercept \(\frac{2}{\sqrt{3}}\) from y-axis.

It intersects x-axis at (2, 0).

Solveing (1) and (2),

ge wet (2 – 3√y)^{2} + y^{2} = 4

4 + 3√y^{2} – 4√3y + y^{2} = 4

4y^{2} – 4√3y = 0

y(y – √3) = 0

y = 0 or y = √3

When y = 0, x = 2

When y = √3, x = -1

Thus the straight line intersects the circle at (2, 0) and (-1, √3).

Area of the portion ACBA.

(ii) Determine the area of the region between the curves y = cos x and y = sin x, bounded by x = 0.

Solution:

The curves y = cos x and y = sin x are shown in the above figure. The region included between these two curves in [0, \(\frac{\pi}{4}\)] is OABO.

(iii) Find the area enclosed by the two parabolas y^{2} = 4 ax and x^{2} = 4ay.

Solution:

The given parabolas are y^{2} = 4ax and x^{2} = 4ay.

The graphs of the two parabolas are shown in the figure.

⇒ x^{4} = 64 a^{4}

⇒ x^{4} – 64 a^{3}x = 0

⇒ x (x^{3} – (4a)^{3}) = 0

⇒ x (x – 4a) (x^{2} + 4ax + 16a^{2}) = 0

⇒ x = 0, 4a

When x = 0, y = 0 and

when x = 4a, y = 4a

Thus the two parabolas intersect at (0, 0) and (4a, 4a).

Area between two parabolas

(iv) Determine the area common to the parabola y^{2} = x and the circle x^{2}+ y^{2} = 2x.

Solution:

Gien parabola is y^{2} = x

Given circle is

x^{2} + y^{2} = 2x ⇒ (x – 1)^{2} + y^{2} = 1

The centre is at (1, 0) and radius is 1.

Solving (1) and (2) we get

x^{2} + x = 2x ⇒ x^{2} – x = 0 ⇒ x(x – 1) = 0

⇒ x = 0, x = 1

When x = 0, y = 0 and when x = 1, y = 1.

Thus both the parabola and circle intersect at (0, 0) and (1, 1).

Required Area