Class 12 RD Sharma Solutions - Chapter 21 Areas of Bounded Regions - Exercise 21.3

In this article, we will delve into the solutions for Exercise 21.3 from Chapter 21: Areas of Bounded Regions of the RD Sharma textbook for Class 12. This chapter is crucial for understanding how to calculate the area enclosed by the curves and lines a fundamental concept in integral calculus. The solutions provided here aim to guide students in solving the exercise problems accurately and efficiently helping them build a solid foundation in this topic.

Areas of Bounded Regions

The concept of Areas of Bounded Regions involves finding the area enclosed between the curves, lines or both using integration. This method is essential for calculating areas that are not easily measurable using basic geometric formulas. By setting up integrals based on the given curves and their intersections students can determine the area of the complex regions in a systematic manner.

Question 1: Calculate the area of the region bounded by the parabolas y² = 6x, x² = 6y.

Solution:

Given, 

y² = 6x, x² = 6y

y² = 6x 

⇒ y = √(6x)

x² = 6y 

⇒ y = √(x²/6)

Area of the region bounded by curve = Area under the curve y²=6x along x-axis - Area under the curve x²=6y along the x-axis

Required Area =\int_0^6(\sqrt{6x} - \frac{x^2}{6}) dx

                       = [\frac{\sqrt{6}x^\frac{3}{2}}{\frac{3}{2}}-\frac{x^3}{18}]^6_0

Putting the upper and lower limits we get,

Required Area =  [\frac{\sqrt{6}(6^\frac{3}{2})}{\frac{3}{2}}-\frac{6^3}{18}]  

                       =[\frac{2\times6^\frac{1}{2}(6^\frac{3}{2})}{3}-\frac{6^3}{18}] \\

                       =[\frac{2\times6^2}{3}-\frac{6^3}{18}]

                       = 24 - 12

                       = 12 sq. units

Question 2. Find the area of the region common to the parabolas 4y² = 9x, 3x² = 16y.

Solution:

Given,

4y² = 9x, 3x² = 16y

4y² = 9x ⇒ y = √(9x/4)

3x²=16y ⇒ y = √(3x²/16)

Area of the region bounded by curve = Area under the curve 4y²=9x along x-axis - Area under the curve 3x²=16y along x-axis

Required area = \int_0^4(\sqrt{\frac{3x}{2}} - \frac{3x^2}{16}) dx 

= [x^\frac{3}{2}-\frac{x^3}{16}]^4_0                  

Putting the upper and lower limits we get,

Required Area = 4^3/2-4³/16

                       = 8 - 64/16

                       = 4 sq. units

Question 3. Find the area of the region bounded by y=√x and x=y.

Solution:

Given, 

y = √x, x = y 

Area of the shaded region = Area under the curve along the x-axis - Area under the line along the x-axis

= \int^1_0(\sqrt{x}-x)dx

= [\frac{x^{3/2}}{3/2}-\frac{x^2}{2}]_0^1

Putting the integration limits, we get:

Area of shaded region = = [\frac{1^{3/2}}{3/2}-\frac{1^2}{2} - 0]

                                     = 2/3 -1/2

                                     = 1/6 sq. units

Question 4. Find the area bounded by the curve y=4-x² and the lines y=0 and y=3.

Solution:

Given, 

y = 4-x², y = 0, y = 3 

Area of shaded region = Area of region OCBEO + Area of region ODAEO

As the Y-axis divides the parabola into two symmetrical regions, the area of OCBEO = area of ODAEO

So, 

Area of shaded region = 2 Area of ODAEO

                                       = 2\int^2_0(4-x^2)dx 

                                       = 2[4x-x^3/3]^2_0

Putting the limits, we get

Area of shaded region = 2[4(2)-2^3/3-0]

                                   = 2[8-8/3]

                                   = 32/3 sq. units

Question 5. Find the area of the region [(x,y):\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1 \leq \frac{x}{a}+\frac{y}{b}]

Solution:

Given

 \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \implies y = \frac{b}{a}\sqrt{a^2-x^2}

\frac{x}{a}+\frac{y}{b}=1 \implies y=\frac{b}{a}(a-x)

Area of shaded region = Area under the sector OAB - Area of triangle OAB

= \int^a_0(\frac{b}{a}\sqrt{a^2-x^2}-\frac{b}{a}(a-x))dx

= \frac{b}{a}\int^a_0(\sqrt{a^2-x^2}-(a-x))dx

\frac{b}{a}[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}(\frac{x}{a})-ax+\frac{x^2}{2}]^a_0

Putting the upper and lower limits, we get:

Area of shaded region = \frac{b}{a}[\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}sin^{-1}(\frac{a}{a})-a(a)+\frac{a^2}{2}]

                                    =\frac{b}{a}[\frac{a}{2}*0+\frac{a^2}{2}(\frac{\pi}{2})-a^2+\frac{a^2}{2}]

                                    =\frac{b}{a}[\frac{a^2\pi}{4}-\frac{a^2}{2}]

                                    = \frac{b}{a}[\frac{a^2\pi}{4}-\frac{a^2}{2}]

                                    = \frac{b}{a}*a^2[\frac{\pi}{4}-\frac{1}{2}]

                                    =\frac{ab}{4}[\pi-2] sq. units

Question 6. Using integration, find the area of the region bounded by the triangle whose vertices are (2,1), (3,4), and (5,2).

Solution:

Given, A(2,1), B(3,4) and C(5,2). We need to find the area of triangle ABC. To calculate the area we need to find the equations of the lines AB, BC, and CA.

The equation of a line can be calculated as:

y-y_1=[\frac{y_2-y_1}{x_2-x_1}](x-x_1)

Equation of AB:

y-1=[\frac{4-1}{3-2}](x-2)

y-1=[\frac{3}{1}](x-2)

y-1 = 3x-6 ⇒ y=3x-5

Equation of BC:

y-4=[\frac{2-4}{5-3}](x-3)

y-4=[\frac{-2}{2}](x-3)

y-4= -x+3 ⇒ y=-x+7

Equation of CA:

y-1=[\frac{2-1}{5-2}](x-2)

y-1=[\frac{1}{3}](x-2)

  y-1=\frac{x}{3}-\frac{2}{3}

y = x/3 + 1/3 ⇒ 3y = x+1

Required Area = Area of Region I + Area of Region II..................(1)

Area of Region I = Area under AB from x=2 to x=3 - Area under AC from x=2 to x=3

= \int^3_2(3x-5-\frac{1}{3}(x+1))dx

=\int^3_2(\frac{8x}{3}-\frac{16}{3})dx

=\int^3_2\frac{8}{3}(x-\frac{2}{3})dx

=\frac{8}{3}[\frac{x^2}{2}-2x]^3_2

=\frac{8}{3}[\frac{3^2}{2}-2*3 -\frac{2^2}{2}+2*2]

= \frac{8}{3}[\frac{9-4}{2}-2(3-2)]

= \frac{8}{3}[\frac{5}{2}-2] = \frac{4}{3}

Area of Region I = 4/3 sq. units .....................(2)

Area of Region II = Area under BC from x=3 to x=5 - Area under AC from x=3 to x=5

= \int^5_3(-x+7-\frac{1}{3}(x+1))dx

=\int^5_3(\frac{-4x}{3}+\frac{20}{3})dx

=[\frac{-4x^2}{6}+\frac{20x}{3}]^5_3

= [\frac{-100}{6}+\frac{100}{3}-(\frac{-36}{6}+\frac{60}{3})]

= 8/3 sq. units

Area of Region II = 8/3 sq. units ............................(3)

Putting the value of (2) and (3) in (1), we get:

Required Area = Area of region I + Area of region II 

= 4/3 + 8/3 = 4 sq. units

Question 7. Using integration, find the area of the region bounded by the triangle whose vertices A, B, and C are (-1,1), (0,5), and (3,2).

Solution:

Given, A (2,1), B (3,4) and C (5,2). We need to find the area of triangle ABC. To calculate the area we need to find the equations of the lines AB, BC, and CA.

The equation of a line can be calculated as:

y-y_1=[\frac{y_2-y_1}{x_2-x_1}](x-x_1)

Equation of AB:

y-1=[\frac{5-1}{0+1}](x+1)

y-1=[\frac{4}{1}](x+1)

y-1 = 4x+4 

y = 4x+5

Equation of BC:

y-5=[\frac{2-5}{3-0}](x-0)

y-5=[\frac{-3}{3}](x-0)

y-5=-1(x-0)

y-5= -x 

y = -x+5

Equation of CA

y-1=[\frac{2-1}{3+1}](x+1)

y-1=[\frac{1}{4}](x+1)

y-1=[\frac{x}{4}+\frac{1}{4}]

y = x/4 + 5/4

4y = x+5

Required Area = Area of Region I + Area of Region II   ..................(1)

Area of Region I = Area under AB from x=-1 to x=0 - Area under AC from x=-1 to x=0

= \int^0_{-1}(4x+5-\frac{1}{4}(x+5))dx

=\int^0_{-1}(\frac{15x}{4}+\frac{15}{4})dx  =\frac{15}{4}\int^0_{-1}(x+1)dx

= \frac{15}{4}[\frac{x^2}{2}+x]_{-1}^0

= \frac{15}{4}[0-(\frac{1}{2}-1)] = \frac{15}{8}

Area of Region I = 15/8 sq. units .....................(2)

Area of Region II = Area under BC from x=0 to x=3 - Area under AC from x=0 to x=3

= \int^3_0(-x+5-\frac{1}{4}(x+5))dx

=\int^3_0(\frac{-5x}{4}+\frac{15}{4})dx

=\frac{5}{4}\int^3_0(-x+3)dx

= \frac{5}{4}[\frac{-x^2}{2}+3x]^3_0

 \frac{5}{4}[\frac{-3^2}{2}+3(3)]

= \frac{5}{4}[\frac{-9}{2}+9-(0)]

 = \frac{45}{8} sq.~units

Area of Region II = 45/8 sq. units ............................(3)

Putting the value of (2) and (3) in (1), we get:

Required Area = 15/8+ 45/8 

                       = 15/2 sq. units

Question 8. Using integration, find the area of triangular the region, the equations of whose sides are y = 2x+1, y = 3x+1, and x = 4.

Solution:

Area of the shaded region in the above graph can be calculated as:

Required area = Area under line y=3x+1 from x=0 to x=4 - Area under the line y=2x+1 from x=0 to x=4 

= \int_0^4(3x+1-2x-1)dx

= \int^4_0xdx

= [\frac{x^2}{2}]^4_0

= [\frac{4^2}{2}]

= 8 sq. units

Question 9. (x - a)² + (y - b)² = r². Find the area of the region {(x, y): y²≤8x, x²+y²≤9}

Solution:

Given,

y² = 8x ⇒ y = √8x

x² + y²=9 ⇒ y = √(9-x²)

As the circle is divided by the parabola into two symmetrical halves,

Area of the shaded region = 2(Area under the parabola y²=8x from x=0 to x=1 + Area under the circle x²+y²=9 from x = 1 to x = 3)

= 2(\int_0^1\sqrt{8x}dx + \int^3_1\sqrt{9-x^2}dx)

= 2(\int_0^12\sqrt{2x}dx + \int^3_1\sqrt{3^2-x^2}dx)

= 2([\frac{2\sqrt{2}x^\frac{3}{2}}{3/2}]^1_0 + [\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}sin^{-1}(\frac{x}{3})]^3_1)

= 2(\frac{4\sqrt{2}}{3}+\frac{3}{2}\sqrt{9-9}+\frac{9}{2}sin^{-1}(1)-\frac{1}{2}-\sqrt{9-1}+\frac{9}{2}sin^{-1}(\frac{1}{3}))

= 2(\frac{4\sqrt{2}}{3}+\frac{3}{2}(0)+\frac{9}{2}(\frac{\pi}{2})-\frac{1}{2}\sqrt{9-1}+\frac{9}{2}sin^{-1}(\frac{1}{3}))

= 2[\frac{4\sqrt{2}}{3}+\frac{9\pi}{4}-\sqrt{2}-\frac{9}{2}sin^{-1}(\frac{1}{3})]

Required Area = 2[\frac{\sqrt{2}}{3}+\frac{9\pi}{4}-\frac{9}{2}sin^{-1}(\frac{1}{3})]~sq.~units

Question 10. Find the area of the region bounded by the circle x²+y²=16 and the parabola y²=6x.

Solution:

Given 

y² = 6x ⇒ y = √6x

x² + y²=16 ⇒ y = √(16-x2)

As the circle is divided by the parabola into two symmetrical halves,

Area of the shaded region = 2(Area under the parabola y²=6x from x=0 to x=2 + Area under the circle x²+y²=16 from x = 2 to x = 4)

= 2(\int_0^2\sqrt{6x}dx + \int^4_2\sqrt{16-x^2}dx)

= 2([\frac{\sqrt{6}x^\frac{3}{2}}{3/2}]^2_0 + [\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}sin^{-1}(\frac{x}{4})]^2_4)

= 2([\frac{\sqrt{6}x^\frac{3}{2}}{3/2}]^2_0 + [\frac{x}{2}\sqrt{16-x^2}+8sin^{-1}(\frac{x}{4})]^2_4)

= 2(\frac{4\sqrt{12}}{3}+\frac{16}{2}\sqrt{16-16}+8\sin^{-1}(1)-\frac{1}{2}-\sqrt{12}-8\sin^{-1}(\frac{1}{2}))

= 2(\frac{4\sqrt{12}}{3}+\frac{16}{2}(0)+8(\frac{\pi}{2})-\frac{1}{2}-\sqrt{12}-8(\frac{\pi}{6}))

= 2[\frac{8\sqrt{3}}{3}+4\pi-2\sqrt{3}-\frac{4\pi}{3}]

= 2[\frac{2\sqrt{3}}{3}+\frac{8\pi}{3}]

= \frac{4}{3}(4\pi+\sqrt{3})

Required Area = \frac{4}{3}(4\pi+\sqrt{3})    sq.units

Read More:

• Area Between Two Curves: Formula, Definition and Examples
• Area Under Curve