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

Last Updated : 29 Mar, 2023
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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)

Areas of Bounded Regions: Exercise 21.3 Question 1

 

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)

Areas of Bounded Regions: Exercise 21.3 Question 2

 

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                 [Tex] [/Tex]

Putting the upper and lower limits we get,

Required Area = 43/2-43/16

                       = 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 

Areas of Bounded Regions: Exercise 21.3 Question 3

 

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 

Areas of Bounded Regions: Exercise 21.3 Question 4

 

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)

Areas of Bounded Regions: Exercise 21.3 Question 5

 

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.

Areas of Bounded Regions: Exercise 21.3 Question 6

 

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.

Areas of Bounded Regions: Exercise 21.3 Question 7

 

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:

Areas of Bounded Regions: Exercise 21.3 Question 8

 

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)

Areas of Bounded Regions: Exercise 21.3 Question 9

 

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



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