Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 2 - Areas Of Bounded Regions

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Page No 20.14:

Question 1:

Using integration, find the area of the region bounded between the line x = 2 and the parabola y² = 8x.

Answer:

$y^{2} = 8 x represents a parabola with vertex at origin and axis of symmetry along the + ve direction of x - axis x = 2 is line parallel to y - axis Let (x, y) be a given point on the parabola, y^{2} = 8 x Since parabola y^{2} = 8 x is symmetric about x - axis, ∴ Required area = 2 (area OCAO) On slicing the area above x - axis into vertical strips of length = |y| and width = d x \Rightarrow area of rectangular strip = |y| d x The approximating rectangle moves between x = 0 and x = 2 . So, area A = 2 (area OCAO) \Rightarrow A = 2 \int_{0}^{2} |y| d x = 2 \int_{0}^{2} y d x as y > 0 \Rightarrow A = 2 \int_{0}^{2} \sqrt{8 x} d x \Rightarrow A = 2 \times 2 \int_{0}^{2} \sqrt{2 x} d x = 4 \sqrt{2} \int_{0}^{2} \sqrt{x} d x \Rightarrow A = 4 \sqrt{2} {[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]}_{0}^{2} = \frac{8}{3} \sqrt{2} [2^{\frac{3}{2}} - 0] = \frac{8}{3} \times 2^{2} = \frac{32}{3} sq . units ∴ Area A = \frac{32}{3} sq . units$

x	0	$\frac{π}{6}$	$\frac{π}{2}$	$\frac{5 π}{6}$	$π$
sin x	0	$\frac{1}{2}$	1	$\frac{1}{2}$	0
$y = \frac{π}{2} + 2 \sin^{2} x$	1.57	2.07	3.57	2.07	1.57

y = \frac{π}{2} + 2 \sin^{2} x is an arc cutting y - axis at (1.57, 0) and x = π at (π, 1.57) x = π is a line parallel to y - axis Consider, a vertical strip of length = |y| and width = d x in the first quadrant ∴ Area of the approximating rectangle = |y| d x The approximating rectangle moves from x = 0 to x = π \Rightarrow Area of the shaded region = \int_{0}^{π} |y| d x \Rightarrow A = \int_{0}^{π} y d x \Rightarrow A = \int_{0}^{π} (\frac{π}{2} + 2 \sin^{2} x) d x \Rightarrow A = \int_{0}^{π} (\frac{π}{2} + 2 (\frac{1 - \cos 2 x}{2})) d x \Rightarrow A = \frac{π}{2} \int_{0}^{π} d x + \int_{0}^{π} (1 - \cos 2 x) d x \Rightarrow A = \frac{π}{2} {[x]}_{0}^{π} + {[x - \frac{\sin 2 x}{2}]}_{0}^{π} \Rightarrow A = \frac{π}{2} (π) + [π - \frac{\sin 2 π}{2} - 0] \Rightarrow A = π (\frac{π}{2} + 1) \Rightarrow A = π (\frac{π + 2}{2}) \Rightarrow A = \frac{π}{2} (π + 2) sq . units ∴ Area of curve bound by x = 0 and x = π is \frac{π}{2} (π + 2) sq . units

Page No 20.15:

Question 22:

Draw a rough sketch of the curve $y = \frac{x}{π} + 2 \sin^{2} x$ and find the area between the x-axis, the curve and the ordinates x = 0 and x = π.

Answer:

The table for different values of x and y is

x	$\frac{π}{6}$	$\frac{π}{2}$	$\frac{5 π}{6}$	$π$
sinx	$\frac{1}{2}$	1	$\frac{1}{2}$	0
$y = \frac{x}{π} + 2 \sin^{2} x$	$\frac{2}{3}$	$\frac{5}{2}$	$\frac{4}{3}$	1

y = \frac{x}{π} + 2 \sin^{2} x, is an arc cutting y - axis at O (0, 0) and cutting x = π at (π, 1) Consider a vertical strip of length = |y| and width = d x in the first quadrant ∴ Area of approximating rectangle = |y| d x The approximating rectangle moves from x = 0 to x = π \Rightarrow Area of the shaded area = \int_{0}^{π} |y| d x \Rightarrow A = \int_{0}^{π} y d x \{As, y > 0 \Rightarrow |y| = y\} \Rightarrow A = \int_{0}^{π} (\frac{x}{π} + 2 \sin^{2} x) d x \Rightarrow A = \frac{1}{π} \int_{0}^{π} x d x + 2 \int_{0}^{π} \sin^{2} x d x \Rightarrow A = \frac{1}{π} {[\frac{x^{2}}{2}]}_{0}^{π} + 2 {[\frac{x}{2} - \frac{1}{2} \sin x \cos x]}_{0}^{π} \Rightarrow A = \frac{π^{2}}{2 π} + \frac{2}{2} [π - \frac{1}{2} \sin π \cos π - 0] \Rightarrow A = \frac{π}{2} + π \Rightarrow A = \frac{3 π}{2} sq . units ∴ Area of the curve enclosed between x = 0 and x = π is \frac{3 π}{2} sq . units

x	0	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$	$π$
$y = \cos^{2} x$	1	0.5	0.25	0	0.25	0.75	1
$y = \sin^{2} x$	0	0.5	0.75	1	0.75	0.25	0

Let A₁ be the area of curve

y = \cos^{2} x between x = 0 and x = π

Let A₂ be the area of curve

y = \sin^{2} x between x = 0 and x = π

Consider, a vertical strip of length

= |y|

and width

= d x

in the shaded region of both the curves

The area of approximating rectangle

= |y| d x

The given region is the intersection of
          $y^{2} \leq 3 x and 3 x^{2} + 3 y^{2} \leq 16$
Clearly ,y² = 3x is a parabola with vertex at (0, 0) axis is along the x-axis opening in the positive direction.
Also 3x² + 3y² = 16 is a circle with centre at origin and has a radius $\sqrt{\frac{16}{3}}$ .
Corresponding equations of the given inequations are
$y^{2} = 3 x . . . . . (1) and 3 x^{2} + 3 y^{2} = 16 . . . . . (2)$
Substituting the value of y² from (1) into (2)
$3 x^{2} + 9 x = 16$
$\Rightarrow 3 x^{2} + 9 x - 16 = 0$
$\Rightarrow x = \frac{- 9 \pm \sqrt{81 + 192}}{6}$
$\Rightarrow x = \frac{- 9 \pm \sqrt{273}}{6}$
By figure we see that the value of x will be non-negative.
$∴ x = \frac{- 9 + \sqrt{273}}{6}$
Now assume that x-coordinate of the intersecting point, $a = \frac{- 9 + \sqrt{273}}{6}$
The Required area A = 2(Area of OACO + Area of CABC)
Approximating the area of OACO the length $= |y_{1}|$ width = dx
    Area of   OACO $= \int_{0}^{a} |y_{1}| d x$
                            $= \int_{0}^{a} y_{1} d x$
                            $= \int_{0}^{a} \sqrt{3 x} d x (∵ {y^{2}}_{1} = \sqrt{3 x} \Rightarrow y_{1} = \sqrt{3 x})$
                            $= {[\frac{2 \sqrt{3} x^{\frac{3}{2}}}{3}]}^{a}_{0}$
                        $= \frac{2 \sqrt{3} a^{\frac{3}{2}}}{3}$
Therefore, Area of OACO $= \frac{2 \sqrt{3} a^{\frac{3}{2}}}{3}$
Similarly approximating the are of CABC the length $= |y_{2}|$    and the width = dx
              Area of CABC $= \int_{a}^{\frac{4}{\sqrt{3}}} |y_{2}| d x$
                                     $= \int_{a}^{\frac{4}{\sqrt{3}}} y_{2} d x$
                               $= \int_{a}^{\frac{4}{\sqrt{3}}} \sqrt{\frac{16}{3} - x^{2}} d x (∵ 3 x^{2} + 3 {y_{2}}^{2} = 16 \Rightarrow y_{2} = \sqrt{\frac{16}{3} - x^{2}})$
                                     $= {[\frac{x}{2} \sqrt{\frac{16}{3} - x^{2}} + \frac{8}{3} \sin^{- 1} (\frac{x \sqrt{3}}{4})]}_{a}^{\frac{4}{\sqrt{3}}}$
                                $= [\frac{4}{2 \sqrt{3}} \sqrt{\frac{16}{3} - \frac{16}{3}} + \frac{8}{3} \sin^{- 1} (\frac{4 \sqrt{3}}{4 \sqrt{3}}) - \frac{a}{2} \sqrt{\frac{16}{3} - a^{2}} - \frac{8}{3} \sin^{- 1} (\frac{a \sqrt{3}}{4})]$

   $= - \frac{a}{2} \sqrt{\frac{16}{3} - a^{2}} + \frac{8}{3} \sin^{- 1} (1) - \frac{8}{3} \sin^{- 1} (\frac{a \sqrt{3}}{4})$
Area of CABC $= - \frac{a}{2} \sqrt{\frac{16}{3} - a^{2}} + \frac{4 π}{3} - \frac{8}{3} \sin^{- 1} (\frac{a \sqrt{3}}{4})$
Thus the required area A = 2(Area of OACO + Area of CABC)
                                $= 2 [\frac{2 \sqrt{3} a^{\frac{3}{2}}}{3} - \frac{a}{2} \sqrt{\frac{16}{3} - a^{2}} + \frac{4 π}{3} - \frac{8}{3} \sin^{- 1} (\frac{a \sqrt{3}}{4})]$
               $= \frac{4 a^{\frac{3}{2}}}{\sqrt{3}} - a \sqrt{\frac{16}{3} - a^{2}} + \frac{8 π}{3} - \frac{16}{3} \sin^{- 1} (\frac{a \sqrt{3}}{4})$
$Hence, the reguired area is, \frac{4 a^{\frac{3}{2}}}{\sqrt{3}} - a \sqrt{\frac{16}{3} - a^{2}} + \frac{8 π}{3} - \frac{16}{3} \sin^{- 1} (\frac{a \sqrt{3}}{4}) square units, where a = \frac{- 9 + \sqrt{273}}{6}$

Page No 20.51:

Question 14:

Draw a rough sketch of the region {(x, y) : y² ≤ 5x, 5x² + 5y² ≤ 36} and find the area enclosed by the region using method of integration.

Answer:

The given region is intersection of
$y^{2} \leq 5 x and 5 x^{2} + 5 y^{2} \leq 36$
Clearly, $y^{2} \leq 5 x$ is a parabola with vertex at origin and the axis is along the x-axis opening in the positive direction.Also $5 x^{2} + 5 y^{2} \leq 36$ is a circle with centre at the origin and has a radius $\sqrt{\frac{36}{5}} or \frac{6}{\sqrt{5}}$ .
Corresponding equations of given inequations are
$y^{2} = 5 x . . . . . (1) 5 x^{2} + 5 y^{2} = 36 . . . . . (2)$
Substituting the value of y² from (1) into (2), we get

5 x^{2} + 25 x = 36

\Rightarrow 5 x^{2} + 25 x - 36 = 0

\Rightarrow x = \frac{- 25 \pm \sqrt{625 + 720}}{10}

\Rightarrow x = \frac{- 25 \pm \sqrt{1345}}{10}

From the figure we see that x-coordinate of intersecting point can not be negative.

∴ x = \frac{- 25 + \sqrt{1345}}{10}

Now assume that x-coordinate of intersecting point,

a = \frac{- 25 + \sqrt{1345}}{10}

The Required area,
A = 2(Area of OACO + Area of CABC)
Approximating the area of OACO the length =

|y_{1}|

and a width = dx=|y1=dx

Area of O A C O = \int_{0}^{a} |y_{1}| d x

= \int_{0}^{a} y_{1} d x

= \int_{0}^{a} \sqrt{5 x} d x (∵ {y_{1}}^{2} = 5 x \Rightarrow y_{1} = \sqrt{5 x})

\sqrt{5} {[\frac{2 x^{\frac{3}{2}}}{3}]}^{a}_{0}

Therefore, Area of OACO

= \frac{2 \sqrt{5} a^{\frac{3}{2}}}{3}

Similarly approximating the area of CABC the length

= |y_{2}|

and the width = dx

Area of C A B C = \int_{a}^{\frac{6}{\sqrt{5}}} |y_{2}| d x

= \int_{a}^{\frac{6}{\sqrt{5}}} y_{2} d x

= \int_{a}^{\frac{6}{\sqrt{5}}} \sqrt{\frac{36}{5} - x^{2}} d x

= {[\frac{x}{2} \sqrt{\frac{36}{5} - x^{2}} + \frac{18}{5} \sin^{- 1} (\frac{x \sqrt{5}}{6})]}_{a}^{\frac{6}{\sqrt{5}}}

= \frac{6}{2 \sqrt{5}} \sqrt{\frac{36}{5} - \frac{36}{5}} + \frac{18}{5} \sin^{- 1} (1) - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

= 0 + \frac{18}{5} \sin^{- 1} (1) - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

= \frac{18}{5} \times \frac{π}{2} - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

Area of CABC

= \frac{9 π}{5} - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

Thus the Required area, A = 2(Area of OACO + Area of CABC)

A = 2 [\frac{2 \sqrt{5} a^{\frac{3}{2}}}{3} + \frac{9 π}{5} - \frac{a}{2} \sqrt{\frac{36}{5} - a^{2}} - \frac{18}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})]

= \frac{4 \sqrt{5} a^{\frac{3}{2}}}{3} + \frac{18 π}{5} - a \sqrt{\frac{36}{5} - a^{2}} - \frac{36}{5} \sin^{- 1} (\frac{a \sqrt{5}}{6})

Where, a = \frac{- 25 + \sqrt{1345}}{10}

Page No 20.51:

To find: area of the region bounded by the curve y = x + 1, x-axis and the lines x = 2 and x = 3

The required area of the region = $\int_{2}^{3} y d x$
Thus,
$Required area = \int_{2}^{3} y d x = \int_{2}^{3} (x + 1) d x = {(\frac{x^{2}}{2} + x)}_{2}^{3} = (\frac{3^{2}}{2} + 3) - (\frac{2^{2}}{2} + 2) = (\frac{9}{2} + 3) - (2 + 2) = (\frac{9 + 6}{2} - 4) = \frac{15}{2} - 4 = \frac{15 - 8}{2} = \frac{7}{2} Thus, Area = \frac{7}{2} sq . units$

Hence, the area of the region bounded by the curve y = x + 1, x-axis and the lines x = 2 and x = 3 is $\frac{7}{2} sq . units .$

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