Rd Sharma XII Vol 2 2021 Solutions for Class 12 Science Maths Chapter 2 Areas Of Bounded Regions are provided here with simple step-by-step explanations. These solutions for Areas Of Bounded Regions are extremely popular among Class 12 Science students for Maths Areas Of Bounded Regions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2021 Book of Class 12 Science Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2021 Solutions. All Rd Sharma XII Vol 2 2021 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

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:



y2=8x  represents a parabola with vertex at origin and axis of symmetry along the +ve direction of x-axisx=2 is line parallel to y-axisLet (x, y)  be a given point on the parabola , y2=8xSince parabola   y2=8x   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 =dx area of rectangular strip=y dxThe approximating rectangle moves between x=0 and x=2. So, area A=2 area OCAOA=202y dx=202y dx           as y>0  A=2028x dx A=2×2022x dx= 4202x dxA= 42 x323202=832 232-0 =83×22=323  sq. units Area A=323 sq. units

Page No 20.14:

Question 2:

Using integration, find the area of the region bounded by the line y − 1 = x, the x − axis and the ordinates x = −2 and x = 3.

Answer:




 y-1 =x  is a straight line cutting x-axis at D(-1, 0) and y-axis at A( 0, 1) And, x=-2 and x=3 are straight  lines parallel to y-axis Also, since y =-1+x,       for x-1And             y = 1+x,                x>-1 Area of region bound by line y-1=x , x axis and the ordinates x=-2 and x=3 isArea A = area FED+area DOA +area OABCA=-23y dxA=-2-1y dx   +-10y dx  +03y dxA=-2-1-1+x dx +-101+x dx+031+x dxA=-x+x22-2-1+x+x22-10+x+x2203A=1-12-2+42+0+1-12+3+92A=-1+32+1-12+3+92=3+3-1+92=3+112=172  sq. unitsArea of the bound region =172  sq. units



Page No 20.15:

Question 3:

Find the area of the region bounded by the parabola y2 = 4ax and the line x = a.

Page No 20.15:

Question 4:

Find the area lying above the x-axis and under the parabola y = 4xx2.

Answer:




The equation y=4x-x2 represents a parabola opening downwards and cutting the x axis at O(0, 0) and B(4, 0)Slicing the region above x axis in vertical strips of length= y and width =dx , area of corresponding rectangle is = y dxSince the corresponding rectangle can move from x=0 to x=4,Required area OABO isA=04y dx= 04y dx                    As, y>0 for 0 x 4  y =y A=044x-x2 dx   A=4x22-x3304 A=32-643 A=323 square units

Page No 20.15:

Question 5:

Draw a rough sketch to indicate the region bounded between the curve y2 = 4x and the line x = 3. Also, find the area of this region.

Answer:



y2=4x  represents a parabola with vertex at (0, 0) and axis of symmetry along the +ve direction of x axis x=3 is a line parallel to y axis and cutting x axis at (3, 0)Since  y2=4x  is symmetrical about x axis , Required area A=OACO=2× area OABOSlicing the area above x axis into vertical strips of length=y and width = dx    Area of corresponding rectangle= y dxThe corresponding rectangle moves from x=0 to x=3A=2× area OABOA=203 y dx=  203 y dx                   As, y=y, y>0A=  2034x dx A=403x dx A=4x323203= 83x3203=83×33=83  sq. unitsArea of region bound by curve y2=4x and x=3  is  83  sq. units

                   

         

     

Page No 20.15:

Question 6:

Make a rough sketch of the graph of the function y = 4 − x2, 0 ≤ x ≤ 2 and determine the area enclosed by the curve, the x-axis and the lines x = 0 and x = 2.

Answer:




y=4-x2, 0x2 represents a half parabola with vetex at (4, 0) x=2 represents a line parallel to y-axis and cutting x-axis at (2, 0) In quadrant OABO, consider a vertical strip of length= y, width=dxArea of approximating rectangle= y dx  The approximating rectangle  moves from x=0 to x=2A=Area OABO  =02 y dx A=02 y dx                            As, y>o, y =yA=024-x2 dx   A=4x-x3302A=8-83A=163 sq. unitsThe area enclosed by the curve and x-axis and given lines =163 sq. units

Page No 20.15:

Question 7:

Sketch the graph of y = x+1 in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the lines x = 0, x = 4.

Answer:



y=x+1  in 0, 4 represents a curve which is part of a parabola x=4 represents a line parallel to y-axis and cutting x-axis at (4, 0)Enclosed area bound by the curve and lines x=0 and x=4 is OABCOConsider a vertical strip  of lenght =y and width=dxArea of approximating rectangle= y dxThe approximating rectangle moves from x=0 to x=4A=Area OABCO=04y dxA=04y dx                  y>0y =yA=04x+1 dxA=04x+112 dxA=x+1323204A=23532-1 sq. units Enclosed area between the curve and given lines =23532-1  sq. units

Page No 20.15:

Question 8:

Find the area under the curve y = 6x+4 above x-axis from x = 0 to x = 2. Draw a sketch of curve also.

Answer:




y=6x+4  represents a parabola, with vertex V(-23, 0) and symmetrical about x-axis x=0 is the y-axis . The curve cuts it at A(0, 2 ) and A'(0, -2)x=2 is a line parallel to y-axis, cutting the x-axis at  C(2, 0)The enclosed area of the curve between x=0 and x=2 and above x-axis =area OABCConsider, a vertical strip of length =y  and width =dx Area of approximating rectangle =y dxThe approximating rectangle moves from x=0  to x=2  area OABC =02y dxA=02y dx                As, y>0 , y =yA=026x+4 dx  A=026x+412 dxA=166x+4323202A=2181632-432A=21843-23A=21864-8 =218×56 =569 sq. unitsEnclosed area =569 sq. units

Page No 20.15:

Question 9:

Draw the rough sketch of y2 + 1 = x, x ≤ 2. Find the area enclosed by the curve and the line x = 2.

Answer:


y2+1=x, x2  is a parabola with vertex (1, 0) and symmetrical about +ve side of x-axis x=2 is a line parallel to y-axis and cutting x-axis at (2, 0)Consider, a vertical section of height= y  and width =dx in the first quadrant area of corresponding rectangle =  y dxSince, the corresponding rectangle is moving from x=1 to x=2 and the curve being symmetricalA=area ABCA=2×area ABDAA=212y dx A=212y dx                          y=y as y>0A=212x-1 dx                y2+1=x  y=x-1 A=2x-1323212  A=43132-0A=43  sq. unitsEnclosed area=43 sq. units

Page No 20.15:

Question 10:

Draw a rough sketch of the graph of the curve x24+y29=1 and evaluate the area of the region under the curve and above the x-axis.

Answer:


Since in the given equation  x24+y29 =1, all the powers of both x and y are even, the curve is symmetrical about both the axis .Area encloed by the curve and above x axis = area A'BA =2×area enclosed by ellipse and x-axis in first quadrant(2, 0 ), (-2, 0) are the points of intersection of curve and x-axis(0, 3), (0, -3) are the points of intersection of curve and y-axisSlicing the area in the first quadrant into vertical stripes of height =y  and width =dxArea of approximating  rectangle =y dxApproximating  rectangle can move between x=0 and x=2 A=Area of  enclosed curve above x-axis =202y dxA=202y dxA=202324-x2 dxA=3024-x2 dxA=312x 4-x2 +12 4 sin-1x02=30+12×4 sin-11=3×12×4×π2= 3π sq. unitsArea of enclosed region above x-axis = 3π sq. units

Page No 20.15:

Question 11:

Sketch the region {(x, y) : 9x2 + 4y2 = 36} and find the area of the region enclosed by it, using integration.

Answer:



We have,9x2+4y2=36     .....14y2=36-9x2y2=944-x2y=324-x2     .....2From 1 we getx24+y29 =1Since in the given equation  x24+y29 =1, all the powers of both x and y are even, the curve is symmetrical about both the axes.Area enclosed by the curve and above x axis = 4×area enclosed by ellipse and x-axis in first quadrant(2, 0 ), (-2, 0) are the points of intersection of curve and x-axis(0, 3), (0, -3) are the points of intersection of curve and y-axisSlicing the area in the first quadrant into vertical stripes of height =y  and width =dxArea of approximating  rectangle =y dxApproximating  rectangle can move between x=0 and x=2A=Area of  enclosed curve=402y dxA=402324-x2dx    From 2=4×32024-x2dx=60222-x2dx=6x222-x2+1222sin-1x202=60+124sin-11=612×4π2A=6π sq units

Page No 20.15:

Question 12:

Draw a rough sketch of the graph of the function y = 21-x2, x ∈ [0, 1] and evaluate the area enclosed between the curve and the x-axis.

Answer:



We have,y=21-x2y2=1-x2y24=1-x2x2+y24=1x21+y24=1Since in the given equation  x21+y24 =1, all the powers of both x and y are even, the curve is symmetrical about both the axes .Required area=area enclosed by ellipse and x-axis in first quadrant(1, 0 ), (-1, 0) are the points of intersection of curve and x-axis(0, 2), (0, -2) are the points of intersection of curve and y-axisSlicing the area in the first quadrant into vertical stripes of height =y  and width =dxArea of approximating  rectangle =y dxApproximating  rectangle can move between x=0 and x=1 A=Area of  enclosed curve above x-axis =01y dxA=0121-x2 dx=2011-x2 dx=212x1-x2+12sin-1x01=212sin-11=212π2-0A=π2 sq.units

Page No 20.15:

Question 13:

Determine the area under the curve y = a2-x2 included between the lines x = 0 and x = a.

Answer:



We have,y=a2-x2y2=a2-x2x2+y2=a2Since in the given equation  x2+y2=a2, all the powers of both x and y are even, the curve is symmetrical about both the axis .Required area = area enclosed by circle in first quadrant(a, 0 ), (-a, 0) are the points of intersection of curve and x-axis(0, a), (0, -a) are the points of intersection of curve and y-axisSlicing the area in the first quadrant into vertical stripes of height =y  and width =dxArea of approximating  rectangle =y dxApproximating  rectangle can move between x=0 and x=aA=Area of  enclosed curve in first quadrant =0ay dxA=0aa2-x2 dx=12xa2-x2+12a2sin-1xa0a=12a2sin-11=12a2π2 =a2π4 sq units

Page No 20.15:

Question 14:

Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8.

Answer:



We have,
Straight line 2y = 5x + 7 intersect x-axis and y-axis at ( −1.4, 0) and (0, 3.5) respectively.
Also x = 2 and x = 8 are straight lines as shown in the figure.
The shaded region is our required region whose area has to be found.
When we slice the shaded region into vertical strips, we find that each vertical strip has its lower end on x-axis and upper end on the line
2y = 5x + 7
So, approximating rectangle shown in figure has length = y and width = dx and area = y dx.
The approximating rectangle can move from x = 2 to x = 8.
So, required is given by,

A=28y dx=285x+72 dx=1228(5x+7) dx=1252x2+7x28=1252×64+56-52×4-14=12×192=96 sq units

Page No 20.15:

Question 15:

Using definite integrals, find the area of the circle x2 + y2 = a2.

Answer:


Area of the circle x2 + y2 = a2 will be the 4 times the area enclosed between x = 0 and x = a in the first quadrant which is shaded.

A=40ay dx=40aa2-x2 dx=412xa2-x2+12a2sin-1xa0a=40+12a2sin-11=412a2π2           sin-11=π2=a2π sq units

Page No 20.15:

Question 16:

Using integration, find the area of the region bounded by the following curves, after making a rough sketch: y = 1 + | x + 1 |, x = −2, x = 3, y = 0.

Answer:



We have,

y = 1 + | x + 1 | intersect x = − 2 and at ( −2, 2) and x = 3 at (3, 5).
And y = 0 is the x-axis.
The shaded region is our required region whose area has to be found

y=1+x+1=1-x+1    x<-11+x+1      x1=-x       x<-1x+2      x1

Let the required area be A. Since limits on x are given, we use horizontal strips to find the area:
A=-23ydx=-2-1ydx+-13ydx=-2-1-x dx+-13x+2dx=-x22-2-1+x22+2x-13=-12-42+92+6-12+2=32+8+82=32+8+4=272 sq. units     

Page No 20.15:

Question 17:

Sketch the graph y = | x − 5 |. Evaluate 01x-5 dx. What does this value of the integral represent on the graph.

Answer:

We have,
y = | x − 5 | intersect x = 0 and x = 1 at (0, 5) and (1, 4)
Now,
y=x-5=-x-5     For all x0, 1

Integration represents the area enclosed by the graph from x = 0 to x = 1
A=01ydx=01x-5 dx=01-x-5 dx=-01x-5 dx=-x22-5x01=-12-5-0-0=--92=92 sq. units


Page No 20.15:

Question 18:

Sketch the graph y = | x + 3 |. Evaluate -60x+3 dx. What does this integral represent on the graph?

Answer:



We have,
y = | x + 3 | intersect x = 0 and x = −6 at (0, 3) and (−6, 3)
Now,
y=x+3=x+3     For all x>-3-x+3     For all x<-3


Integral represents the area enclosed between x = 6 and x = 0
A=-60y dx=-6-3y dx+-30y dx=-6-3-x+3 dx+-30x+3 dx=-x22+3x-6-3+x22+3x-30=-92-9-362+18+0+0-92+9=-92+9+362-18-92+9=9 sq. units

Page No 20.15:

Question 19:

Sketch the graph y = | x + 1 |. Evaluate -42x+1 dx. What does the value of this integral represent on the graph?

Answer:

We have,
y = | x + 1 | intersect x = −4 and x = 2 at (−4, 3) and (2, 3) respectively.
Now,
y=x+1=x+1     For all x>-1-x+1     For all x<-1


Integral represents the area enclosed between x = −4 and x = 2
A=-42ydx=-4-1ydx+-12ydx=-4-1-x+1dx+-12x+1dx=-x22+x-4-1+x22+x-12=-12-1-162+4+42+2-12+1=-3-152+5-12=-3+152+5-12=9 sq. units

Page No 20.15:

Question 20:

Find the area of the region bounded by the curve xy − 3x − 2y − 10 = 0, x-axis and the lines x = 3, x = 4.

Answer:

We have,
xy-3x-2y-10=0xy-2y=3x+10yx-2=3x+10y=3x+10x-2
Let A represent the required area:
A=34ydx=343x+10x-2dx=343x-6+16x-2dx=343+16x-2dx=3x+16 log x-234=12+16 log 2 -9 -16 log 1=3+16 log 2 sq. units

Page No 20.15:

Question 21:

Draw a rough sketch of the curve y = π2+2 sin2 x and find the area between x-axis, the curve and the ordinates x = 0, x = π.

Answer:



 

x 0 π6 π2 5π6 π
sin x 0 12 1 12 0
y=π2+2sin2x 1.57 2.07 3.57 2.07 1.57


y=π 2+2 sin2 x is an arc cutting y-axis at (1.57, 0 ) and x=π at π, 1.57x=π  is a line parallel   to y-axis Consider, a vertical strip of length =y  and width =dx  in the first quadrantArea of the approximating rectangle =y dx   The approximating rectangle moves from  x=0 to x=π Area of the shaded region =0πy dx A=0πy dx A=0ππ 2+2 sin2x dxA=0ππ 2+21-cos 2x2 dxA=π 20πdx +0π1-cos 2x dxA=π 2x0π +x-sin 2x20πA=π 2π +  π -sin 2π 2-0A=π π 2+1A=π π+2 2A=π2 π+2  sq. unitsArea  of curve bound by x=0 and x=π is π2 π+2  sq. units

Page No 20.15:

Question 22:

Draw a rough sketch of the curve y=xπ+2 sin2 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 0 π6 π2 5π6 π
sinx 0 12 1 12 0
y=xπ+2sin2 x 0 23 52 43 1


y=xπ+2 sin2 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 = dx in the first quadrant Area of approximating rectangle =  y  dxThe approximating rectangle moves from  x=0  to x=πArea of the shaded area =0πy dxA=0πy dx                                                            As, y>0   y =yA=0πxπ+2 sin2 x dxA=1π0πx dx  +2 0πsin2 x dxA=1πx220π+2x2-12sin x cos x 0πA=π22π+22π-12sin π cos π  -0 A=π2+πA=3π2 sq. units Area of the curve enclosed between x=0 and x=π   is  3π2 sq. units 



Page No 20.16:

Question 23:

Find the area bounded by the curve y = cos x, x-axis and the ordinates x = 0 and x = 2π.

Answer:




 The shaded region is the required area bound  by the curve  y=cos x  , x axis   and x=0  , x=2πConsider  a vertical strip of length= y    and   width= dx  in the first quadrant Area of the approximating rectangle = y dxThe approximating rectangle moves from x=0 to x=2πNow ,0xπ2and 3π2x2π ,   y>0y =yπ2x3π2, y<0y =-yArea of the shaded region =02π y dxA=0π2 y dx +π23π2 y dx +3π22π y dxA=0π2 y dx +π23π2-y  dx +3π22π y dxA=0π2 cos x dx +π23π2-cos x dx +3π22πcos x dxA=sin x0π2+-sin xπ23π2+-sin x3π22πA=1+1+1+0--1A=4 sq. units Area bound  by the curve y=cos x, x-axis and x=0, x=2π=4 sq. units 

Page No 20.16:

Question 24:

Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = π3 are in the ratio 2 : 3.

Answer:


  
The shaded area A1 is the required area bound by the curve y=sin x and x=0, x=π3The shaded area A2 is the required area bound by the curve y=sin 2x and x=0, x=π3 A1 =0π3y dxA1=0π3y dx               As, y>0 , y=yA1=0π3sin x  dxA1=-cos x0π3  A1=-cos π3+cos 0A1=-12+1=12                                          ... 1And, A2=0π3y dxA2=0π3y dx                    As, y>0 , y=yA2=0π3sin 2x  dxA2=0π32 sin x cos x dxA2=2-14cos 2x0π3A2=12-cos 2π3+cos 0A2=121+12 =12×32                                           ... 2From 1 and 2A1A2=12 12×32=23 Thus, the area of curve y=sin x and y=sin 2x for x=0 and x=π3  are in the ratio 2  :  3 

Page No 20.16:

Question 25:

Compare the areas under the curves y = cos2 x and y = sin2 x between x = 0 and x = π.

Answer:



 

Consider the value of  y  for different values of   x
 

  x 0 π4 π3 π2 2π3 5π6 π
y=cos2 x 1 0.5 0.25 0 0.25 0.75 1
y =sin2x 0 0.5 0.75 1 0.75 0.25 0


Let A1 be the area of curve y=cos2x  between x=0   and   x=π
Let A2 be the area of curve y=sin2 x between x=0  and x=π

Consider, a vertical strip of   length  =y   and width =dx in the shaded region of both the curves

The area of approximating rectangle =y dx

The approximating rectangle moves from  x=0 to x=πA1=0πy dxA1=0πy dx                                 0xπ ,  y>0 y=yA1=0πcos2 x dxA1=0π1+cos 2x dx                           cos2 x=1+cos 2x A1=12x+sin 2x20πA1=12π+sin 2π2-0A1=π2 Sq. unitsAlso,A2=0πy dxA2=0πy dx                                 0xπ ,  y>0 y=yA2=0πsin2 x dxA2=x2-12sin 2x20πA2=π2-12sin 2π2A2=π2  sq. unitsArea of curves  y=cos2 x  and area of curve  y=sin2 x  are both equal to  π2sq. units

Page No 20.16:

Question 26:

Find the area bounded by the ellipse x2a2+y2b2=1 and the ordinates x = ae and x = 0, where b2 = a2 (1 − e2) and e < 1.

Answer:



x2a2+y2b2=1  represents a parabola , symmetrical about both the axis .It cuts x axis at Aa, 0 and A'-a, 0It cuts y axis at B0, b and B'0, -bx=ae is a  line parallel to y axis Consider a vertical strip of length =y  and width=dx  ,in the first quadrantArea of approximating rectangle in first quadrant  = y dxApproximating rectangle moves from  x=0 to  x=aeArea of the shaded region =2 area in the first quadrant A=20aey dxA=20aey dx                                     As, y>0 , y=yA=20aebaa2-x2 dx                     x2a2+y2b2=1 y=baa2-x2 A=2ba0aea2-x2 dx A=2ba12xa2-x2  +12a2sin-1xa0aeA=2ba12aea2-a2e2  +12a2sin-1aea-0A=baa2e1-e2  +12a2sin-1eA=baa2e1-e2  +12sin-1eA=abe1-e2  +12sin-1e   sq. unitsRequired area  of the ellipse  bound by x=0 and x=ae  is= abe1-e2  +12sin-1e   sq. units

Page No 20.16:

Question 27:

Find the area of the minor segment of the circle x2+y2=a2 cut off by the line x=a2.

Answer:


The equation of the circle is x2+y2=a2.

Centre of the circle = (0, 0) and radius = a.

The line x=a2 is parallel to y-axis and intersects the x-axis at a2,0.


Required area = Area of the shaded region

                      = 2 × Area of the region ABDA

                      =2×a2aycircledx=2a2aa2-x2dx=2x2a2-x2+a22sin-1xaa2a=20+a22sin-11-a4×a2-a24+a22sin-112
                      =2a22×π2-a4×3a2-a22×π6=a2π2-3a24-a2π6=6a2π-33a2-2a2π12=a2124π-33 square units

Page No 20.16:

Question 28:

Find the area of the region bounded by the curve x=at2,y=2at between the ordinates corresponding t = 1 and t = 2.         [NCERT EXEMPLAR]

Answer:


The curve x=at2,y=2at represents the parametric equation of the parabola.

Eliminating the parameter t, we get

y2=4ax

This represents the Cartesian equation of the parabola opening towards the positive x-axis with focus at (a, 0).



When t = 1, x = a

When t = 2, x = 4a

∴ Required area = Area of the shaded region

                           = 2 × Area of the region ABCFA

                          =2a4ayparaboladx=2a4a4axdx=2×2a×x3232a4a=8a34a32-a32=8a38aa-aa=8a3×7aa=563a2 square units

Page No 20.16:

Question 29:

Find the area enclosed by the curve x = 3cost, y = 2sint.                    [NCERT EXEMPLAR]

Answer:


The given curve x = 3cost, y = 2sint represents the parametric equation of the ellipse.

Eliminating the parameter t, we get

x29+y24=cos2t+sin2t=1

This represents the Cartesian equation of the ellipse with centre (0, 0). The coordinates of the vertices are ±3,0 and 0,±2.



∴ Required area = Area of the shaded region

                          = 4 × Area of the region OABO

                         =4×03yEllipsedx=40341-x29dx=4×23039-x2dx=83x29-x2+92sin-1x303=830+92sin-11-0+0=83×92×π2=6π square units

Page No 20.16:

Question 30:

If the area between the curves x=y2 and x=4 is divided into two equal parts by the line x=a, then find the value of a by using integration.

Answer:

Curves x=y2 and x=4 is divided into two equal parts by the line x=a

Now, 0ax dx=a4x dx   (given)x320a32=x32a432a32-0=432-a322a32=8a32=4a=423    a=4213    a=1613


 



Page No 20.24:

Question 1:

Find the area of the region in the first quadrant bounded by the parabola y = 4x2 and the lines x = 0, y = 1 and y = 4.

Answer:




y=4x2 represents a parabola , openeing upwards, symmetrical about +ve y-axis  and having vertex at O(0, 0)y=1 is a line parallel to x-axis , cutting parabola at -12 , 1 and 12, 1y=4    is a line parallel to x axis , cutting parabola at -1, 1 and 1, 1x=0 is the y-axis Consider a horizontal strip of  length= x and width=dy in the first quadrantArea of approximating rectangle =x dyApproximating rectangle moves from y=1 to y=4 Area of the curve in the  first quadrant enclosed by y=1 and y=4 is the required area of the shaded region Area of the shaded region =04x dyA=14x dy                As, x>0, x =xA=14y4 dy A=   1214y dy  A=12y323214A=12×23432-132A=138-1A=73 sq. unitsThe area  enclosed by parabola in the first quadrant  and y=1, y=4 is 73  sq. units

Page No 20.24:

Question 2:

Find the area of the region bounded by x2 = 16y, y = 1, y = 4 and the y-axis in the first quadrant.

Answer:



x2=16 y is a parabola, with vertex at O0, 0  and symmetrical about +ve y-axis y=1  is line parallel to x-axis cutting the parabola at -4, 1 and 4, 1 y=4  is line parallel to x axis cutting the parabola at -8, 1 and 8, 1 Consider a horizontal strip of length= x   and width =dy Area of approximating rectangle = x  dy The approximating rectangle moves from  y=1  to  y=4 Area of the curve in the first quadrant  enclosed by y=1 and y=4  is the shaded areaArea of the shaded region =14x  dyA=14x  dy                     As,  x>0,   x=xA=1416 y  dyA=414y  dyA=4y323214A=83432-132A=83×7=563 sq. unitsArea enclosed by parabola  in the first quadrant and y=1 and y=4  is 563 sq. units

Page No 20.24:

Question 3:

Find the area of the region bounded by x2 = 4ay and its latusrectum.

Answer:



x2=4ay represents a parabola with vertex O(0, 0) opening upwards and symmetrical about y-axis. F(0, a ) is the focus of the parabola and y=a its latus rectumConsider a horizontal strip of length =x and width dy in the first quadrant Area of approximating rectangle=x dyThe approximating rectangle moves from y=0 to y=a Area OAB =0ax dyArea OAA'O=2×Area OAB A=20ax dy =0ax dy               As, x>0 x=xA=20a4ay  dyA=2×2a0ay dyA=4a y32320aA=4a 23a32-0A=83a2 sq. units 

Page No 20.24:

Question 4:

Find the area of the region bounded by x2 + 16y = 0 and its latusrectum.

Answer:



x2+16 y=0 x2 =-16 yComparing it with equation of parabola x2=4aya=-4Thus, x2+16 y=0 represents a parabola, opening downwards, with vertex at O(0, 0) and -ve y-axis being its axis of symmetry Focus of the parabola is F(0, -4)y=-4 is the latus rectum of the parabolaThe latus rectum cuts the parabola at B(8, -4) and B'(-8, -4)x=8 cuts the x-axis at A(8, 0)Area of the curve bound by latus rectum= Shaded area BOB'B= 2 Area OBF                                                           ...1Consider a vertical strip of length= y and width=dx in shaded area OAB such that point P(x, y ) lies on the parabolaThe area of the approximating rectangle = y dxBut the approximating rectangle moves from x=0 to x=8Area of the shaded region OAB=08y dx A=08-x2 16 dx                     x2=-16 yy=-x2 16  Area of the shaded region OAB=08 x2 16 dx =116×13x308=8×8×816×3=323 sq. units                                       ...2So, area of rectangle OABF = OA×AB =8×4 =32 sq. units                                                                                               ...3 From 2   and 3Area of OBF=Area of rectangle OABF -Area of the shaded region OAB=32  -323=643  sq. units Shaded area BOB'B= 2 Area OBF=2×643=1283  sq. units Thus, area of the curve x2+16 y =0  bound by its latus rectum =1283  sq. units

Page No 20.24:

Question 5:

Find the area of the region bounded by the curve ay2=x3, the y-axis and the lines y = a and y = 2a.

Answer:


The equation of the given curve is ay2=x3.

The given curve passes through the origin. This curve is symmetrical about the x-axis.

The graph of the given curve is shown below.



The lines y = a and y = 2a are parallel to the x-axis and intersects the y-axis at (0, a) and (0, 2a), respectively.

∴ Required area = Area of the shaded region

                          =a2axcurvedy=a2aay213dy=a13a2ay23dy=a13×y5353a2a=35a132a53-a53=35253a2-a2=35253-1a2 square units



Page No 20.51:

Question 1:

Calculate the area of the region bounded by the parabolas y2=6x and x2=6y.

Answer:

Ans

Page No 20.51:

Question 2:

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

Answer:




3x2=16 y     ... 1 is a parabola with vertex at (0, 0) opening upwards and symmetrical about +ve y-axis 4y2=9x        ... 2 is a parabola with vertex at (0, 0) opening sideways and symmetrical about +ve x-axis Solving the equations 1 and 2, we get the points of intersection of the two parabolas O(0, 0) and A(4, 3 )   Consider a vertical strip of length= y2-y1 and width = dx such that P(x, y1) lies on 1 and Q(x, y2) lies on 2area of approximating rectangle =y2-y1  dx Approximating rectangle  moves from x=0  to x=4Area  of the shaded region =04y2-y1 dx =04y2-y1 dx                      As, y2-y1 =y2-y1 for y2-y1>0 A=0494x   -316x2 dx A=3204x dx  -31604x2 dx A=32x323204-316x3304A=432-116×43A=8-4=4 sq. units Area bound by the two curves = 4  sq. units  

Page No 20.51:

Question 3:

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

Answer:



y=x             ... 1  is a parabola opening side ways, with vertex at O(0, 0) and +ve x-axis as axis of symmetry x=y                  ...2 is a straight line passsing through O(0, 0) and at angle 45o with the x-axisSolving  1 and 2  y2=x=y y2=y y(y-1)=0 y=0 or y=1 and x=0 or x=1Thus, the line intersects the parabola at O(0, 0 ) and A(1, 1)Consider a approximating rectangle of length =y2-y1  and width= dx Area of approximating rectangle= y2 -y1  dxApproximating rectangle  moves from x=0 to x=1  Area of the shaded region=01 y2 -y1  dx =01 y2 -y1 dx              As, y2 >y1, y2 -y1=y2-y1 A=01 x  -x dx  A=x3232-x2201A=13232-122-0A=23-12=16  sq. units Area bound by the parabola and straight line = 16 sq. units

Page No 20.51:

Question 4:

Find the area bounded by the curve y = 4 − x2 and the lines y = 0, y = 3.

Answer:



y=4-x2 is a parabola, with vertex (0, 4), opening downwars and having axis of symmetry as -ve y-axis y=0   is the x-axis, cutting the parabola at A(2, 0) and A'(-2, 0)y=3 is a line parallel to x-axis, cutting the parabola at B(1, 3 ) and B'(-1, 3) and y-axis at C(0, 3)  Required area is the shaded area ABB'A =2 area ABCOConsider a horizontal strip of length= x2-x1 and width=dy in the shaded region Area of approximating rectangle= x2-x1 dyThe approximating rectangle moves from y=0 to y=3 Area of shaded region =  230x2-x1 dy A=203x2-x1 dy                       As,  x 2-x1 =x2-x1 , x2>x1 A=2034-y  -0 dyA=-24-y3232 03A=-24-y3232 03A=2 ×23432-132A=43×7  A=283  sq. unitsArea bounded by the two parabolas =283 sq. units

Page No 20.51:

Question 5:

Find the area of the region x, y:x2a2+y2b21xa+yb.

Answer:




Let R=x, y  :  x2 a2+y2b2 1xa+ybR1=x, y  :  x2 a2+y2b2 1and R2 =x, y  : 1xa+ybThen, R =R1R2Consider   x2 a2+y2b2=1. This represents  an ellipse, symmetrical about both axis and cutting  x-axis at  A(a, 0) and A'(-a, 0) and y-axis at B(0, b), B'(0, -b) R1= x2 a2+y2b2 1   represents the area inside the ellipse xa+yb=1 = represents a straight line cutting x-axis at  A(a, 0) and y-axis at B(0, b)R2=xa+yb1represents the area  above the straight line R =R1R2 represents the smaller shaded area bounded by the line and the ellipse In the shaded region, consider a vertical strip with length=y2-y1  and width =dx, such that P(x, y2) lies on ellipse and Q(x, y1) lies on the straight line Area of approximating rectangle=y2 -y1  dx The  approximating rectangle moves from  x=0 to x=aArea of the shaded  region =0ay2 -y1  dx =0ay2 -y1 dx                      As,  y2>y1, y2 -y1= y2-y1 A= 0abaa2-x2  -baa-x dxA=0abaa2-x2dx  - 0abaa-x  dxA=bax2a2-x2+12a2 sin-1xa0 a    -baax-x22A=ba0+12a2 sin-11-a2-a22A=ba12a2×π2-a22A=ab2π2-1A=ab4π-2 sq. units 

Page No 20.51:

Question 6:

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

Answer:



Consider the points A(2, 1), B(3, 4) and C(5, 2) We need to find area of shaded triangle ABCEquation of AB is y-1=4-33-2 x-23x-y-5=0                                                         ... 1Equation of BC isy-4=2-45-3 x-3x=y-7=0                                                           ... 2Equation of CA is y-2=2-15-2 x-5x-3y+2=0                                                           ... 3Area of  ΔABC =Area of  ΔABD+Area of ΔDBC In ΔABD,  Consider point P(x, y2) on AB and Q(x, y1) on AD Thus, the area of approximating rectangle with length=y2-y1 and width=dx is y2-y1 dxThe approximating rectangle moves from x=2 to x=3 Area of ΔABD =23y2-y1 dx =23y2-y1  dx A=233x-5-x+13 dxA=239x-15-x-13 dx A=238x-163 dxA=138x22-16x23A=134×32-16×3-4×22+16×2A=1368-64A=43  sq. units Similarly, for S(x, y4) on AB and R(x, y3)  on DC Area of approximating rectangle of length  y4-y3 and width dx= y4-y3 dxApproximating rectangle  moves from x=3 to x=5Area BDC=35y4-y3  dxA=357-x -x+13 dxA=133520-4x dxA=1320 x-4x2235A=13100-50-60-18A=1350-42=83 sq. units Area of  ΔABC =Area of  ΔABD +Area of  ΔDBC=43+83=123 =4 sq. units

Page No 20.51:

Question 7:

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

Answer:




Equation of line AB isy-1=5-10+1x--1y=4x+5  Area under the line AB = area ABDO=-104x+5 dx=4x22+5x-10=0-2-5Area ABDO=3 sq. units                                             ... 1Equation of line BC is  y-5=2-53-0x-0y=-x+5 Area under line BC  =Area OBCP=03-x+5 dx=-x22+5x03=-92+15 -0Area OBCP=212  sq.  units                                     ... 2Equation of line CA isy-2=2-13--1x-3 4y =x+5Area under line AC  =Area ACPAA=-13x+54dxA=14x22+5x-13A=14322+5×3--122+5-1A=1492+15-12+5Area ACPA=244=6 sq. units                                    ... 3From 1, 2 and 3Area Δ ABC =Area ABDO+Area OBCP-Area ACPAA=3+212-6A=212-3 =21-62=152  sq. units Area Δ ABC=152  sq. units 

Page No 20.51:

Question 8:

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

Answer:


Solving the given equations The point of intersection of the three lines are A(0, 1), B(4, 13) and C(4, 9). We need to find the area of ABCArea under line AB=area OABCLArea OABCL=043x+1 dx                    Equation of BC is y=3x+1 and x moves from A, x=0 to B, x=4 =3x22+x04 =3422+4=24+4=28 sq. unitsArea under line BC =Area OACLArea OACL=042x+1dx                       Equation of BC is y=2x+1 and x moves from A, x=0 to C, x=4 =2x22+x04=16+4=20 sq. unitsArea  ΔABC= Area OABCL-Area OACLArea  ΔABC=28-20 =8 sq. units Area of triangle formed by the three given lines=8 sq. units 

Page No 20.51:

Question 9:

Find the area of the region {(x, y) : y2 ≤ 8x, x2 + y2 ≤ 9}.

Answer:



Let R=x, y: y28x, x2+y29R1=x, y: y28xR2=x, y: x2+y29Thus, R=R1R2 Now, y2=8x  represents a parabola with vertex O(0, 0) and symmetrical about x-axis Thus, R1 such that y28x is the area inside the parabola Also, x2+y2=9 represents with circle with centre O(0, 0) and radius 3 units. The circle cuts the x axis at C(3, 0) and  C'(-3, 0 ) and Y-axis at B(0, 3) and B'(0, -3 )Thus, R2 such that x2+y29 is the area inside the circleR =R1R2 =Area OACA'O=2 shaded area OACO                       ... 1The point of intersection between the two curves  is obtained by solving the two equationsy2=8x  and x2+y2=9 x2+8x=9 x2+8x -9 =0x+9x-1=0x=-9  or  x=1Since, parabola is symmetric about +ve x-axis,  x=1  is the correct solution y2=8 y=±22Thus, A1, 22 and A' 1, -22  are the two points of intersectionArea OACO =area OADO +area DACD                       ... 2Area OADO  =018x dx                Area bound by curve y2=8x between x=0 and x=1=22x323201Area OADO=423                                                       ... 3Area DACD = area bound by x2+y2=9  between x=1 to x=3A=139-x2 dx  =12x9-x2+129 sin-1x313=0+92 sin-133 -129-12 -92 sin-113=92 sin-11 -128 -92 sin-113=92 π2-1222 -92 sin-113Area DACD=9 π4-2 -92 sin-113                            ... 4From 1, 2, 3 and 4R=Area OACA'O =2423+9 π4-2 -92 sin-113  =2423-2 +9 π4-92 sin-113  Area OACA'O =223 +9π4-92 sin-113 sq. units 

Page No 20.51:

Question 10:

Find the area of the region common to the circle x2 + y2 = 16 and the parabola y2 = 6x.

Answer:




 

Points of intersection of the parabola and the circle is obtained by solving the simultaneous equations

x2+y2=16 and y2=6xx2+6x=16 x2+6x-16 =0x+8x-2=0x=2  or  x=-8 , which is not the possible solution.  When x=2, y=±6×2 =±12=±23B2 ,23 and B'2 ,-23  are points of intersection of the parabola and circle. Now,Required area= areaOBAB'O                               =2areaOBAO                           =2areaOBDO+areaDBAD                           =2×026xdx +2416-x2 dx                           =2×6x323202+12x16-x2 +12×16 sin-1x424                           =2× 6×23×232-0 +12416-42 +12×16 sin-144-12×216-22 -12×16 sin-124                           =2 ×  6×23×22+0+8 sin-11-12 -8 sin-112                           =2×833+8×π2-23-8π6                           =2 83-633+8π2-π6                           =2233+82π6                           =433+16π3

Page No 20.51:

Question 11:

Find the area of the region between the circles x2 + y2 = 4 and (x − 2)2 + y2 = 4.

Answer:



x2+ y2 = 4   .......(1)       represents a circle with centre O(0,0) and radius  2
x-22+y2 = 4 ......(2)   represents a circle with centre A(2 ,0) and radius 2
 
Points of intersection  of two circles is given by solving the equations

x-22=x2x2-4x+4 =x2x=1 y2=3 y =±3Now, B1,3 and B'1,-3  are two points of intersection of the two circlesWe need to find shaded area= 2×areaOBAO     ...1AreaOBAO=areaOBPO+areaPBAP =01y1dx+12y2dx                                                   y1>0 y1=y1 and y2>0 y2=y2=01y1 dx+12y2dx=014-x-22 dx+124-x2 dx=12x-24-x-22 +12×4×sin-1x-2201+12x4-x2 +12×4×sin-1x212=-32+2sin-1-12 -0+2sin-1-1+0-123+2sin-11-sin-112=-32+2sin-1-12 -0-2sin-1-1+0-123+2sin-11-2sin-112=-3-4sin-112 +4sin-11=-3-4×π6 +4×π2=-3-2π3 +2π=4π3-3Now, From equation 1Shaded area= 2×areaOBAO=24π3-3=8π3-23  sq units 

Page No 20.51:

Question 12:

Find the area of the region included between the parabola y2 = x and the line x + y = 2.

Answer:



We have, y2=x and x+y=2
To find the intersecting points of the curves ,we solve both the equations.
 
y2+y-2=0y+2y-1=0y=-2 or y=1x=4 or 1Consider a horizantal strip of length x2-x1  and width dy where Px2,y lies on straight line and Qx1,y lies on the parabola.Area of approximating rectangle =x2-x1 dy , and it moves from y=-2 to y=1Required area = areaOADO =-21x2-x1 dy=-21x2-x1 dy             x2-x1=x2-x1 as x2>x1=-212-y-y2 dy=2y-y22-y33-21=2-12-13--4-2+83=2-12-13+6-83=92 sq units Area  enclosed by the line and given parabola =92 sq units             
                             
                                          
                                    

Page No 20.51:

Question 13:

Draw a rough sketch of the region {(x, y) : y2 ≤ 3x, 3x2 + 3y2 ≤ 16} and find the area enclosed by the region using method of integration.

Answer:



The  given region is the intersection of
          y23x   and  3x2+3y216
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 163.
Corresponding equations of the given inequations are
y2=3x               .....1and3x2+3y2=16     .....2
 Substituting the value of y2 from (1) into (2)
  3x2+9x=16
3x2+9x-16=0
x=-9±81+1926
x=-9±2736
By figure we see that the value of x will be non-negative.
x=-9+2736
Now assume that x-coordinate of the intersecting point, a=-9+2736
The  Required area A  = 2(Area of OACO + Area of CABC)
Approximating the area of OACO  the length =y1  width = dx
    Area of   OACO  =0ay1 dx  
                            =0ay1 dx
                            =0a3xdx         y21=3x   y1=3x
                            =23 x323a0
                            =23a323 
Therefore, Area of  OACO =23a323
Similarly approximating the are of CABC the length =y2   and the width = dx
              Area of  CABC  =a43y2 dx     
                                     =a43y2 dx
                                     =a43163-x2 dx        3x2+3y22=16 y2=163-x2
                                     =x2163-x2 +83sin-1x34a43
                                    =423163-163+83sin-14343-a2163-a2-83sin-1a34
                        
                                   =-a2163-a2+83sin-11-83sin-1a34
 Area of  CABC =-a2163-a2+4π3-83sin-1a34
Thus the required area A = 2(Area of OACO + Area of CABC)
                                    =223a323-a2163-a2+4π3-83sin-1a34
                                   =4a323-a163-a2+8π3-163sin-1a34 
 Hence, the reguired area is,4a323-a163-a2+8π3-163sin-1a34 square units , where a=-9+2736
              
                             

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
y25x  and  5x2+5y236
Clearly, y25x is a parabola with vertex at origin and the axis is along the x-axis opening in the positive direction.Also 5x2+5y236 is a circle with centre at the origin and has a radius 365 or  65.
Corresponding equations of given inequations are
         y2=5x                           .....1  5x2+5y2=36                .....2
Substituting the value of y2 from (1) into (2), we get

5x2+25x=36
5x2+25x-36=0
x=-25±625+72010
 x=-25±134510
From the figure we see that x-coordinate of intersecting point can not be negative.
x=-25+134510
Now assume that x-coordinate of intersecting point,  a=-25+134510
The  Required area,
A  = 2(Area of OACO + Area of CABC)
Approximating the area of OACO  the length =y1 and a width = dx=|y1=dx
Area of OACO=0ay1dx
               =0ay1 dx
               =0a5xdx         y12=5x   y1=5x
               = 52x323a0
Therefore, Area of OACO  =25a323
Similarly approximating the area of CABC the length =y2 and the width = dx
Area of CABC=a65y2dx 
               =a65y2 dx
               =a65365-x2dx
               =x2365-x2 +185sin-1x56a65
               =625365-365+185sin-11-a2365-a2-185sin-1a56
              =0+185sin-11-a2365-a2-185sin-1a56
              =185×π2-a2365-a2-185sin-1a56
Area of CABC  =9π5-a2365-a2-185sin-1a56
Thus the  Required area, A  = 2(Area of OACO + Area of CABC)
 A=2 25a323+ 9π5-a2365-a2-185sin-1a56
 =45a323+18π5-a365-a2-365sin-1a56 ,       Where,  a=-25+134510  .
  
   

Page No 20.51:

Question 15:

Using integration, find the area of the region enclosed by the parabola y=3x2 and the line 3x-y+6=0.

Answer:

Curves y=3x2 and y=3x2+6.