Areas of Bounded Regions Solved Problems

AREAS OF BOUNDED REGIONS SOLVED PROBLEMS Ex.1 Sol. Using integration, find the region bounded by the line 2y + x = 8 or, (2y = –x + 8) the x-axis and the lines x = 2 and x = 4. Here, the given line is Y 1 (8 – x) ....(1) 2  the required area (shaded region) bounded by the line, the x-axis and the line x = 2 and x = 4 2y + x = 8 4 y= B x=2 4 1  ydx  (8  x )dx [From (1)] 2 2 2   x=4 X’ 4  Ex.2 Sol. P(x,y) C 2y + x = 8 1 x2  1 8 x   =  [(32 – 8) – (16 – 2)] = 5 sq. units. 2 2 2 2 A O X D Y’ Find the area of the region bounded by the parabola y2 = 4x and y = 2x The points of intersection of y2 = 4x and y = 2x and O(0, 0) and A(1, 2) Here, we take y2 = 4x or y = 2 x as f(x) and y = 2x as g(x), where f(x)  g(x) in [0, 1] Therefore, Area of the shaded region Y y=2x A(1,2) 1   [ f ( x )  g( x )]dx O 0 (1,0) X’ 1 X   (2 x  2x ) dx 0 1 2 x2  2  2 x 3 / 2    2  3 2 3  0 Ex.3 Sol. y2=4x 1 1 = sq. units 2  3 Y’ Find the area of the smaller region enclosed between the circle x2 + y2 = 4 and the line x + y = 2. The points of intersection of x2 + y2 = 4 and the line x + y = 2 are A(2, 0) and B(0, 2) Here, we take x2 + y2 = 4 i.e., y  4  x 2 as f(x) and x + y = 2 i.e., y = 2 – x as g(x), where f(x)  g (x) in [0, 2]. Therefore, Area of the shaded region Y B(0,2) 2   [ f ( x )  g( x )]dx X’ 0 2 A(2,0) O   4  x 2  (2  x ) dx    0 2  x 4  x2 x2  x   2 sin 1   2x   2 2  2 0 = 2 sin (1) – 4 + 2 = ( – 2) sq. units –1 Y’ X x+y=2 AREAS OF BOUNDED REGIONS Ex.4 Sol. Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y. Given equations to the curves are x2 + y 2 = 9 4 ...(1) x2 = 4y ...(2) 3 . 2 Equation (2) represents a parabola whose vertex is O(0, 0) and x-axis is y-axis. Equation (1) represents a circle whose centre is (0, 0) and radius From (2), we have y= x2 4 Y x2=4y 2 Put y = x in (1), we have 4 A(– 2 ,1/ 2) x4 9  16 4 x4 + 16x2 – 36 = 0 2 (x + 18) (x2 – 2) = 0 x2 = 2 or – 18 B( 2 ,1/ 2) 4x2 + 4y2 = 9 x2       X’ O – 2 X 2 x=± 2 (x2 cannot be – 18) y Y’ 1 x2 = 2 4 Taking y = 9  x2 4 Required area = on f(x) and y =  2 – 2 x2 as g(x), where f(x)  g (x) in [– 2 , 2 ] we have ve 4  2 [ f ( x )  g( x )] dx = 2 [ f ( x )  g( x )] dx = 2 0  2 0  9 x2   x 2   dx  4   4 2    9  9 2  x2  2 3  x  9  2x  x  9 2  2x  4 4  sin 1   –  sin 1  = 2 = 2 –  2 8   2 8 8 6  8  12         0  0 2 2 2   2 9  2  2 2  2 9  2 2  2 9   =  sin 1   = 2 = 2  sin 1    sin 1   3  sq. units     6 4 12 8 3 8 8 3    4         Ex.5 Sol. Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1. Given equations to the curves are x2 + y 2 = 1 ...(1) (x – 1)2 + y2 = 1 ...(2) (x – 1)2 + (1 – x2) = 1  x2 – 2x + 1 + 1 – x2 = 1  2x = 1  x = Requried area 1 1/ 2   2 y of circle (2) dx  y of circle (1) dx   0  1/ 2   1 1/ 2   2 1  ( x  1)2 dx  1  x 2 dx   0  1/ 2   Y 1 2 X’ O (1,0) Y’ x=1/2 x=1 X AREAS OF BOUNDED REGIONS 1/ 2 1  2   x 1 x2 1   1 1  ( x  1) 1  ( x  1) 1  2  sin ( x  1)    sin x )  2 2 2 2       0 1/ 2    1   1 1 1 1 1 1 1  1  1     2   1   sin 1     0  sin 1( 1)  0  sin 1(1)  1   sin 1     4 2 2 2 4 4 2  2  2     4    1 3 1 3  1  1  sin 1  –   sin 1( 1)  sin 1(1)   sin 1   2 4 2 4  2 2  3      3     – –     4  6  2 2 4 6  3 2   –   2 3  sq. units.  Ex.6 Sol. Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and x = 3. Given equations are y = x2 + 2 ...(1) y=x ...(2) x=0 ...(3) x=3 ...(4) Y y = x2 + 2 Equation (1) represetns a parabola with vertex A(0, 2) and axis is y-axis. y=x Equation (2) represents a line passing through (0, 0) and (1, 1) Equation (3) represents y-axis Equation (4) represent a line parallel to y-axis and passing through B(3, 0) Required area =  3 0  A(0,2) X’ B(3,0) O x=3 3 y(of parabola )dx  y (of line y  x ) dx x=0 0 3 X Y’ 3  x3   x2  9  21 2  2 x  ( x  2 ) dx  xdx     = (9  6 )    0  = = 0 = 3 sq. units. 0 2 2   2  0  0  3  3 Ex.7 Find the area of the region | (x,y) : x2 + y2  2ax, y2  2ax, a > 0, x  0, y  0| Sol. Required area    a 0 a   0 Y A(a,a) a 2  ( x  a)2  ax  dx  a  2 1 2 2 a 3/2  2 1  x  a  x    a  ( x  a)  a sin  2 3  a    0 X’ C X Y’  a 2 2 2  2 2 1 2 – 2 2 1 2    1 2  1 1  a  sq. units.   a sin (0 )  a    a sin ( 1)  a  a   =  3  3  2 3 2  2  2   4 Ex.8 Sol. Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A (2, 0), B(4, 5) and C (6, 3). Required area = area of the region ABC Y B(4,5) = area of the region APB + area of the region PQCB – Area of the region AQC = 6 5 x  10 dx  (9  x )dx  2 4 2  4  4 6 C(6,3)  3x  6   dx 2 4  6   6    5  x2 x2  3  x2 = 2  2  2x   9 x  2   4  2  2x   2  4  2 = 7 sq. units. X’ O A(2,0) Y’ P Q X