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Must Do Problems Of Areas Bounded by Regions For NDA

Must Do Problems Of Areas Bounded by Regions

Must Do Problems
Q 2815045869

The area bounded by the circle `x^2 + y^2 = 8`, the parabola `x^2 = 2y` and the line `y = x` in `y >= 0`
I. Has area `= ( 2 pi + 4/3)` sq units
II. Has area `= ( 2 pi + 2/3)` sq units
III. The point of intersection of circle, parabola and line in 1st quadrant is `(2,2)`.

(A)

Only I

(B)

Only II

(C)

I and III

(D)

II and III

Solution:

The required area

` = int_(-2)^2 sqrt ( 8 - x^2) - int_(-2)^0 1/2 x^2 dx - int_0^2 x dx`

` = 2 int_0^2 sqrt ( 8 - x^2) dx - [ x^3/6]_(-2)^0 - [ x^2/2 ]_0^2`

` = 2 [ x/2 sqrt ( 8 - x^2) + 8/2 sin^(-1) x/(2 sqrt2) ]_0^2 - 4/3 - 2`

` = 2 [ 2 + 4 . pi/4 ] - (10)/3 = 2/3 + 2 pi`
Correct Answer is `=>` (D) II and III
Q 2835045862

The area enclosed by the curves `y = sin x + cos x` and `y = | cos x - sin x |` over the interval ` ( 0 , pi/2)` is
I. ` 2 sqrt2 (2 - sqrt2)` II. `2 sqrt2 (sqrt 2 - 1 )` III. `2( sqrt 2 - 1)`

(A)

Only I

(B)

Only II

(C)

Only III

(D)

None of these

Solution:

Given, `y = sin x + cos x`

`(dy)/(dx) = cos x - sin x ,`

`y = | cos - sin x | `

` = [ tt (( cos x - sin x , x in [ 0 , pi//4]) , ( sin x - cos x , x in [ pi//4 , pi//2 ] )) ]`

Thus, required area

` = int_0^(pi//4) | ( sin x + cos x) - (cos x - sin x ) | dx`

` + int_(pi//4)^(pi//2) | 2 cos x | dx`

` = int_0^(pi//4) | 2 sin x| dx + int_(pi//4)^(pi//2) | 2 cos x | dx`

`= 2[-cosx]_0^(pi//4) + 2 [sin x]_(pi//4)^(pi//2)`

` = 2 [ (-1)/sqrt2 + 1 + 1 - 1/sqrt2 ]`

` = 2 (2 - sqrt2 ) = 2 sqrt2 ( sqrt2 - 1)`
Correct Answer is `=>` (B) Only II
Q 2815745669

Find the area of the region bounded by ` x^2 = 4y , x = 4y - 2`.

(A)

`3/8` sq unit

(B)

`5/8` sq unit

(C)

`7/8` sq unit

(D)

None of these

Solution:

The curves intersect each other, if

`x^2/4 = ( x + 2)/4 => x^2 - x - 2 = 0`

`=> x = - 1, 2`

Hence, the points of intersection are

`(- 1, 1//4)` and `(2, 1)`. The region is

plotted in following figure.

Since, the straight line `x = 4 y - 2` is

always above the parabola `x^2 = 4 y` in

the interval `[- 1, 2]`. The required area

is given by

` = int_1^2 [ (x + 2)/4 - x^2/4 ] dx`

` = 1/4 [ 1/2 x^2 + 2x - 1/3 x^3 ]_(-1)^2`

` = 1/4 [ ( 2 + 4 - 8/3 ) - ( 1/2 - 2 + 1/3 ) ]`

` = 9/8` sq units
Correct Answer is `=>` (D) None of these
Q 2815645569

Area bounded by the curve `xy^2 = a^2 (a-x)` and the Y -axis is

(A)

`pi a^2 // 2` sq units

(B)

`3 pi a^2 ` sq units

(C)

`pi a^2 ` sq units

(D)

None of these

Solution:

`xy^2 = a^2 (a - x) , x = a^3/(y^2 + a^2)`

The given curve is symmetrical about

`X`-axis and meets it at `(a, 0)`.

The line `x = 0` i.e., Y -axis is an

asymptote (tangent at infinity).

Area `= 2 int_0^oo a^3/(y^2 + a^2) dy`

`= 2a^3 . 1/a [ tan^( -1) y/a ]_0^oo = 2 a^2 . pi/2 = pi a^2`
Correct Answer is `=>` (C) `pi a^2 ` sq units
Q 2845645563

What is the area bounded by the regions `y = e^x`, `y = e^(-x)` and the straight line `x = 1`?

(A)

`( e + 1/e)` sq units

(B)

`( e - 1/e)` sq units

(C)

`( e + 1/e - 2)` sq units

(D)

`( e - 1/e - 2)` sq units

Solution:

The given curves, `y = e^x , y = e^(-x)` and `x = 1`

Area of `EACOE = int_0^1 y dx`

` = int_0^1 e^x dx = [ e^x ] _0^1 = ( e - 1)`

Area of `EACOE = int_0^1 e^(-x) dx`

` = [ - e^(-x) ]_0^1 = - [ e^(-1) - 1 ] = (1 - 1//e)`

So, the area of `EABE =` Area of

`EACOE` - Area of `EBCOE`

` = ( e - 1) - (1 - 1// e)`

` = ( e + 1/e - 2)` sq units
Correct Answer is `=>` (C) `( e + 1/e - 2)` sq units
Q 2835245162

Area bounded by the curves `y = x sin x` and X - axis between `x = 0` and `x = 2 pi` is

(A)

`2 pi`

(B)

`3 pi`

(C)

`4 pi`

(D)

`6 pi`

Solution:

Area `= int_0^(pi) dx + int_(pi)^( 2 pi) y dx`

` = | int_o^pi x sin x dx | + | int_pi^(2 pi) x sin x dx |`

` = | [ x (- cos x) + sin x ]_0^(pi)`

` + | [ x (- cos x) + sin x ]_pi^(2 pi) |`

` = | - ( - pi - 0 ) + ( 0 - 0 ) |`

` + | [ - 2 pi + pi (-1) ] + ( 0 - 0) |`

` = pi + 3 pi = 4 pi`
Correct Answer is `=>` (C) `4 pi`
Q 2835145062

Area lying in the first quadrant and bounded by the circle `x^2 + y^2 = 4` and the line `x = y sqrt3` equals
to

(A)

`pi`

(B)

`pi//2`

(C)

`pi//3`

(D)

`pi//4`

Solution:

Line and the curve meet at `P ( sqrt3, 1)`

in I st quadrant. Draw perpendicular PM.

`:.` Area `= Delta OPM + int_(sqrt3)^2 y dx`

Now, `x = 2 cos theta , y = 2 sin theta `, then the

limits changes

`= 1/2 sqrt3 . 1 + int_(pi//6)^0 (2 sin theta ) (- 2 sin theta ) d theta`

`= sqrt3/2 + 4 int_0^(pi//6) (1 - cos 2 theta)/2 d theta`

` = sqrt3/2 + 2 [ theta - ( sin 2 theta)/2 ] _0^(pi//6)`

` = sqrt3/2 + 2 [ pi/6 - 1/2 . sqrt3/2 ] = pi/3`
Correct Answer is `=>` (C) `pi//3`

Set - 2

Q 2486612577

The area of the portion of the circle
`x^2 + y^2 = 64` which is exterior to the parabola
`y^2 = 12x`, is
UPSEE 2014
(A)

`(8pi - sqrt(3))` sq units

(B)

`(16)/3 ( 8 - sqrt3)` sq units

(C)

`(16)/3 ( 8 pi - sqrt3)` sq units

(D)

None of these

Solution:

Required shaded area =

= Area of circle` -2 [ int_0^4 2sqrt(3) sqrt(x) dx + int_4^0 sqrt(64 - x^2) dx ]`

` = 64 pi - 2 [ ( 2sqrt(3) x^(3//2) xx 2/3)_0^4+ ( x/2 sqrt(64 - x^2) + (64)/2 sin^(-1) \ x/8 )_4^8 ]`

` = 64 pi - (64)/sqrt(3) - 32 pi + 16sqrt3 + (32 pi)/3`

` = (128 pi)/3 - (16 sqrt3)/3`

` = (16)/3 ( 8 pi - sqrt3) `sq units
Correct Answer is `=>` (C) `(16)/3 ( 8 pi - sqrt3)` sq units
Q 2446656573

For which of the following value of `m`, is the
area of the region bounded by the curve
`y = x - x^2` and the line `y = mx` where m>0 quals to `9/2`?
UPSEE 2013
(A)

`-4`

(B)

`-2`

(C)

`2`

(D)

`4`

Solution:

Area `= int_(0)^(1-m) (x-x^2 -mx) dx`

`= [ (1-m) * x^2/2 - x^3/3 ]_(0)^(1-m)`

`= (1-m)^3 (1/2 -1/3) = pm 9/2 `(given)

Taking `+`ve sign,` (1- m)^3 =27 =>m = -2`

Taking `-`ve sign, `(1- m)^3 = -27 => m = 4`
Correct Answer is `=>` (D) `4`
Q 2416623570

For `0 le x le pi`, the area between the curve
`y = sin x` and `x`-axis is
UPSEE 2010
(A)

`1 ` sq unit

(B)

`0` sq unit

(C)

`2` sq unit

(D)

`-1` sq unit

Solution:

The area between `y = sin x` and `x`-axis

`= |int_0^pi y dx |`

`=|int_0^pi sinx dx| `

`= |[(cosx]_0)^pi|`

`= |(-1 - 1 )|`

`= 2 `
Correct Answer is `=>` (C) `2` sq unit
Q 2564234155

The area between the parabola `y = x^2` and the
line `y =x` is :
UPSEE 2008
(A)

`1/6` sq unit

(B)

`1/3 ` sq unit

(C)

`1/2 ` sq unit

(D)

None of these

Solution:

Equation of parabola is

`y =x^2`............(i)

Equation of straight line is

`y =x`.............(ii)

From Eqs. (i) and (ii)

`x^2 - x = 0`

`=> x(x-1)=0`

`=> x= 0, 1`

`=> y =0, 1`


Hence, point of intersections are `(0, 0)` and `(1,1)`


`:.` Required area `= int_(0)^(1) x dx -int_(0)^(1) x^2 dx`

`= [x^2/2 - x^3/3]_(0)^(1)`

`=[1/2- 1/3] =1/6` sq unit.
Correct Answer is `=>` (A) `1/6` sq unit
Q 2554345254

The area bounded by the curve `y^2(2a - x) = x^3`
and the line `x = 2a` is
UPSEE 2008
(A)

`3 pi a^2` sq unit

(B)

`(3 pi a^2)/2` sq unit

(C)

`(3 pi a^2)/4` sq unit

(D)

`(6 pi a^2)/5` sq unit

Solution:

The equation of curve and line are respectively

`y^2(2a- x) = x^3` ... (i)

and `x= 2a` ... (ii)

The given curve is symmetrical about `x`-axis and
passes through origin

Also, `x^3/(2a -x) < 0` for `x > 2a` and `x < 0`


So, curve does not is in `x > 2a` and `x < 0`
therefore curve lies wholly on `0 le x le 2a`.


`:.` Required area `= int_(0)^(2a) (x^(3/2))/(sqrt (2a -x) ) dx`

Put `x = 2a sin^2 theta` and `dx = 4a sin theta cos theta d theta`

`:.`Required area `= int_(0)^(pi/2) 8 a^2 sin^4 theta d theta`

`= 8a^2 [3/4 * 1/2 * pi/2]` (using gamma function)

`=(3 pi a^2)/2` sq unit
Correct Answer is `=>` (B) `(3 pi a^2)/2` sq unit
Q 1685034867

Let the straight line `x = b` divide the area enclosed by
`y = (1- x)^2, y = 0` and `x = 0` into two parts `R_1 (0 le x le b)`

and `R_2 (b le x le 1)` such that `R_1 - R_2 = 1/4`. Then, `b` equals
BITSAT 2016
(A)

`3/4`

(B)

`1/2`

(C)

`1/3`

(D)

`1/4`

Solution:

Clearly,

`R_1 = int_0^b (1 - x)^2 dx` and `R_2 = int_b^1 (1- x)^2 dx`

` => R_1 = [((b-1)^3)/3]_0^b` and `R_2 = [((b-1)^3)/3]_1^b`

` => R_1 = ((x-1)^3)/3 + 1/3` and `R_2 = - ((x-1)^3)/3`

` : . R_1 - R_2 = 1/4`

` => 2/3 (b-1)^3/3 + 1/3 = 1/4 => 2/3 (b-1)^3/3 = 1/(12)`

` => (b-1)^3 = - 1/8 => b-1 = -1/2`

` => b = 1/2`
Correct Answer is `=>` (B) `1/2`
Q 1605012868

The area enclosed between the curve `y = log_e (x + e)`
and the coordinate axes, is
BITSAT 2016
(A)

`4`

(B)

`3`

(C)

`2`

(D)

`1`

Solution:

The area enclosed by `y = log_e (x + e)` and the coordinate
axes is shaded in figure.
Let `A` be the required area. Then,

`A = int_(1-e)^0 y dx = int_(1-e)^0 log_e (x +e) dx`

` => A = [x log_e (x+e)]_(1-e)^0 - int_(1-e)^0 x/(x+e) dx`

We have to find the area of shaded region.

` => A = [x log_e (x+e)]_(1-e)^0 - [x -e log_e (x + e)]_(1-e)^e`

` => A= 0- [(0 -e)- (1- e)+ O]= 1` sq. unit
Correct Answer is `=>` (D) `1`
Q 1653001844

The area of the region `R = {(x,y) : |x| < |y| and x^2 + y^2<1}`
BITSAT 2015
(A)

`(3 pi)/8` sq units

(B)

`(5 pi)/8` sq units

(C)

`pi/2` sq units

(D)

`pi/8` sq units

Solution:

Required area `= 4` (Area of the shaded region in first
quadrant)

` = 4 int_0^(1//sqrt 2) (y_1 - y_2) dx = 4 int_0^(1//sqrt 2) (sqrt(1-x^2)-x)dx`

` = 4 [x/2sqrt(1-x^2)+1/2sin^(-1) x - x^2/2]_0^(1//sqrt 2)`

` = 4 [1/(2sqrt2) xx 1/sqrt2 + 1/2 xx pi/4 - 1/4]`

` = pi/2` sq units
Correct Answer is `=>` (C) `pi/2` sq units
Q 1571645526

The area bounded by the `x-` axis and the
curve `y = sin x` and `x = 0, x = pi` is
BITSAT 2011
(A)

`1 sq` unit

(B)

`2 sq` units

(C)

`0`

(D)

`4 sq` units

Solution:

`:.` .Required area `= int_0^pi sinx dx`

` = [-cosx]_0^pi`

` = [+1+1]`

` = 2sq` units
Correct Answer is `=>` (B) `2 sq` units
Q 1515223169

The area bounded by the ` x`-axis and the curve `y=\sin x` and `x=0,x=\pi` is:
BITSAT 2011
(A)

`1` sq. unit

(B)

`2` sq units

(C)

`0`

(D)

`4` sq units

Solution:

`\therefore` Required area `=\int _{ 0 }^{ \pi } \sin x dx `

` = [ -\cos x ] _{ 0 }^{ \pi }`

` = [ +1+1 ]`

`=2` sq units
Correct Answer is `=>` (B) `2` sq units
Q 1530080812

The area (in sq unit) of the region bounded by the curves `2x = y^2 - 1` and `x = 0` is
BITSAT 2008
(A)

`5/3`

(B)

`2/3`

(C)

`1/3`

(D)

`2/5`

Solution:

Given curve can be rewritten as
`y^2 = 2 x + 1/2`

Required area = `int_-1^1 x dy`

= `2 int_0^1 (y^2-1)/2 dy`

= `y^3/3 -y|_0^1`

=`1/3 -1` = `2/3` sq unit
Correct Answer is `=>` (B) `2/3`
Q 1570480316

Area bounded by curve `y = x ^2` and `y= 2- x^2` is
BITSAT 2005
(A)

`8/3` sq unit

(B)

`3/8` sq unit

(C)

`3/2` sq unit

(D)

None of these

Solution:

Given curves are

`y= x^2 `..........................(i)

or `y = 2- x^ 2` .........................(ii)

On solving Eqs. (i) and (ii), we get

`x =-1,1`

and `y =1,1`

`:.` Required area =Area of curve `OABCO`

`= 2` Area of curve `OABO`

`=2 int_0^1 y dx`

`=2 int_0^1 [ (2-x^2) - (x^2)]dx`

`=2 int_0^1 (2-2x^2)dx`

`=4 [x - x^3/3]_0^1`

`=4[ 1 -1/3]`

`=8/3` sq unit
Correct Answer is `=>` (A) `8/3` sq unit
Q 2488134007

The area enclosed by `y = sqrt(5-x^2)` and

`y = | x-1 |` is
WBJEE 2016
(A)

`((5pi)/4 -2)` sq units

(B)

`((5pi -2)/2)` sq units

(C)

`((5 pi)/2 -1/2)`sq units

(D)

`(pi/2 -5)` sq units

Solution:

The graphs of `y = | x - 1 |` and `y = sqrt(5-x^2)` are shown in
the figure and the shaded region is the required region
bounded by the two curves.

Let A be the area bounded by the given curves. Then,

`A= int_(-1)^(1) ( sqrt (5-x^2) +x-1 ) dx + int_(1)^2 (sqrt(5-x^2) -x +1 ) dx`

`= int_(-1)^(1) sqrt(5-x^2) dx + int_(1)^2 sqrt (5-x^2) dx`

`+ int_(-1)^1 (x-1) dx + int_(1)^2 (-x +1) dx`

`= int_(-1)^2 sqrt(5-x^2) dx + [x^2/2 -x]_(-1)^1 + [-x^2/2 +x]_(1)^2`

`= [1/2 xx sqrt (5-x^2) +5/2 sin^(-1) x/sqrt(5) ]_(-1)^2 -5/2`

`=1+ 5/2 sin^(-1) 2/sqrt(5) +1 +5/2 sin^(-1) 1/sqrt(5) -5/2`

`= -1/2 +5/2 sin^(-1 ) (2/sqrt(5) xx sqrt (1-1/5) + 1/sqrt(5) sqrt (1-4/5) )`

`=-1/2 +5/2 sin^(-1) (1) = (5pi)/4 -1/2` sq units
Correct Answer is `=>` (C) `((5 pi)/2 -1/2)`sq units
Q 2510678510

The area of the region bounded by the curve
`y = x^3` , its tangent at `(1,1)` and `X`-axis, is
WBJEE 2015
(A)

`1/12 ` sq unit

(B)

`1/6 ` sq unit

(C)

`2/17 ` sq unit

(D)

`2/15 ` sq unit

Solution:

We have, `y=x^3` and `A(1,1)`

`:. (dy)/(dx) = 3x^2`.................(i)



On putting `x=1` in Eq. (i), we get

`(dy)/(dx) = 3 (1)^2 =3`

`:.` Equation of tangent at `A(1,1)` is

`y-1 = 3 (x-1) => y =3x -2`

`:. ` Required area

`= int_(0)^(1) x^3 dx- int_(2/3)^1 (3x -2) dx`

`= [x^4/4]_(0)^(1) - [(3x^2)/2 -2x]_(2/3)^(1)`


`= 1/4 - [(3/2 -2) - (2/3 -4/3)]`

`=1/12` sq unit
Correct Answer is `=>` (A) `1/12 ` sq unit
 
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