HOW TO FIND COMMON AREA OF TWO PARABOLAS ,AREA UNDER TWO PARABOLAS , AREA UNDER CURVES

How to find common area of two parabolas  , Area under two parabolas , Area of region bounded by two parabolas .

Let us consider two parabolas whose equations are given by
y² =  4ax  --------------  (1)
x² =  4ay ----------------  (2)

To check whether these parabolas intersect with each others or not And if they intersect then what is/are their point/s of intersection.

How to find Points of Intersection

To find coordinate of points of intersection ,we have to solve equation (1) and (2)

Consider   eq (2) 

 x² =  4ay 

⇒ y  = x² /4a   ---------------(3)

Putting the value of "y" in equation  (1) ,we get

Area under Curves

(x² /4a)² =  4ax

 ⇒    x⁴/16a² =  4ax

⇒    x⁴=  64xa³

 ⇒   x⁴- 64xa³= 0

Taking 'x' common 

x(x³- 64a³) = 0
Either  x = 0    or x³- 64a³ = 0
⇒ x = 0    or    (x)³- (4a)³ = 0 
⇒ x = 0    or    (x)³- (4a)³ = 0 

⇒ x = 0    or    (x-4a)[ (x)²+ (4a)² + (x)(4a)]   = 0 

⇒ x = 0    or    (x-4a) =  0  or  [ (x)²+ (4a)² + (x)(4a)]   = 0 
⇒ x = 0    or    x =  4a  or   (x)²+ (4a)² + (x)(4a)   = 0 

Since   x²+ 4a.x + 16a²   = 0   have no real  roots ,because its discriminant is negative, therefore this quadratic equation have complex roots. And these roots are rejected .

To find values of y

Now putting both  values of  "x"  in eq (3) i. e . y  = x² /4a   ,we get

1st  put x = 0 

y = 0 / 4a = 0         ⇒ when x = 0 then y = 0

and put x = 4 

y =  (4a)² /(4a)

y =  4a                    ⇒ when x = 4a then y = 4a

Hence two points of intersection of (1) and (2)   O(0,0) and A (4a , 4a) .

Now draw two parabolas using their points of intersections as drawn in given picture.

How to Find Required Area

Now to find the area enclosed between two Parabolas.

Required Area = shaded Area =Area OLAMO - Area ONAMO

Watch this video to remove your doubts if any

My Previous Posts

  How to find area bounded by two circles 

 How to integrate integral with square root in numerator     

How to find area of the circles which is interior to the parabola

Final words 

Thanks for visiting this website and spending your valuable time to read this post regarding how to find area bounded by two parabolas  .If you liked this post , do share it with your friends to benefit them also ,we shall meet in next post , till then bye and take care......

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HOW TO FIND SAMPLE SPACE FOR TOSSING OF FIVE COINS

Let us discuss the coin toss probability formula, sample space of tossing 1 coin , sample space of tossing 2 coins , sample space of tossing 5 coins , How to make a tree diagram.

To find the sample space for tossing of one  coin

we know that there are two outcomes in tossing of coin . Out of two outcome one is H and second is T.
So  Put H and T in sample space 'S'
S = {H , T}

To find the sample space for tossing of two coins

1st put  2²   = 4  elements in a set of sample space,

1st character of 1st two elements must be H and 1st character of last two elements must be T .

2nd character of all the  elements must be H and T alternatively

 ^{S={ HH , HT , TH , TT}}

To find the sample space for tossing of three coins

1    Put  2³  = 8  elements in a set of sample space,
2    1st character of 1st four elements must be H and 1st character of last four elements must be T .

3   2nd character of 1st ,2nd ,5th and 6th  elements must be H and  2nd character of 3rd, 4th, 7th and 8th  elements must be T alternatively

4    Last character of all the  elements must be H and T alternatively.

Therefore sample  space  'S'  is given by

S = { HHH , HHT , HTH , HTT ,THH, THT , TTH , TTT }

To find the sample space for tossing of Four coins

1    Put  2⁴  = 16  elements in a set ,
2    1st character of 1st eight elements must be H and 1st character of last eight elements must be T.

3   2nd character of 1st four and 9th to 12th  elements must be H and  2nd character of 5th, 6th , 7th and 8th and last four elements must be T alternatively.

4  3rd character for 1st , 2nd , 5th , 6th , 9th , 10th , 13th and 14th elements must be H and for remaining characters ( 3rd , 4th ,7th ,8th , 11th ,12th ,15th and 16th)  it must be T.

5    Last character of all the  elements must be H and T alternatively.

Hence sample space ' S' is given by 

S = { HHHH , HHHT , HHTH , HHTT , HTHH , HTHT , HTTH , HTTT , THHH , THHT , THTH , THTT , TTHH , TTHT , TTTH , TTTT }

To find the sample space for tossing of Five Coins

1    Put  2⁵  = 32  elements in a set ,
2    1st character of 1st sixteen elements must be H and 1st character of last sixteen elements must be T.

3   2nd character of 1st eight and  17th to 24th  elements must be H and  2nd character of 9th to 16th and 25th to 32nd  elements must be T alternatively.

4  3rd character for 1st four , 9th to 12th , 17th to 20th , 25th to 28th elements must be H , And  3rd character for 5th to 8th , 13th to 16th , 21st to 24th and 29th to 32nd elements  must be T.

5  4th character for 1st two ,5th and 6th ,9th and 10th , 13th and 14th , 17th and 18th, 21st and 22nd , 25th and 26th, 29th and 30th and 32nd must be H  And 3rd and 4th , 7th and 8th , 11th and 12th and 15th and 16th , 19th and 20th , 23rd and 24th , 27th and 28th , 31st and 32nd elements must be T.

6    Last character of all the  elements must be H and T alternatively.

Hence sample space ' S' is given by 


S = { HHHHH , HHHHT , HHHTH , HHHTT , HHTHH , HHTHT , HHTTH , HHTTT , HTHHH, HTHHT, HTHTH , HTHTT, HTTHH, HTTHT, HTTTH, HTTTT, THHHH , THHHT , THHTH , THHTT , THTHH , THTHT , THTTH , THTTT ,TTHHH , TTHHT, TTHTH , TTHTT, TTTHH, TTTHT , TTTTH , TTTTT }

^{Watch this video for finding sample space of coins}

Final Words

Thanks for giving your precious time to read this post regarding  probability formula, sample space of tossing 1 coin ,  sample space of tossing 5 coins , How to make a tree diagram.. If you liked it then share it with your near and dear ones to benefit them. we shall meet in next post with another beneficial article till then bye ,take care.......
 

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USING METHOD OF INTEGRATION ,HOW TO FIND AREA OF TRIANGLE BOUNDED BY THREE LINES

Using method of integration find the area of triangle,using the method of integration find the area of the region bounded by the lines,area of triangle by integration method 

Using Method of Integration , How to find the area of triangle bounded by three lines 

2x + y = 0 , 3x - 2y = 6  and x - 3y  + 5 = 0

Solution

Given lines are

2x + y  = 4   --------  (1)
3x - 2y = 6  ---------  (2) and 
x - 3y  = -5  ---------  (3)


If these lines are intersecting then we have to find their coordinates of points of intersection .

To Find Coordinate of Point A

Multiply (1) by 2 and adding to (2) , we get
4x + 2y + 3x - 2y = 8 + 6
7x = 14 ⇒ x =2 
Putting x = 2 in (1) , we get 

Area under Curve

2(2) + y = 4
4 + y = 4 ⇒ y = 0
∴ (1) and (2) meets at point A(2,0).

To Find Coordinate of Point B

To find point of intersection (2) and (3);
Multiply (3) by -3 and adding to (2) , we get
3x - 2y -3x +9y  = 6 +15 
7y = 21  ⇒ y = 3 

Putting y = 3 in (3) , we get 
x-3(3)  = -5  
⇒ x = -5 +9  ⇒ x = 4 
∴ (2) and (3) meets at point B(4,3).

To Find Coordinate of Point C

To find point of intersection (1) and (3);
Multiply (1) by 3 and adding to (3) , we get
6x + 3y + x - 3y = 12 - 5 
7x = 7  ⇒ x =1 

Putting x = 1 in (1) , we get 
2(1) + y  = 4  
⇒ y =4 - 2  ⇒ y = 2 
∴ (1) and (3) meets at point C(1,2).

we get points of intersection of (1) and (2)  A(2,0),  points of intersection of (2) and (3) B(4,3) and points of intersection of (1) and (3) C(1,2).

Required Area = Shaded Area   =  Area DCBED - Area DCAD -Area ABEA

Also read my previous post  How to find area bounded by two circles 

For better understanding watch this video 

Conclusion 

 Thanks  for visiting this website and spending your precious time to read how to find area of triangle Using method of integration find the area of triangle,using the method of integration find the area of the region bounded by the lines,area of triangle by integration method.

If you are a mathematician Don't forget to visit my Mathematics You tube channel ,Mathematics Website and Mathematics Facebook Page , whose links are given below

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HOW TO SOLVE DETERMINANTS USING ELEMENTARY TRANSFORMATIONS

How to Prove Determinants using elementary transformations 

In this post we shall discuss Short trick of elementary transformation,Solving Determinants using elementary transformations,define elementary transformation, elementary transformation class 12, elementary row transformation questions.

By this method we have  to reduce maximum elements of specific Rows or column to zero, so that we can solve it easily

To solve the determinants using elementary transformations , Let us suppose L H S = △

As we can see that 'a' is common in 1st Row , 'b' is common in 2nd Row and 'c' is common in 3rd row ,
Therefore Taking a ,b ,c common from R_1  , R_2  , and  R₃       respectively

If we add R_1    to   R_2  and   R₁   to  R_{3   then we get zero in 1st column, so}  Operating  R_{1  →} R₁   + R₂   and    R_{3  →} R₁   + R₃  

_{As we have received maximum possible  zero in 1st column Therefore Expanding along} C₁  

△= (-a)×[(0)-(2c×2b)]
△ _{= abc{-a(-4bc)}}

△ = 4a²b²c² 

Hence the proof


Watch this video for Understanding Elementary transformations

PROBLEM

Proof:- Put L H S of determinant to Δ

Operating R₁ ➡️xR₁  , R₂ ➡️ yR₂ and R₃➡️ zR₃

Taking common xyz from C₃

Operating  R₂ ➡️ R₁ - R_2  and  R₃ ➡️ R₁ - R3

Expanding along  C₁ 

Δ = (x² - y² )( x³ - z³ ) - (x³ - y³ )(x² - z² )

Δ = (x- y )(x+ y)(x- z )( x² + z²  + xz ) - (x- y )( x² + y²  + xy ) (x - z )(x +z )

Δ = (x- y )(x- z )[(x+ y)( x² + z²  + xz ) -( x² + y²  + xy ) (x +z )]

Cancelling the same colour terms in the previous line ,then we have 

 Δ = (x- y )(x- z )[xz²   + yz² - y²x - y²z  ]

Arranging  terms in Squared Bracket  in such a way that the term containing z² must be at 1st and 3rd position and the term containing y² must be at 2nd and 4th position .

Δ = (x- y )(x- z )[(yz² - y²z) +( xz²  - y²x)]

Δ = (x- y )(x- z )[yz(z - y) + x(z²  - y²)]

Δ = (x- y )(x- z )[yz(z - y) + x(z - y)(z+ y)]

Δ = (x- y )(x- z )(z - y)[yz + x(z+ y)]

Δ = (x- y )(x- z )(z - y)[yz + xz+ xy]

Taking -1 common from (x- z )(z - y) in previous line ,

Δ = (x- y )(y- z )(z - x)[yz + xz+ xy]

Hence the   proof

Also Read   HOW TO FIND THE SLOPE OF LINE  AX + BY = C ,  SLOPE OF A  LINE

Final Words

Thanks for investing your precious time to read this post containing  Solving Determinants using elementary transformations,Short trick of elementary transformation  , elementary row transformation questions. If you liked it then share it with your near and dear ones to benefit them. we shall meet in next post with another beneficial article till then bye ,take care.......

If you are a mathematician Don't forget to visit my Mathematics You tube channel ,Mathematics Website and Mathematics Facebook Page , whose links are given below

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HOW TO FIND AREA BOUNDED BY TWO CIRCLES , INTEGRATION OF AREA UNDER CURVE

HOW TO FIND AREA OF TWO CIRCLES INTERSECTING EACH OTHERS,

Here we are going to discuss how to find common area of two circles which are overlapping or intersecting at two points with the help of an example

 Let us consider two circles whose equations are  given below
x² + y² =  1^{2                   .......................(1)}
( x - 1 )² + y² = 1²          .....................(2)

Let us draw these circles in coordinate planes, We can compare these equations with standard form of circle to find  the coordinate of  centre of both the circles are (0,0) and (1,0) respectively and radius of both the circles are 1.

If these two circles intersect with each other then we have to find their point/s of intersection.
To find points of intersection subtracting equation (1) from eq (2) , we get 

    ( X - 1 )² + Y² - X² - Y² =  1^2 - 1² 
⇒( X - 1 )²  - X² =  0
 ⇒X ²  +1 ²  - 2×(1)×(X) - X² =  0

⇒1 - 2X = 0

⇒ X = 1/2 ,
Now to find the values of y put the value of x in equation # 1

(1/2)² + Y² =  1²     
   Y² = 1- (1/4) = 3/4

   Y = 土⇃(3/4)
Therefore two points of intersection are B(1/2 , ⇃(3/4)) , C ( (1/2 , -⇃(3/4))

To understand better the solution of  this problem watch this  video 

Required area = shaded  Area , 

we can divide shaded area into four equal parts , As each parts is symmetrical , Therefore to find shaded area  it is sufficient to find the area of any one of four part and then then multiply it with 4.

Hence  Required area = 4 area OBLO = 4 Area BALB
To avoid tedious calculations choose 2nd part to integrate
I= Required area = 4 Area BALB    ------------- (3)

After simplification , we have

    ALSO READ        HOW TO INTEGRATE INTEGRAL WITH SQUARE ROOT IN NUMERATOR     

My previous post HOW TO FIND AREA OF THE CIRCLE  WHICH IS INTERIOR TO THE PARABOLA

  Final words 

Thanks for visiting this website and spending your valuable time to read this post regarding how to find area bounded by two circles .If you liked this post , do share it with your friends to benefit them also we shall meet in next post , till then bye and take care....

..

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