UNIT–2

Multivariable Calculus

2.1. Limit, continuity and partial derivatives

Limits-

The function f(x,y) is said to tend to limit ‘l’ , as x →a and  y→b Iff the limit is dependent on point (x,y) as  x →a and  y→b

We can write this as,

Example-1: evaluate the

Sol. We can simply find the solution as follows,

Example-2:  evaluate

Sol.

-6.

Example-3: evaluate

Sol.

Conitinuity –

At point (a,b) , a function f(x,y) is said to be continuous if,

Working rule for continuity-

Step-1: f(a,b) should be well defined.

Step-2: should exist.

Step-3:

Example-1: Test the continuity of the following function-

Sol. (1) the function is well defined at (0,0)

(2) check for the second step,

That means the limit exists at (0,0)

Now check step-3:

So that the function is continuous at origin.

Example-2: check for the continuity of the following function at origin,

Sol. (1) Here the function is well defined at (0,0)

(2) check for second step-

Limit f is not unique for different values of m

So that the limit does not exists.

Therefore the function is not continuous at origin.

Steps to check for existence of limit-

Step-1: find the value of f(x,y) along x →a and  y→b

Step-2: find the value of f(x,y) along x →b and  y→a

Note- if the values in step -1 and step-2 are same then we can say that the limits exist otherwise not.

Step-3: if a →0 and  b→0 then find the limit along y =mx , if the value does not contain m then limit exist,  If it contains m then the limit does not exist.

Note-1- put x = 0 and y = 0 in f , then find f1

2 - Put y = 0 and x = 0 In f  then find x2

If f1 and f2 are equal then limit exist otherwise not.

3- put y = mx  then find f3

If f1 = f2 ≠f3, limit does not exist.

4- put y = mx² and find f4,

If f1 = f2 = f3 ≠ f4 , limit does not exist

If f1 = f2 = f3 = f4 , limit exist.

Example-1:  Evaluate 

Sol . 1.     

2.

Here f1 = f2

3. Now put y = mx, we get

Here f1 = f2 = f3

Now put y = mx²

4.

Therefore ,

F1 = f2 = f3 =f4

We can say that the limit exists with 0.

Example-2: evaluate the following-

Sol.  First we will calculate f1 –

Here we see that f1 = 0

Now find f2,

Here , f1 = f2

Therefore the limit exists with value 0.

Partial derivatives-

First order partial differentiation-

Let f(x , y) be a function of two variables. Then the partial derivative of this function with respect to x can be written as and defined as follows:

=

Now the partial derivative of f with respect to f can be written as and defined as follows:

=

Note: a. While calculating partial derivatives treat all independent variables, other than the variable with respect to which we are differentiating , as constant.

b. We apply all differentiation rules.

Higher order partial differentiation-

Let f(x , y) be a function of two variables. Then its second-order partial derivatives, third order partial derivatives and so on are referred as higher order partial derivatives.

These are second order four partial derivatives:

(a)     =            

(b)  =                 

(c)   =                 

(d)     =

b and c are known as mixed partial derivatives.

Similarly we can find the other higher order derivatives.

Example-1: - Calculate    and   for the following function

f(x , y) = 3x³-5y²+2xy-8x+4y-20

Sol. To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 9x² - 0 + 2y – 8 + 0 – 0

= 9x² + 2y – 8

Similarly partial derivative of f(x,y) with respect to y is:

= [3x³-5y²+2xy-8x+4y-20]

= 3x³] - 5y²] + [2xy] -8x] +4y] - 20]

= 0 – 10y + 2x – 0 + 4 – 0

= 2x – 10y +4.

Example-2: Calculate    and   for the following function

f( x, y) = sin(y²x + 5x – 8)

Sol. To calculate   treat the variable y as a constant, then differentiate f(x,y) with respect to x by using differentiation rules,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= (y² + 50) cos(y²x + 5x – 8)

Similarly partial derivative of f(x,y) with respect to y is,

[sin(y²x + 5x – 8)]

= cos(y²x + 5x – 8)(y²x + 5x – 8)

= 2xy cos(y²x + 5x – 8)

Example-3: Obtain all the second order partial derivative of the function:

f( x, y) = ( x³y² - xy⁵)

Sol.    3x²y² - y⁵,      2x³y – 5xy⁴,

  = = 6xy²

  = 2x³ - 20xy³

= = 6x²y – 5y⁴    

= = 6x²y -   5y⁴

Example-4: Find

Sol.  First we will differentiate partially with repsect to r,

Now differentiate partially with respect to θ, we get

Example-5:  if,

Then find.

Sol-

Example-6: if , then show that-

Sol. Here we have,

u =     …………………..(1)

Now partially differentiate eq.(1) w.r to x and y , we get

=

Or

                   ………………..(2)

And now,

=

         ………………….(3)

Adding eq. (1) and (3) , we get

  =  0

Hence proved.

2.2. Total derivative

When we measure the rate of change of the dependent variable owing to any change in a variable on which it depends, when none of the variable is assumed to be constant.

Let the function,     u = f( x, y),  such that    x = g(t)  ,     y = h(t)

ᵡ  Then we can write,

=

=

This is the total derivative of u with respect to t.

Change of variable-

If w = f (x, y) has continuous partial variables fx and fy and if x = x (t), y = y (t) are

Differentiable functions of t, then the composite function w = f (x (t), y (t)) is a

Differentiable function of t.

In this case, we get,

fx (x (t), y (t)) x’ (t) +  fy (x(t), y (t)) y’ (t).

Example-:1 let q = 4x + 3y      and    x = t³ + t² + 1    , y = t³ - t² - t

Then find  .

Sol. :            . =

Where, f1 =         ,  f2 = 

In this example f1 = 4    ,      f2 = 3

Also,              3t² + 2t    ,                      

4(3t² + 2t) + 3(

=  21t² + 2t – 3

Example-2: Find      if u = x³y⁴    where   x = t³     and    y = t².

Sol. As we know that by definition,    =

3x²y⁴3t² + 4x³y³2t = 17t¹⁶.

Example-3: if w = x² + y – z + sint and  x + y = t, find

(a)    y,z

(b)   t, z  

Sol. With x, y, z independent, we have

t = x + y, w = x² + y - z + sin (x + y).

Therefore,

y,z = 2x + cos(x+y)(x+y)

=  2x + cos (x + y)

With x, t, z independent, we have

Y = t-x,   w=  x² + (t-x) + sin t

Thus     t, z   =  2x - 1

Example-4: If u = u( y – z , z - x , x – y)  then prove that  = 0

Sol. Let,

Then,

By adding all these equations we get,

= 0    hence proved.

Example-5:   if φ( cx – az ,  cy – bz) = 0 then show that ap + bq = c

Where p =      q = 

Sol.  We have,

φ( cx – az ,  cy – bz) = 0   

φ( r , s) = 0

Where,

We know that,

Again we do,

By adding the two results, we get

Example-6: If z is the function of x and y , and x =   , y = , then prove that,

Sol. Here , it is given that,  z is the function of x and y & x , y are the functions of u and v.

So that,

  ……………….(1)

And,

     ………………..(2)

Also there is,

x =   and  y = ,

Now,

   ,       ,      ,    

From equation(1) , we get

  ……………….(3)

And from eq. (2) , we get

      …………..(4)

Subtracting eq. (4) from (3), we get

= ) – ( 

= x

Hence proved.

Differentiability-

Let f(x) be a single valued function of the variable x,  then,

f’(x) =

Provided that the limit exists and is dependent of the path along which

Example: suppose that the function,

f(z) = 4x + y + i( -x + 4y)

Discuss df/dz.

Sol. Here,

Example-2: if

Then df/dz , z = 0.

Sol.

2.3. Tangent plane and normal line

Equation of a tangent plane-

And

Equation of the normal to the plane-

Example: Find the equation of the tangent plane and normal line to the surface

Sol.

Here,

At point (1, 2, -1)-

Therefore the equation of the tangent plane at (1, 2, -1) is-

Equation of normal line is-

2.4. Maxima, minima and saddle point

If  f(x) is a single valued function defined in a region R then

Maxima is a maximum point if and only if

Minima is a minimum point if and only if   

Maxima and Minima of a function of two independent variables

Let be a defined function of two independent variables.

Then the point is said to be a maximum point of if

Or   =

For all positive and negative values of h and k.

Similarly the point is said to be a minimum point of if

Or   =

For all positive and negative values of h and k.

Saddle point:Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.

A point is a saddle point of a function of two variables if

At the point.

Stationary Value

The value is said to be a stationary value of if

i.e. the function is a stationary at (a , b).

Rule to find the maximum and minimum values of

1. Calculate.
2. Form and solve , we get the value of x and y let it be pairs of values
3. Calculate the following values :

4.     (a) If

(b) If

(c) If

(d) If

Example1 Find out the maxima and minima of the function

Given     …(i)

Partially differentiating (i) with respect to x we get

    ….(ii)

Partially differentiating (i) with respect to y we get

    ….(iii)

Now, form the equations

Using (ii) and (iii) we get

          using above two equations

Squaring both side we get

Or 

This show that

Also we get

Thus we get the pair of value as

Now, we calculate

Putting above values in

At point (0,0)  we get

So, the point (0,0) is a saddle point.

At point we get

So the point is the minimum point where

In case 

So the point is the maximum point where

Example2 Find the maximum and minimum point of the function

Partially differentiating given equation with respect to and x and y then equate them to zero                 

On solving above we get 

Also

Thus we get the pair of values (0,0), (,0) and (0,

Now, we calculate

At the point (0,0)

So function has saddle point at (0,0).

At the point (

So the function has maxima at this point (.

At the point (0,

So the function has minima at this point (0,.

At the point (

So the function has an saddle point at (

Example3 Find the maximum and minimum value of

Let

Partially differentiating given function with respect to x and y and equate it to zero

  ..(i)

    ..(ii)

On solving (i) and (ii) we get

Thus pair of values are

Now, we calculate

At the point (0,0)

So further investigation is required

On the x  axis y  = 0  , f(x,0)=0

On the line y=x,

At the point

So that the given function has maximum value at

Therefore maximum value of given function

At the point

So that the given function has minimum value at

Therefore minimum value of the given function

2.5. Method of Lagrange’s multiplier

Let be a function of x, y, z which to be discussed for stationary value.

Let be a relation in x, y, z

for stationary values we have,

i.e.    … (1)

Also from we have

    … (2)

Let ‘’ be undetermined multiplier then multiplying equation (2) by and adding in equation (1) we get,

     … (3)

     … (4)

      … (5)

Solving equation (3), (4) (5) & we get values of x, y, z and .

1. Decompare a positive number ‘a’ in to three parts, so their product is maximum

Solution:

Let x, y, z be the three parts of ‘a’ then we get.

    … (1)

Here we have to maximize the product

i.e.

By Lagrange’s undetermined multiplier, we get,

       … (2)

       … (3)

        … (4)

i.e.

        … (2)’

        … (3)’

        … (4)

And

From (1)

Thus .

Hence their maximum product is  .

2.     Find the point on plane nearest to the point (1, 1, 1) using Lagrange’s method of multipliers.

Solution:

Let be the point on sphere which is nearest to the point . Then shortest distance.

Let

Under the condition    … (1)

By method of Lagrange’s undetermined multipliers we have

       … (2)

       … (3)

i.e. &

       … (4)

From (2) we get

From (3) we get

From (4) we get

Equation (1) becomes

i.e.

y = 2

2.6. Multiple integration: Double and triple integrals (Cartesian and polar)

Double integral –

Before studying about multiple integrals , first let’s go through the definition of definition of definite integrals for function of single variable.

As we know, the integral

Where is belongs to the limit a ≤ x ≤ b

This integral can be written as follows-

Now suppose we have a function f(x , y) of two variables x and y in two dimensional finite region R in xy-plane.

Then the double integration over region R can  be evaluated by two successive integration

Evaluation of double integrals-

If A is described as 

Then,

]dx

Let do some examples to understand more about double integration-

Example-1: Evaluate , where dA is the small area in xy-plane.

Sol.     Let ,     I = 

=

= 

= 

=    84 sq. Unit.

Which is the required area.

Example-2:  Evaluate

Sol. Let us suppose the integral is I,

I =

Put c = 1 – x in I, we get

I =

Suppose , y = ct

Then dy = c

Now we get,

I =

I =

I =

I =

I  =

As we know that by beta function,

Which gives,

Now put the value of c, we get

Example-3:  Evaluate the following double integral,

Sol.  Let ,

I = 

On solving the integral, we get

Double Integral over Rectangular and general regions

Consider a function f (x, y) defined in the finite region R of the x-y plane. Divide R into n elementary areas   A1, A2,…,An. Let (xr, yr) be any point within the rth elementary are Ar

                               Fig. 6.1

 f (x, y)  dA  =  f (xr, yr)  A 

Evaluation of Double Integral when limits of Integration are given(Cartesian Form).

Ex. 1 : Evaluate  ey/x dy dx.

Soln. : 

Given :  I  =   ey/x dy dx

Here limits of inner integral are functions of y therefore integrate w.r.t y,

I  =  dx

=  

=

I  =  

=            =

ey/x dy dx = 

Ex. 2 : Evaluate   xy (1 – x –y) dx dy.

Soln. : 

Given :  I  =  xy (1 – x –y) dx dy.

Here the limits of inner integration are functions of y therefore first integrate w.r.t y.

I  =   xdx

Put      1 – x =  a (constant for inner integral)

    I  =   xdx

Put  y = at   dy = a dt

   I  =  xdx

   I  =    xdx

I  =    xa dx 

I  =   x (1 – x) dx =   (x– x4/3) dx

I  =

=  

I  =  =

 xy (1 – x –y) dx dy    =  

Ex. 3 : Evaluate

Soln. : 

Let, I  = 

Here limits for both x and y are constants, the integral can be evaluated first w.r.t any of the variables x or y.

I  =  dy 

I =  

=          

=            

=          

=  

=

 = 

Ex. 4:  Evaluate  e–x2 (1 + y2) x dx dy.

Soln. : 

Let  I  =   e–x2 (1 + y2)  x dy  =   dy   e–x2 (1 + y2)  x dy

=  dy    e– x2 (1 + y2)  dx

=  dy   [∵ f (x)  ef(x) dx = ef(x) ]

=   (–1) dy                (∵ e– = 0)

=  =  =

 e–x2 (1 + y2) xdx dy   = 

NOTES:

Type II: Evaluation of Double Integral when region of Integration is provided (Cartesian form)

Ex.1: Evaluate  y dx dy over the area bounded by x= 0 y = and x + y = 2 in the first quadrant  

Soln. : 

The area bounded by y = x2 (parabola) and x + y = 2 is as shown in Fig.6.2

The point of intersection of y = x2 and x + y = 2.

x + x2  =  2     x2 + x – 2 = 0

    x   = 1, – 2

At   x  =  1, y = 1 and at x  =  –2, y = 4

Fig. 6.2

 (1, 1) is the point of intersection in Ist quadrant. Take a vertical strip SR, Along SR x constant and y varies from S to R i.e. y = x2 to y = 2 – x.

Now slide strip SR, keeping IIel to y-axis, therefore y constant and x varies from x = 0 to x = 1.

    I  =  

=

= 

=  (4 – 4x + – ) dx

=  =  

    I  =  16/15

Ex. 2  : Evaluate          over x  1, y 

Soln. : 

 Let      I =            over x  1, y 

The region bounded by x  1 and y 

Is as shown in Fig. 6.3.

Fig. 6.3

Take a vertical strip along strip x constant and y varies from y =

To y = . Now slide strip throughout region keeping parallel to y-axis. Therefore y constant and x varies from x = 1 to x = .

I =               

=             

=                   [ ∵ dx =  tan–1 (x/a)]

=  =

= –  = (0 – 1)

    I  = 

Ex. 3 : Evaluate  (+ ) dx dy through the area enclosed by the curves y = 4x, x + y = 3 and   y =0, y = 2.

Soln. : 

Let   I  =   (+ ) dx dy

The area enclosed by the curves y = 4x, x + y =3, y = 0 and y = 2 is as shown in Fig. 6.4.

(find the point of intersection of  x + y = 3 and y = 4x)

Fig. 6.4

Take a horizontal strip SR, along SR y constant and x varies from x =  to x = 3 – y. Now slide strip keeping IIel to x axis therefore x constant and y varies from y = 0 to y = 2.

I  =   dy (+ ) dx

=      

=  +dy

   I = 

=  

=  

=   +  – 6 + 18

    I = 

Triple integral-

Definition: Let f(x,y,z) be a function which is continuous at every point of the finite region (Volume V) of three dimensional space. Divide the region V into n sub regions of respective volumes. Let () be a point in the    sub region then the sum:

=

Is called triple integration of f(x, y, z) over the region V provided limit on R.H.S of above Equation exists.

Spherical Polar Coordinates

Where the integral is extended to all positive values of the variables subjected to the condition

Ex.1: Evaluate

Solution: Let

I     =    

=    

(Assuming m = )

= dxdy

=

=

= dx

=  dx

= 

=

I =

Ex.2: Evaluate        Where V is annulus between the spheres 

And      ()

Solution:  It is convenient to transform the triple integral into spherical polar co-ordinate by putting

,                      ,   

,    dxdydz=sindrdd,   

   and  

For the positive octant,  r  varies from  r =b to r =a  ,     varies from

And      varies from

I= 

= 8

=8

=8

=8

=8 log

= 8 log

I= 8 log           I =  4 log

Ex.3: Evaluate

Solution:-

Ex.4:Evaluate

Solution:-

NOTES:

The volume of solid is given by

Volume =

In Spherical polar system 

In cylindrical polar system

Ex.1: Find Volume of the tetrahedron bounded by the co-ordinates planes and the plane

Solution: Volume =                                                 ………. (1)   

Put          ,  

From     equation (1) we have

V =

=24

=24           (u+v+w=1)  By Dirichlet’s theorem.

=24

=  = =  4

Volume =4

Ex.2:  Find volume common to the cylinders,  .

Solution:    For given cylinders,

,    .

Z varies from

Z=-   to  z =

Y varies from

y= -   to  y =

x varies from   x= -a to x = a

By symmetry,

Required volume= 8 (volume in the first octant)

=8

=8

= 8dx

=8

=8

=8

Volume   =   16

Ex.3:  Evaluate   

1. Solution:-                  

Ex.4:

Ex.5: Evaluate

Solution:-   Put

2.7. Change of order of integration in double integrals

Ex. 1 : Change the order of integration for the integral  and evaluate the same with reversed order of integration.             

Soln. :  

Given,  I  =    …(1)

In the given integration, limits are

y = ,  y = 2a – x and x = 0,   x = a

The region bounded by x2 = ay, x + y = 2a                                   Fig.6.5

And x = 0, x = a  is as shown in Fig. 6.5

Here we have  to change order of Integration. Given the strip is vertical.

Now take horizontal strip SR.

To take total region, Divide region into two parts by taking line y = a.  

1 st Region  :

Along  strip, y constant and x varies from x = 0 to x = 2a – y. Slide strip IIel to  x-axis therefore y varies from y = a to y = 2a.

     I1 =  dy  xy dx …(2)

2nd Region :

Along strip, y constant and x varies from x = 0 to x = . Slide strip IIel to x-axis therefore x-varies from y = 0 to y = a.

   I2 =  dy   xy dx    …(3)

From Equation (1), (2) and (3),

   =   dy  xy dx +  dy   xy dx

= +  y dy

=dy  +  (ay) dy =  y (4a2 – 4ay + y2) dy +  ay2 dy

=  (4a2 y – 4ay2 + y3) dy +   y2 dy

= 

= +

=            a4

Ex. 2 : Evaluate

I =

Soln. :  

Given :  I  =      …(1)

In the given integration, limits are

x = 0,  x = a,   y = 0,   y =

The bounded region is as shown in Fig. **.

In the given, strip is vertical. Now take horizontal strip SR. Along strip y constant and x varies from x = 0 to

x = . Slide strip IIel to X-axis therefore y varies from y = 0 to y = a.

 I =  dy

=          

Put  a2 – y2 = b2

 I = 

=  =

=  dy = dy

= 

=

= 

=  [∵ a = a loge]

  I =  dy = 

Ex.3 : Express as single integral and evaluate  dy dx +  dy  dx.

Soln. : 

Given :  I  =   dy dx +  dy  dx

I  =  I1 + I2

The limits of region of integration I1 are
x = – ; x =  and y = 0, y = 1 and I2 are x = – 1,
x = 1 and y = 1, y = 3.

The region of integration are as shown in Fig. 6.7

To consider the complete region take a vertical strip SR along the strip y varies from y = x2 to
y = 3 and x varies from x = –1 to x = 1. Fig. 6.7

I  =

2.8. Change of variable (Cartesian and polar)

Evaluation of Double Integral by Changing Cartesian to Polar co-ordinates (when limits are given).

Ex. 1 : Evaluate

Soln. : 

The region of integration bounded by

y = 0, y = and x = 0, x = 1

y  =   x2 + y2 = x

The region bounded by these is as shown in Fig. 6.8.

Convert the integration in polar co-ordinates by using x = r cos , y = r sin  and dx dy = r dr d

 x2 + y2 =  x becomes r = cos 

y  =  0 becomes r sin  = 0  = 0

x = 0 becomes r cos  = 0  =

And x = 1 becomes r = sec                     

Take a radial strip SR with angular thickness , Along strip  constant and r varies from r = 0 to r = cos . Turning strip throughout region therefore  varies from = 0 to  =

   I  =   r dr d

=  4 cos  sin  d r dr

=  4 cos  sin  d [–]

=  – 2  cos  sin  [+1] d

=  – 2  [cos  sin  – cos  sin ] d

=   –2 + 2  cos  sin  d

=   –  + 2

=   + 1 =

Ex. 2: Sketch the area of double integration and evaluate

dxdy

Soln. : 

Let  I  = dxdy

The region of integration is bounded by the curves

x = y, x = and y = 0, y =    Fig. 6.9

i.e. x  =  y, x2 + y2 = a2 and y = 0, y =

The region bounded by these is as shown in Fig. 6.9. 

The point of intersection of x = y and x2 + y2 = a2 is x =

Convert given integration in polar co-ordinates by using polar transformation x = r cos , y = r sin  and dx dy = r dr d

 x  =  y gives  r cos  = r sin  tan  = 1  =

x2 + y2 = a2   r2 = a2 r =  a

y  =  0 gives r sin  = 0  = 0.

y  =  gives   r sin  =   r =  cosec 

Take a radial strip SR, along SR  constant and r varies from r = 0 to r = a. Turning this strip throughout region therefore  varies from  = 0 to  =

 I  =  log r2 r dr d = 2  d r log r dr

I  =

=  2 d

=  2 d

=  2 d

=  2 []

I   =  [/4] =  

Evaluation of Double Integral when region of Integration is provided (Polar form)

Ex. 1 : Evaluate  r4 cos3 dr d over the interior of the circle r = 2a cos 

Soln. : 

The region of the integration is as shown in Fig. 6.10.Take a radial strip SR, along strip  constant and r varies from r = 0 to r = 2a cos. Now turning this strip throughout region therefore  varies from  =  to  =

  I  =   r4 cos3 dr d

=  cos3 d

=  cos3 cos5 d

=   cos8 d    Fig. 6.10

=  

=   2 

I  =  

 r4 cos3 dr d =

Ex. 2 : Evaluate  r sin  dr d over the cardioid r = a (1 – cos ) above the initial line.

Soln. :  

The cardioid r = a (1 – cos ) is as shown in Fig. 6.11. The region of the integration is above the initial line.

Take a radial strip SR, along strip  constant and r Varies from r = 0 to r = a (1 – cos ).

New turning the strip throughout region therefore  varies from  = 0 to  = .

   I =  r sin  dr d

=  sin  d

Fig.6.11

= sin  [a2 (1 – cos )2]

=   

I =  (sin   – 2 sin  cos  + sin ) d

=     2 (sin   – sin2  + sin ) d

= +

I =  a2=  a2

I =  

NOTES:

Area enclosed by plane curves expressed in Cartesian coordinates:

Y                             (x,y)   P        Q (x+dx,y+dy)        y=f(x)

                                         V      U

G       H                     dxdy

                  A                R S B

x=a             y=0             x=b                          X

Consider the area enclosed by the curves

And and the ordinates

Area =

Consider the area enclosed by the curves

And and the ordinates

Area =

Area enclosed by plane curves expressed in Polar coordinates:

Consider the area enclosed by the polar curves

And and the line

Area =

Ex.1: Find the area between the curves and its asymptote.

Solution: The curve is symmetrical about x axis not passing through origin. Also is the asymptote to the curve and intersect x axis at for    curve doesn’t exists. And for and. Because of symmetry

Put

Ex: 2: Show that the area of curve   is

Solution: The curve is symmetrical about Y - axis passing through origin and there exists a cusp at origin. It intersects Y-axis at (0 , 0) , (0 , 2a) .

Also   .

Putting

.

Ex.3: Find the Area included between the two cardioids

Solution: - Area = 

=

=

=

Ex.4: Find the Area common to the two circle 

Solution: - By converting the given circle into polar form we get

Area =

2.9. Applications: Area and volume by double integration

Area in Cartesian coordinates-

Example-1: Find the area enclosed by two curves using double integration.

y = 2 – x and y² = 2 (2 – x)

Sol.  Let,

y =  2 – x  ………………..(1)

And                y² = 2 (2 – x) ………………..(2)

On solving eq. (1) and (2)

We get the intersection points  (2,0) and (0,2) ,

We know that,

Area =

Here we will find the area as below,

Area =

Which gives,

= ( - 4 + 4 /2 ) + 8 / 3  = 2 / 3.

Example-2: Find the area between the parabola y ² = 4ax and another parabola x² = 4ay.

Sol. Let,

y ² = 4ax ………………..(1)

And

x² = 4ay…………………..(2)

Then if we solve these equations , we get the values of points where these two curves intersect

x varies from y²/4a to and y varies from o to 4a,

Now using the conceot of double integral,

Area = 

Area in polar coordinates-

Example-3:  Find the area lying inside the cardioid r = a(1+cosθ) and outside the circle r = a, by using double integration.

Sol. We have,

r = a(1+cosθ)   …………………….(1)

And

r = a ……………………………….(2)

On solving these equations by eliminating r , we get

a(1+cosθ)    =   a

(1+cosθ) = 1

Cosθ = 0

Here a θ varies from – π/2  to    π/2 

Limit of r will be a and 1+cosθ)   

Which is the required area.

Example-4: Find the are lying inside a cardioid r = 1 + cos θ and ouside the parabola r(1 + cos θ) = 1.

Sol. Let,

r = 1 + cos θ   ……………………..(1)

r(1 + cos θ) = 1……………………..(2)

Solving these equestions , we get

(1 + cos θ )( 1 + cos θ ) = 1

(1 + cos θ )² = 1

1 + cos θ    = 1

Cos θ = 0  

θ = ±π / 2  

So that, limits of r are,

1 + cos θ    and 1 / 1 + cos θ  

The area can be founded as below,

2.10. Centre of mass and gravity

The centre of gravity of a lamina is the point where it balances perfectly which means the lamina’s centre of mass.

If and are the coordinates of the centroid C of area A, then-

And

Example: Determine the coordinates of the centroid of the area lying between the curve y = 5x - x² and the x-axis.

Sol.

Here y = 5x - x² when y = 0, x = 0 or x = 5

Therefore the curve cuts the x-axis at 0 and 5 as in the figure-

Now

= 2.5

Therefore the centroid of the area lies at (2.5, 2.5)