(a) yx² - y = x² + 1

|| to x axis

          X²=0

          X = 0                    No asymptote || to x axis

|| to y axis

        X²-1 = 0

            X² = 1

            X = 1

(b) Volume will be given by integrating the terms

             =  

             =  

(d) let x and y be semi major and semi minor axes of an ellipse.

                                        Area of ellipse = A=

                                        Log A = Logπ + Log x +Log y

 1+1=2

                                         Error in area = 2 %

Yes, Degree = 2

=

  =  

  =    

  =    

  =   2

    =

       =

        = -2y + 2y

           [  ]

For solenoidal function

div 

div f =   

= -2x + 4 + 2x -4

= 0

So it is solenoidal

Green’s theorem in plane:-

Let R bed closed region of x-y plane bounded by a simple closed cense c and let M and N be continuous function of x and y having continuous partial derivative  and  in R then:

(a) Symmetry:  Symmetrical about x-axis

Origin: passes through origin and tangent at origin

       y= x and y =-x   therefore origin is a node.

      Asymptotes x = a

 Points:  it crones the Ares at (0,0) and  (-a,0 )

      When x >a  or  <-a    y is imaginary

Shape of curve is strophoid      

(b) Given figure is a cardioid and symmetrical about q = 0 i.e. initial line.

q  varies from (0-π/2 ) or (0-180).

(a) Given curve is asteroid. Which is symmetrical about x-axis and y-axis therefore entries length of curve is four times the length of one part.

   Diff given curve w.r.t x

2/3x^-1/3+ 2/3y^-1/3dy/dx =0

          2/3y^-1/3dy/dx = -2/3x^-1/3

dy/dx=-(y/x)^1/3

total length = 4    = 4 

 = 4 

= 6a

(b)   

      Therefore area of ellipse = 4x area of ellipse in I quadrant.

        =

        =

=

Therefore required area = area of region PQRS

 = 4 x area of region OFPB in the first quadrant

  =

  = 

  = 

(a) 

       =

                                        = 

                                        =

                                        =

(b)  Euler’s theorem  

              If H= f (x, y, z)is ahomogeneous function of x, y and z of degree n, then

U= sin^-1(x²+y²/x+y)

       Sin u = x²+y²/x+y

      F =   x²+y²/x+y

Which is homogeneous of degree 1.

 By Euler’s theorem  

          =

(a) Temperature at any point on sphere is given by

              T= 400 xy (1-x²-y²)

  = 400y -400y³-1200x²y

400x – 1200xy²-400x³

 For critical points

400y-400y³-1200x²y

  400x(1-3y²-x²)=0

critical points after solution above equation are

(0,0),(±1,0),(0, ±1),( ±1/2, ±1/2) out of which first three given the value of T to be zero. For other four point (±1/2, ±1/2). Now A = =  2400 xy, B =   = 400-1200y²-1200x², C= = -2400xy

At (1/2,1/2,) A= -600 <  0, B= -200, C = -600 so that AC-B²= 320000> 0

∴  T is maximum at(1/2,1/2)

                                      Whose maximum value will be given by

                               = 400 ¼(1 -1/4 - 1/4) = 100 x ½ =50

(b)  f(x,y) = tan^-1xy

   Let x=0.9  , y=-1

X+δx= 0.9      1+δx=0.9                =δx=-1            =   

     -y+ δy=-1.2       δy=-1.2+1            = δy = -0.2

df=                        = y/1+x²y²  dx + x/1+x²y²dy = ydx+xdy/1+x²y²

    When x =1

               Y = -1

df =

                       =   =   = 

(a) Given region of integration

         0

it is a circle of radius 1

we will introduce a vertical strip which will vary from

order to change the order into integration

= 2

(b)   =

                 = 

                 = 

                 = 

                = 

                = 

                = 

(a) f(x,y,z) = x²+y²+z² = 9

=2x

 = 2y

 = 2z

                  Vector =

     A point (2,-1,2)

= 

=

 Unit vector normal to it =      

(b) u= x²+y²+z²

(by identity)

)

         =

                [ ]

(a)    where

              y =

   and  

(b)   

Equation of given plane x/2+y/z+z/4=1

Let R be orthogonal projection of x+y +z/2=1in XoY plane i.e. z=0 given by x+y=1

Therefore R=0{(x.y):0≤ y≤ 1-x;0≤ x≤ 1}

=x +yx

=(x )

ds =

                                 [Put x=2,y=2,z=neglected ]

Surface contains six faces

 (i) OABC faces Z = 0,

 .

(ii) On face DEFG Z = 1,

(iii) On face OAFG y = 0,

(iv) On face DEBC y = 0,

(v) On face ABEF x = 1,

(vi) On face OCDG x = 0,

Gauss Divergence theorem

If   be a vector point function having continuous first partial derivative in reason v bounded by surface (s).

Then