UNIT 4

MULTIVARIABLE CALCULUS (INTEGRATION)

Question Bank

Question-1: 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

Questiom-2: Evaluate the following double integral,

Sol.  Let,

                           I = 

On solving the integral, we get

Question-3: 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=

Question-4: 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=

                                     =          

=          

                                     =          

                                     = 

                                    =

=

Question-5: 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   = 

Question-6: 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.

Fig.

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 =

Question-7: 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

And x = 0, x = a  is as shown in Fig.

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

Question-8: 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] =

Question-9: 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. 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

=sin  [a2 (1 – cos )2]                                                 Fig.

                                      =   

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

=     2 (sin   – sin2  + sin ) d

=+

I= a2= a2

I=

Queestion-10: Find the Area common to the two circle 

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

                   Area =

Question-11: Evaluate

Solution: Let

     I     =    

            =    

                                            (Assuming m = )

             = dxdy

           =

             =

           = dx

           =  dx

           = 

           =

         I =

Question-12: 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