Unit 5 | unit 5 multiple integrals

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Unit - 5

Multiple Integrals

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 =

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

y	0	a
t	0	1

 I = xdx

I = xa dx 

I =  x (1 – x) dx =  (x– x4/3) dx

I =

I = =

 xy (1 – x –y) dx dy =

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 =

= 

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 =

5. 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

6. Evaluate  over x  1, y 

Soln. :

1a1 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 =

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.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

8. 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  Fig. 6.8

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 =

9. 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 =

10. 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