Change of order of Integration

 

CHANGE OF ORDER OF INTEGRATION

 

Change of order of Integration:

1. Problems under Type I

2. Problems under Type II

3. Problems under Type III

4. Problems with two sub regions

 

By changing the order of integration a double integral may be sometimes conveniently evaluated. If the limits of double integral are constants, we can change the order of integration however the limits of integration will not undergo any change. But when the limits are variables, the change in the limits also change like when the order of integration is changed to fix up the new limits, it is always advisable to draw a rough sketch of the region of integration.

 

TYPE ‒ I

 will take the form 

If all the limits of the integral are constant, change of order is done just by interchanging the limits

TYPE ‒ II

 will take the form 

Since the order of integration is dydx, draw a vertical strip PQ.

TYPE ‒ III

 will take the form 

Since the order of integration is dxdy, draw a horizontal strip PQ.

 

Hint: If the inner limit has x term, draw horizontal strip. If the inner limit has y term, draw vertical strip.

 

 

PROBLEMS UNDER TYPE – I

 

Example 26. Change the order of integration for ₀∫¹₀∫² f(x, y)dx dy.

Solution:

Since the limits of the given integral are constant, change the order is nothing but the interchanging the limits for the corresponding variables. Changing the order is

 I = ₀∫²₀∫¹ f(x, y)dx dy

 

Example 27. Change the order of integration for ₀∫²₁∫³ f(x, y)dy dx.

Solution:

Since the limits of the given integral are constant, change the order is nothing but the interchanging the limits for the corresponding variables. Changing the order is

 I = ₀∫³₀∫² f(x, y)dy dx

 

PROBLEMS UNDER TYPE ‒ II

 

Example 28. Change the order of integration in I = ₀∫^a_y∫^a x/(x²+y²) dxdy and evaluate.

Solution:

To plot the region:

The region of integration is bounded by x = y,

x = a, y = 0, y = a.

x=y, a line passing through the origin intersect XOY plane.

x = a, a line perpendicular to x‒axis.

y = 0 is x‒axis; y = a, a line perpendicular to y‒axis.

The region of integration is OABO (shaded region).

To evaluate the integral:

After changing the order, first integrate with respect to y, then x.

Since the order of integration is dydx, draw a vertical strip PQ.

 

Example 29. Evaluate by changing the order of the integral ₀∫^a_y∫^a x²/√(x²+y²) dxdy.

Solution:

To plot the region

The region of integration is bounded by x = y,

x = a, y = 0, y = a

x = y, a line passing through the origin intersect XOY plane

x = a, a line perpendicular to x‒axis

y = 0 is x‒axis; y = a, a line perpendicular to y‒axis

The region of integration is OABO (shaded region)

To evaluate the integral:

After changing the order, first integrate with respect to y then x

Since the order of integration is dydx, draw a vertical strip PQ

 

Example 30. Change the order of integration   y dxdy and hence evaluate it

Solution:

To plot the region:

The region of integration is bounded by

y=0, y = a, x = a‒y

⇒ x + y = a

 x = √[a²‒y²]

i.e., x² = a² – y²

 x² + y² = a²

y = 0 is x‒axis; y = a is a straight line perpendicular to y‒axis

x + y = a, a line intersect the XOY plane at (a, 0) and (0, a)

 x²+y² = a² is a circle with centre at the origin and radius is a

The region of integration is ABA (shaded region)

To evaluate the integral:

After changing the order, first integrate with respect to y then x

Since the order of integration is dydx, draw vertical strip PQ

At P, y = a ‒ x;  At Q, y = √[a² ‒ x²]

At B, x = 0;  At A, x = a

 

Example 31. Evaluate ∫∫ (x²y + xy²)dxdy over the area between y = x² and y = x.

Solution:

 I = ∫∫ (x²y + xy²) dxdy

This problem is easily evaluated by changing the order of integration

y=x is a straight line passing through the origin.

y= x² is a parabola symmetrical about y axis.

Common region is OAO.

To find the intersecting point A:

We have y = x        ... (1) and y = x²            ... (2)

From (1) and (2)

x² = x

⇒ x²‒x = 0

⇒ x(x‒1)= 0

⇒x=0 or x=1

When x = 1, (1)

⇒ y=1

The point A is (1,1)

To evaluate the integral:

After changing the order, first integrate with respect to y then x

Since the order of integration is dydx, draw vertical strip PQ

At P, y = x²; At Q, y = x

At 0, x = 0; At A, x = 1

 

Example 32. Evaluate ∫∫_D (x+2y) dA, where D is the region bounded by the parabolas y = 2x² and y = 1 + x².

Solution:

 y = 2x² ... (1) is a parabola symmetrical about y‒axis

 y = 1 + x² ... (2) is a parabola symmetrical about y‒axis and the vertex at (0,1).

To find the intersecting points A and B:

From (1) and (2)

2x² = 1 + x², that is, x² = 1 and so x = ±1.

To evaluate the integral:

Here dA = dydx is more convenient.

After changing the order, first integrate with respect to y then x

Since the order of integration is dydx, draw vertical strip PQ

At P, y = 2x²; At Q, y=1+x²

At A, x = ‒1; At B, x = 1

 

PROBLEMS UNDER TYPE ‒ III

 

Example 33. Change the order of integration for ₀∫^a₀∫^x f(x, y) dy dx.

Solution:

To plot the region:

y = 0, is x‒axis

y = x, passing through the origin and bisect the XOY plane

x = 0 is y‒axis; x= a, a line perpendicular to x‒axis

The region of integration is OABO (shaded region)

To change the order:

After changing the order, first integrate with respect to x, then y

Since the order of integration is dxdy, draw a horizontal strip PQ

At P, x = y;

At Q, x = a

At O, y = 0;

At B, y = a

Change of order is I = ₀∫^a_y∫^a f(x, y)dx dy

 

Example 34. By changing the order of integration, evaluate ₀∫^a_x∫^a (x² + y²)dydx

Solution:

To plot the region: The region of integration is bounded by

y = x, y = a, x = 0, x = a.

y = x, a line passing through the origin and intersect XOY plane

y = a, a line perpendicular to y‒axis

x = 0 is y‒axis; x = a, a line perpendicular to x‒axis

The region of integration is OABO (shaded region)

To evaluate the integral:

After changing the order, first integrate with respect to x, then y

Since the order of integration is dxdy, draw a horizontal strip PQ

At P, x = 0;

At Q, x = y;

At 0, y = 0;

At B, y = a;

 

Example 35. Change the order of integration and evaluate ₀∫^∞_x∫^∞ e^‒y/y dydx.

Solution:

To plot the region:

The region of integration is bounded by

y = x, y = ∞, x = 0, x = ∞

y = x, a line passing through the origin intersect XOY plane

y= ∞, a straight line perpendicular to y‒axis at infinity

To evaluate the integral:

After changing the order, first integrate with respect to x, then y

Since the order of integration is dxdy, draw a horizontal strip PQ

At P, x=0;

At Q, x = y

At O, y = 0;

At B, y=∞

 

Example 36. Change the order of integration in ₀∫^∞₀∫^y ye^‒y2/x dxdy and then evaluate it.

Solution:

To plot the region:

The region of integration is bounded by y = 0

 (x‒axis); y = x, x = 0 x = ∞

 y = x, a line passing through the origin intersect XOY plane

 x = ∞, a straight line perpendicular to x‒axis at infinity

To evaluate the integral:

After changing the order, first integrate with respect to x, then y

Since the order of integration is dxdy, draw a horizontal strip PQ

At P, x = y;

At Q, x=∞;

At 0, y = 0;

At B, y= ∞

 

Example 37. Change the order of integration in  and hence evaluate it.

Solution:

To plot the region:

The region of integration is bounded by

x = 0, x = 4a

x= x²/4a

⇒ x² = 4ay          ... (1)

y = 2√(ax) = y² = 4ax         ... (2)

x = 0 is y‒axis; x = 4a is a straight line perpendicular to y‒axis

x² = 4ay is a parabola symmetrical about y‒axis with vertex at the origin

y² = 4ax is a parabola symmetrical about x‒axis with vertex at the origin

To find the point of intersection:

From (2), x = y²/4a     ………(3)

Substitute (3) in (1)

(y² / 4a)² = 4ay

y⁴ / 16a² = 4ay

y⁴ = 64a³y

y⁴‒64a³y = 0

y(y³‒64a³)=0

y = 0, y³ – 64a³ = 0

y³ = 64a³

⇒ y = 4a

y= 0, y = 4a. The points of intersection are (0,0), (4a, 4a).

To evaluate the integral:

Change of order of integration is dxdy

After changing the order, first integrate with respect to x, then y

Since the order of integration is dxdy, draw a horizontal strip PQ.

 

Example 38. By changing the order of integration evaluate .

Solution:

To plot the region:

The region of integration is bounded by

x = 0, x = a, y = √(ax)

⇒ y² = ax, y = a

x = 0 is y‒axis; x = a is a line perpendicular to x‒axis

y² = ax, a parabola symmetrical about x‒axis with vertex at origin

y = a, a line perpendicular to y‒axis

The region of integration is OABO (shaded region)

To evaluate the integral:

After changing the order, first integrate with respect to x, then y

Since the order of integration is dxdy, draw horizontal strip PQ

At P, x=0

At 0, y = 0

At Q, x = y²/a

At B, y = a

 

Example 39. Evaluate the iterated integral ₀∫¹_x∫¹ sin(y²) dydx.

Solution:

If we try to evaluate the integral as it stands, we are faced with the task of first evaluating ∫sin(y²)dy. But it's impossible to do so in finite terms since is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral.

To plot the region: The region of integration is bounded by

y=x, y= 1, x = 0, x = 1.

y=x, a line passing through the origin and intersect XOY plane

y= 1, a line perpendicular to y‒axis

x = 0 is y‒axis; x = 1 a line perpendicular to x‒axis

The region of integration is OABO (shaded region)

To evaluate the integral:

After changing the order, first integrate with respect to x, then y

Since the order of integration is dxdy, draw a horizontal strip PQ

At P, x = 0;

At Q, x=y

At 0, y = 0;

At B, y = 1

 

PROBLEMS WITH TWO SUB‒REGIONS AFTER CHANGING THE ORDER OF INTEGRATION

 

If there are two slopes for the strip, the region is splited in to two parts R₁ and R₂

 

Example 40. Change the order of integration ₀∫¹_x2∫^2‒x xy dydx and hence evaluate.

Solution: To plot the region:

The region of integration is bounded by

y = x², y = 2 − x ⇒ x + y = 2,

x = 0, x = 1

y = x² is a parabola symmetrical about y‒axis.

x+y= 2 straight line intersect the coordinate axes at (2,0) and (0,2).

x = 0 is y‒axis, x = 1, a line perpendicular to x‒axis.

The region of integration is OCBAO (shaded region).

To evaluate the integral:

After changing the order, first integrate with respect to x, then y

Since there are two slopes for horizontal strip, the region is spitted into two parts R₁

& R₂

Draw horizontal strips P₁Q₁, P₂Q₂

 

Example 41. Change the order of integration I = ₀∫^a_x2/a∫^2a‒x xy dxdy and hence evaluate it.

Solution:

Since inner limit is a function of x, first integrate with respect to y then x

I = ₀∫^a_x2/a∫^2a‒x xy dxdy

To plot the region;

The region of integration is bounded by

 y = x²/a, y = 2a ‒ x

x+y=2a, x = 0, x = a

 y = x² is a parabola symmetrical about y‒axis. x + y = 2a a straight line intersect the coordinate axes at (2a, 0) and (0,2a). x = 0 is y‒axis, x = a, a line perpendicular to x‒axis.

The region of integration is OCBAO (shaded region).

To evaluate the integral:

After changing the order, first integrate with respect to x, then y

Since there are two slopes for horizontal strip, the region is spitted into two parts R₁ & R₂

Draw horizontal strips P₁Q₁, P₂Q₂.

 

 

EXERCISE

 

25. Evaluate the following integral by changing the order of integration