Evaluation of Double integrals in a region

EVALUATION OF DOUBLE INTEGRALS IN A REGION

Example 17. Evaluate the double integral ∫∫_R (x‒3y²)dA where

R = {(x, y)|0 ≤ x ≤ 2, 1 ≤ y ≤ 2}.

Solution:

If we first integrate with respect to x, we get

 = [2y ‒ 2y³]₁² = ‒12

Example 18. Evaluate ∫∫_R y sin(xy) dA, where R= [1,2] × [0, π].

Solution:

If we first integrate with respect to x, we get

Example 19. If R = [0,π/2]×[0,π/2], then, find ∫∫_R sinx cosy dA

Solution:

= 1.1 = 1

Example 20. Evaluate ∫∫_R xy dydx, where R is the domain bounded by x‒axis, ordinate x = 2a and the curve x² = 4ay.

Solution:

To plot the region:

x² = 4ay       ... (1) is a parabola symmetrical about y‒axis & passing through the origin

x = 2a          …….. (2) is a line perpendicular to x‒axis

To find the point of intersection:

Substitute (1) in (2) we get

(2a)² = 4ay

 4a² = 4ay

⇒  y = a

The point of intersection is (2a, a)

The common region is OABO (shaded region)

To evaluate the integral: I = ∫∫_R xy dydx

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

 I = a⁴ / 3

Example 21. Evaluate ∫∫_R xy dxdy, where R is the positive quadrant of the circle x² + y² = a².

Solution:

To plot the region:

x²+y² = a² is circle with center at the origin.

The region of integration is the region of the circle lies in the first quadrant

(given in the question)

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

Example 22. Find the limits of the integration ∫∫_R f(x, y) dxdy where R is the region bounded by the lines x = 0, y = 0 and x + y = 2.

To plot the region:

 x = 0 is y‒axis, y = 0 is x‒axis

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

The region of integration in the region is the shaded region OABO.

To evaluate the integral:

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

Example 23. Calculate ∫∫_R sinx / x dA where R is the triangle in the xy‒plane bounded by the x‒axis, the line y = x, and the line x = 1.

Solutionu

To plot the region:

The region of integration is bounded by y = x, x = 1, y= 0. (x‒axis)

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

x = 1, a line perpendicular to x‒axis.

y= 0 is x‒axis.

The region of integration is OABO (shaded region).

To evaluate the integral:

we integrate first with respect to y and then with respect to x

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

Example 24. Calculate ∫∫ r² drdθ over the area included between the circles r = 2sinθ and r = 4sinθ.

Solution:

Given: r = 2 sin θ            …………(1)

is a circle with diameter 2 passing through the origin symmetric about θ = π/2.

 r = 4 sin θ                   ………….(2)

is a circle with diameter 4 passing through the origin symmetric about θ = π/2.

The shaded area between these circles is the region of integration.

To evaluate the integral: I = ∫∫ r² dr dθ

 I = 2 × Region of integration in the first quadrant.

Example 25. Evaluate ∫∫_R r sinθ dr dθ over the cardioid r = a(1‒cos θ) above the initial line.

Solution:

To plot the region:

 r = a(1 − cos θ) is a cardioid passes through the origin.

To evaluate the integral I = ∫∫_R r sinθ dr dθ

Order = dr dθ

At 0, r = 0:  At P,r = a(1 cos 0)

EXERCISE

16. Evaluate ∫∫_R xy dxdy, where R is the positive quadrant of the circle x² + y² = a².

Ans: a⁴/3

17. Evaluate ∫∫_R e^2x+3y dxdy over the triangle bounded by x = 0, y = 0 and x + y = 1.

Ans: 1/6 (2e³ − 3e² + 1)

18. Evaluate ∫∫_R y dydx, where R is the region in the first quadrant bounded by the ellipse x²/a² + y²/b² = 1.

Ans: ab²/3

19. Find ∫∫ dx dy over the region bounded by x ≥ 0, y ≥ 0, x + y ≤ 1.

Ans: 1/2

20. Find the value of the double integral ∫∫ xy dxdy taken over the positive quadrature of the ellipse x²/a² + y²/b² = 1.

Ans: a²b² / 8

21. Evaluate ∫∫_R xy dx dy, where R is the region bounded by the line x + 2y = 2, lying in the first quadrant.

Ans: 1/6

22. Evaluate ∫∫(x² + y²)dx dy over the area bounded by the curves y = 4x, x + y = 3, y = 0, y = 2.

Ans: 463/48

23. Evaluate ∫∫ r² sinθ dr dθ over the cardioids r = a(1 + cosθ) above the initial line.

Ans: 4/3 a²

24. Calculate ∫∫ r³ dr dθ over the area included between the circles r = 2 sinθ and r = 4 sinθ. Ans:45π/2.