In calculus, the integral is a mathematical concept that is essential for evaluating the total value of a given function in the 2-d region. It is a further sub-branch of a single integral that is helpful for calculating the area under the curve in 1-d.
The double integral covers a 2-D region in the xy plane that defines the sum of infinitely small rectangles to find the volume under the surface of a region. It is used in a wide range of real-world applications, including physics, engineering, economics, and computer science.
In this article, we’ll explain the sub-branch of calculus with the help of formulas and solved examples.
In mathematics, the double integral is a fundamental concept of calculus that deals with the two-dim regional functions f(x, y). It is the advanced form of single integral that is used to find the volume under the curve in an XY-plane.
A single integral of a function “f(x)” is a 1-d regional concept that evaluates the area under the curve while the double integral of a function “f(x, y)” is a 2-d regional that calculates the volume under the curve.
The double integral of a function f(x,y) over a region R in the xy-plane is given by:
∫∫R f(x, y) dA
The limits of integration in double integrals determine the boundaries of the region over which the function is being integrated. These boundaries can be defined by specifying the values of x and y that define the edges of the region. Such as if the boundary values of x are (x1, x2) and the boundary values of y are (y1, y2) then,
y1∫y2x1∫x2 f(x, y) dx dy
Change of variables in double integrals is another important concept that allows us to simplify integrals by changing the variables to a new set of coordinates. This is useful in cases where the integral is difficult to evaluate in its original form. Such as:
x1∫x2y1∫y2 f(x, y) dy dx
Here are some properties of a double integral.
| Properties name | Statement | Expression |
| Linearity of Double Integrals | The linearity of double integrals states that if we have two functions f(x, y) and g(x, y), and two constants a and b, then we can add or subtract the integrals of these functions, multiplied by the constants a and b
This property is helpful while dealing with the sum and difference of two functions. |
∫∫R [a f(x, y) ± b g(x, y)] dA
= a ∫∫R f(x, y) dA ± b ∫∫R g(x, y) dA
|
| Fubini’s Theorem for Double Integrals | Fubini’s theorem is a helpful property of double integrals that is used to change the order of integration. The theorem states that if the function f(x,y) is continuous over a rectangle R = [x1, x2] × [y1, y2], then we can evaluate the double integral in any order. | ∫∫R f(x, y) dA = y1∫y2x1∫x2 f(x, y) dx dy = x1∫x2y1∫y2 f(x, y) dy dx |
| Change of Variables in Double Integrals | The change of variables in double integrals is a well-known method that is used to simplify the integrands by changing the variables to a new set of coordinates. This method is useful in cases where the integral is difficult to evaluate in its original form. | x = u(x,y)
y = v(x,y) then, ∫∫R f(x, y) dA = ∫∫S f(u, v) |h(u, v)| Da Where, |h(u, v)| is the Jacobian determinant of the transformation |
Follow the below example to understand the calculations of double integral and how to perform them.
Example
Find the double integral of the given function w.r.t u & v if the boundary values of u are (2, 3) and v are (1, 2)
f(u, v) = 3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv
Solution
Step 1: First, take the 2-dimensional function and write it according to the general form of double integral and then apply the variables and boundary values
f(u, v) = 3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv
∫∫R f(u, v) dA = ∫∫R [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] Da
∫∫R [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] dA = 1∫22∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du dv
Step 2: First of all, integrate the function with respect to “u”.
∫∫R [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] dA = 1∫2 {2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du} dv … (1)
Step 3: Now the function of u and apply the sum and difference property of the double integral and take out the constant coefficients.
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 2∫3 [3uv2] du + 2∫3 [4u2] du – 2∫3 [3v4] du + 2∫3 [6u2v2] du – 2∫3 [12uv] du
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2 2∫3 [u] du + 42∫3 [u2] du – 3v4 2∫3 [1] du + 6v2 2∫3 [u2] du – 12v 2∫3 [u] du
Step 4: Now integrate the function with the help of the power property of integral.
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2 [u1+1 / 1 + 1]32 + 4 [u2+1 / 2 + 1]32 – 3v4 [u]32 + 6v2 [u2+1 / 2 + 1]32 – 12v [u1+1 / 1 + 1]32
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2 [u2 / 2]32 + 4 [u3 / 3]32 – 3v4 [u]32 + 6v2 [u3 / 3]32 – 12v [u2
/ 2]322∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2/2 [u2]32 + 4/3 [u3]32 – 3v4 [u]32 + 6v2/3 [u3 ]32 – 12v/2 [u2]32
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2/2 [u2]32 + 4/3 [u3]32 – 3v4 [u]32 + 6v2/3 [u3 ]32 – 12v/2 [u2]32
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2/2 [u2]32 + 4/3 [u3]32 – 3v4 [u]32 + 2v2 [u3 ]32 – 6v [u2]32
Step 5: Now apply the limit values.
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2/2 [32 – 22] + 4/3 [33 – 23] – 3v4 [3 – 2] + 2v2 [33 – 23] – 6v [32 – 22]
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2/2 [9 – 4] + 4/3 [27 – 8] – 3v4 [3 – 2] + 2v2 [27 – 8] – 6v [9 – 4]
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 3v2/2 [5] + 4/3 [19] – 3v4 [1] + 2v2 [19] – 6v [5]
2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du = 15v2/2 + 76/3 – 3v4 + 38v2 – 30v
Step 6: Now place the above value in (1).
∫∫R [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] dA = 1∫2 {2∫3 [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] du} dv
∫∫R [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] dA = 1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv … (2)
Step 6: Now integrate the function with respect to “v”.
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 1∫2 [15v2/2] dv + 1∫2 [76/3] dv – 1∫2 [3v4] dv + 1∫2 [38v2] dv – 1∫2 [30v] dv
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 15/21∫2 [v2] dv + 76/31∫2 [1] dv – 31∫2 [v4] dv + 381∫2 [v2] dv – 301∫2 [v] dv
Integrate the above expression
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 15/2 [v2+1 / 2 + 1]21 + 76/3 [v]21 – 3 [v4+1 / 4 + 1]21 + 38 [v2+1 / 2 + 1]21 – 30 [v1+1 / 1 + 1]21
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 15/2 [v3 / 3]21 + 76/3 [v]21 – 3 [v5 / 5]21 + 38 [v3 / 3]21 – 30 [v2 / 2]21
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 15/6 [v3]21 + 76/3 [v]21 – 3/5 [v5]21 + 38/3 [v3]21 – 30/2 [v2]21
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 15/6 [v3]21 + 76/3 [v]21 – 3/5 [v5]21 + 38/3 [v3]21 – 15 [v2]21
Place the boundary values
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 15/6 [23 – 13] + 76/3 [2 – 1] – 3/5 [25 – 15] + 38/3 [23 – 13] – 15 [22 – 12]
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 15/6 [8 – 1] + 76/3 [2 – 1] – 3/5 [32 – 1] + 38/3 [8 – 1] – 15 [4 – 1]
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 15/6 [7] + 76/3 [1] – 3/5 [31] + 38/3 [7] – 15 [3]
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 105/6 + 76/3 – 93/5 + 266/3 – 45
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 17.5 + 25.33 – 18.6 + 88.67 – 45
1∫2 [15v2/2 + 76/3 – 3v4 + 38v2 – 30v] dv = 67.9
Step 7: Put the above result in 2.
∫∫R [3uv2 + 4u2 – 3v4 + 6u2v2 – 12uv] dA = 67.9
A double integral calculator can be used to evaluate the above example to avoid such lengthy steps of calculating the volume under the curve.
Now you can grab the definition, formulas, properties, and calculations of the double integral from this post. Double integration is a fundamental mathematics concept that is essential to calculate the volume of a 2-d region.
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