Double Integral: Definition, Rules, Properties & Applications

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Jasmine Grover

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Double integrals are primarily used to integrate the surface area of a 2-dimensional figure like circle, square, triangle, pentagon, rectangle and quadrilateral. The mathematical symbol for double integral is ‘∫∫’ and simple integration forms the basis of doing double integrals. It forms the vital ingredient of calculus wherein integral calculus is utilized to determine the volume and area of a 2-dimensional structure on a very large scale via employing formulas and computational analysis. 

Key Takeaways: Integrals, Double Integral, Single Integrals, Differentiation, Definite Integrals, Indefinite Integrals, Polar Coordinates, Integration, Calculus

Read Also: Methods of Integration


What are Integrals

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Integrals are used to assign numbers to the functions in mathematics in such a way that it easily describes area, volume, displacement and other concepts which arise as a result of combination of a huge number of infinitesimal data points. The process used to find the integrals is referred to as integration. In addition to differentiation, integration also forms an essential component of calculus and aids in serving as a tool used to solve various problems in physics and mathematics involving arbitrary shapes, volume of solids, length of a particular curve and so on. 

The two types of integrals are definite and indefinite integrals. In the conventional sense, the areas which are above the horizontal axis of the plane in the graph are positive while the areas which are below the horizontal line are negative.

Definite Integral

Definite Integral

Moreover, the concept of integrals may be generalized depending upon the different types of functions as well as the domain over which the particular integration is performed. For instance, in a surface integral, the curve is generally replaced by a piece of the surface in a 3-dimensional space. Furthermore, a line integral is for functions of two or more than two variables and the interval of the integration is further replaced by a specific curve which ultimately connects the two endpoints of the interval. 

Integrals Detailed Video Explanation

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Definition of Double Integral

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Double integral is the integration of a function using two variables over a defined region in R2 , that is, the real number plane or the region of integration in the xy plane. The double integral of a function containing two variable, for instance, f(x,y) over a particular rectangular region can be denoted as; 

\(\int \)Rf(x,y) dA=\(\int \)Rf(x,y) dx dy

For instance, if the definite integral abf(x)dx of a function of one of the variables where , is the area under a curve f(x) from x=a to x=b, then the double integral is equivalent to the volume under the surface z = f(x,y) and above the xy-plane in the region of the integration R.

Figure depicting the function of the double integral in the xy plane in the region D

Figure depicting the function of the double integral in the xy plane in the region D


Rules of Double Integration

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In order to perform integrations in calculus, a set rules are to be followed. One of the rules is integration by parts which is used to solve integration problems. Other rules are integration by substitution, or using different formulas. 

Particularly for double integration, one of the key rules followed is double integration by parts, which is given by the formula;

∫∫u dv/dx dx.dy = ∫[uv -∫v du/dx dx]dy


Properties of Double Integrals

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The properties of double integrals are as follows:

  1. ∫x=ab ∫y=cd f(x,y)dy.dx = ∫y=cd∫x=ab f(x,y)dx.dy
  2. ∫∫(f(x,y) ± g(x,y)) dA = ∫∫f(x,y)dA ± ∫∫g(x,y)dA
  3. If f(x,y) < g(x,y), then ∫∫f(x,y)dA < ∫∫g(x,y)dA
  4. k ∫∫f(x,y).dA = ∫∫k.f(x,y).dA
  5. ∫∫R∪Sf(x,y).dA = ∫∫Rf(x,y).dA+∫∫sf(x,y).dA

Certain properties of the double integral are parallel to those of single integrals. These are:

  1. For f and g being continuous in the region D along with C as a rational number:

 ∫∫D(f + g) dA = ∫∫D f dA + ∫∫D g dA
∫∫D cf dA = c ∫∫D f dA 

  1. For f being continuous in the said region D, where D = D1 ∪ D2, where D1 and D2 are actually non-overlapping regions whose union is D:
    ∫∫D f dA = ∫∫D1 f dA + ∫∫D2 f dA

Read Also: Increasing and Decreasing Functions in Calculus


Applications of Double Integrals

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Double integrals demonstrate applications in a wide number of areas of science and engineering, including computations of the following areas:


Double Integral Area

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Let z = f(x, y) be defined over a domain D in the xy plane, and there is a requirement to find the double integral of z. If the required region is divided into vertical stripes and the endpoints for x and y are carefully found out, that is, the limits of the region, in that case the double integral formula can be used is;

Double Integral Area

Double Integral Area

Additionally, if on dividing the required region into horizontal stripes and carefully finding the endpoints for x and y, that is, the limits of the region, then in that case, the formula which can be used is;

formula 1

Formula-1

Furthermore, if the function A is a continuous function, then the formula can be written as;

formula 2

Continuous function

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Double Integral in Polar Coordinates

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The double integral \(\iint \)Rf(x,y)dA\(\iint \)Rf(x,y) dA n rectangular coordinates can be converted to a double integral in polar coordinates as follows:

Double Integral in Polar Coordinates formula 1

Double Integral in Polar Coordinates formula-1

This can also be re-written as: 

Double Integral in Polar Coordinates formula 2

Double Integral in Polar Coordinates formula-2

where, f(r cosθ, r sinθ) = f(r, θ)

In this type of double integral, there is a requirement to first integrate f(r,θ) with respect to r between the limits r = r1 and r = r2 treating θ as a constant. The resulting expression is then further integrated with respect to θ from θ1 to θ2. Here r1 and r2 may be constants or functions of θ. In this case, first, there is integration of f(r,θ) with respect to θ between the limits θ = θ1 and θ = θ2 and treating r as a constant and the resulting expression are integrated with respect to r and that time the function of θ will be constant.


Difference Between Simple and Double Integral

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The key differences between simple and double integrals are tabulated below.

Simple Integral Double Integral
Simple integration aids in determining the area under the curve. Double integration aids in determining the volume under the said surface.
simple integration of a positive function of a single variable is representative of the area of the specified region between the x-axis and the function. Double integral of two distinct variables constitute the volume of the said region which is present between the surface to be defined by the function as well as the plain which contains the domain.

Things to Remember

  1. Integration is used in calculus to determine numerous useful quantities such as areas, volumes, displacement and so on.
  2. Double integrals are used to calculate the area of a region, volume under the surface and average value of the function of two variables over a 2-dimensional structure such as a rectangle.
  3. The double integral of a function of two variables f(x, y) over a rectangular region can be denoted as ∫∫Rf(x,y) dA=∫∫Rf(x,y) dx dy
  4. In the case of double integration the rule followed is double integration by parts, which is given by the equation ∫∫u dv/dx dx.dy = ∫[uv -∫v du/dx dx]dy.

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Sample Questions

Ques. Solve the following function \(\iint\)x.logx.dx.dy (5 marks)

Ans. If I = \(\iint\)x.logx.dx.dy

Taking the inner integration on the function ∫x.logxdx and solving this integral by using integration by parts, we get,

∫ x.logx dx = ∫ (logx)x dx

The function x cannot be integrated with the logarithmic function directly. Hence, we can assume it as u= logx and dv= xdx 

du = (1/x) dx.v = x2/2

So, ∫x.logx.dx = (logx)x2/2 − ∫(x2/2)(1/x)dx

= x2/2(logx)−1/2 ∫x.dx

Hence, ∫x.logx.dx = x2/2(logx) − 1/4x2

On operating the outer integral function by the above result, we get:

∫∫ x.logx.dxdy = ∫[x2/2(logx) − 1/4x2]dy

I = ∫[x2/2(logx)]dy − ∫[1/4x2]dy

I = (x2y/2) (log x) – 1/4x2y.dy + c

Ques.  Find the double integral (x+y)dx dy, i.e. \(\iint\)(x+y)dx dy  (2 marks)

Ans. I = \(\iint\)(x+y) dx dy

I = ∫[∫(x+y)dx]dy

I = ∫[x2/2+yx]dy

I = x2y/2+xy2/2

I = (xy/2)(x+y) + C

Ques. Evaluate the double integral (x2+y2)dx dy Or \(\iint\)(x2+y2)dx dy (2 marks)

Ans. Let us say, I = \(\iint\)(x2+y2)dx dy

I = ∫[∫(x2+y2)dx]dy

I = ∫[x3/3+y2x]dy

I = x3y/3+xy3/3

I = [xy(x2+y2)]/3 + C

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