Double Integral

A double integral is a mathematical tool for computing the integral of a function of two variables across a two-dimensional region on the xy plane. It expands the concept of a single integral by integrating the functions of two variables over regions, surfaces, or areas in the plane. In case two variables are present, we need to substitute the value of one variable in terms of the other. This technique becomes very difficult when we deal with multiple variables to calculate the areas and volumes under the curves. A double integral is very useful in such cases. In this article, we will learn about double integrals in detail.

Double integration in mathematics uses integration with respect to two variables. We do not need to convert the complete equation into one variable for double integral. Instead, we can integrate the function with respect to two variables also. This is very helpful in the case of functions where we are provided with only one function and no relationship between the variables is defined. In such cases, we cannot substitute the value of one variable from the relation. Thus, we use double integral to integrate the function. Double integral is mainly used to calculate the area of 2D surfaces or curves in mathematics.

Double Integral Definition

Double integral is defined as the integral of a function over an area represented as R².

A double integral is represented as follows:

[Tex]\int_R\int_R f(x,y) dx.dy [/Tex]

If the region R spans over an area of [a, b] and then the above equation can be written as:

[Tex]\bold{\int^a_b\int^c_d f(x,y)~dy~dx} [/Tex]

Here f(x, y) is a function of x and y, and the two integration signs represent the double integral over areas [a, b] and .

In the case of double integrals the inner integral is solved first and then we proceed to solve the outer integral. The calculation of the integral proceeds in the normal way by treating one variable at a time. In this case, the equation is integrated w.r.t x first, and then the equation is integrated w.r.t y.

Steps to calculate the double integral are as follows:

Step 1: Write down the function to be integrated with a double integral sign and mention the upper and lower limits of integration on the integral.

Step 2: Integrate the function with respect to any one of the variables initially.

Step 3: Now insert the lower and upper limits of the variable with respect to which we integrated the function in order to bring the left function into one variable.

Step 4: Proceed similarly as above, integrate the function again with respect to the other variable.

Step 5: Insert the lower and upper limits of the second variable to get the result.

Example: Consider the integration [Tex]\bold{\int^2_1\int^2_1(x+y)dy~dx} [/Tex].

Solution:

In order to solve this integral, we will solve the inner integral first and integrate the function w.r.t x

Let [Tex]I = \int^2_1\int^2_1(x+y)dy~dx   [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2}{2}+xy]^{x=2}_{x=1} [/Tex]

Putting the upper and lower limits for x, we get

⇒ I [Tex]= \int^2_1[\frac{2^2}{2}+2y-(\frac{1^2}{2}+y)]dy [/Tex]

⇒ I [Tex]= \int^2_1[2+2y-\frac{1}{2}-y]dy [/Tex]

⇒ I [Tex]= \int^2_1(\frac{3}{2}+y)dy [/Tex]

Now integrating w.r.t y, we get

⇒ I [Tex]= [\frac{3y}{2}+\frac{y^2}{2}]^{y=2}_{y=1} [/Tex]

⇒ I [Tex]= [\frac{3(2)}{2}+\frac{2^2}{2}-(\frac{3(1)}{2}+\frac{1^2}{2})] [/Tex]

⇒ I = 3 + 2 – 2 = 3

Volume using Double Integral is the geometric interpretation of the double integral, to calculate the volume using double integral, let’s consider a region R over [a × b] and . A curve S = f(x, y) is drawn such that it projects an area in this region R. The graph for this is shown below:

In order to calculate the volume of the shown region, we will divide the length cd of the area R into m equal parts and the breadth of the area R into n equal parts. Thus the complete area R gets divided into smaller rectangles. Now from each of these rectangles we will draw a box upwards to the point where it meets S. It can be shown as follows:

Consider that the area of the base of each small rectangle is A and its height is given by f(x’, y’). The volume under region S can be calculated using:

[Tex]\bold{V = \Sigma_{i=1}^n \Sigma_{j=1}^n A.f(x’,y’)} [/Tex]

We have made use of two variables m and n because the volume will be added and calculated in both directions or axes i.e. X axis and Y axis. As the value of m and n becomes very large and approaches infinity which means the area R is divided into infinite small rectangular regions, then this equation can be written as:

[Tex]V=\lim_{m,n \rarr\infty}\sum_{i=1}^m\sum_{j=1}^nA.f(x’,y’) [/Tex]

We know that when the variables tend to infinity then summation can be substituted with integration which is also the definition of integration. Thus above equation becomes equal to:

[Tex]V= \int\int_Rf(x,y)dA [/Tex]

Thus we derive the formula to find the volume under the curve using a double integral.

Consider two functions f(x, y) and g(x, y) to be integrated over regions A and B respectively. Also, consider C and D to be sub-regions of A and B, then a double integral satisfies the following properties:

• Linearity: [Tex]\int \int_R[f(x,y)\pm g(x,y)] = \int \int_Rf(x,y)\pm\int \int_Rg(x,y) [/Tex]
• [Tex]\int^b_a\int^d_cf(x,y)dy~dx = \int^d_c\int^b_af(x,y)dy~dx [/Tex]
• Additivity: If [Tex]R = S \cup T, S \cap T = \phi     [/Tex]then, [Tex]\int\int_Rf(x,y)dy~dx = \int\int_Sf(x,y)dy~dx + \int\int_Tf(x,y)dy~dx [/Tex]
• Monotonicity: If f(x, y) ≥ g(x, y), then [Tex]\int \int_R f(x,y)dy~dx \geq \int\int_Rg(x,y)dy~dx [/Tex]
• [Tex]\int \int_R kf(x,y)dy~dx = k \int\int_R f(x,y)dy~dx [/Tex]

The rules of double integration are mentioned below:

• The double integral is a linear operator, meaning that it satisfies the following properties: [Tex]\int \int_R[f(x,y)\pm g(x,y)] = \int \int_Rf(x,y)\pm\int \int_Rg(x,y)[/Tex] and [Tex]\int \int k.f(x, y)dA = k.\int \int f(x, y)dA[/Tex]
• The order of integration can be interchanged under certain conditions. This is known as Fubini’s theorem. For a continuous function f(x, y) defined over a rectangular region R = [a, b] × . According to Fubini’s Theorem [Tex]\int\int_R f(x, y)dA = \int_a^b\int_c^df(x, y)dydx = \int_c^d\int_a^bf(x, y)dxdy[/Tex]
• Double integrals can be evaluated using a change of variables to transform the region of integration.
• Integration by parts can be applied to double integrals to simplify the computation, especially when dealing with products of functions.

The applications of double integral are mentioned below:

• Double integrals are used to calculate the area of regions bordered by curves or surfaces in the xy plane.
• Double integrals can be used to calculate the volume of solid objects or regions in three-dimensional space.
• Double integrals are used to calculate the mass and density distribution of objects with varying density.
• In probability theory, double integrals are used to determine probabilities for events in two-dimensional probability density functions.
• Double integrals are used to calculate electric field strength and electric flux over surfaces in electrostatics.
• Double integrals are used in fluid dynamics to determine fluid flow rates and fluxes over surfaces.

A detailed comparison between Double and Triple integral is given in the table below:

Also, Check

• Integral
• Riemann Integral
• Line Integral

Example 1: Calculate [Tex]\bold{\int^2_1\int^4_3 xy~dy~dx}   [/Tex].

Solution:

Let [Tex]I = \int^2_1\int^4_3xy~dy~dx  [/Tex]

Putting the upper and lower limits for x, we get

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy\\ \Rightarrow I = \int^2_1[\frac{4^2y}{2}-\frac{3^2y}{2}]dy [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = \int^2_1\frac{7y}{2}dy [/Tex]

Now integrating w.r.t y, we get

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = [\frac{7y^2}{4}]^{y=2}_{y=1} [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = [\frac{7(2^2)}{4}-\frac{7(1^2)}{4}] [/Tex]

⇒ I = 7 – 7/4 = 21/4

Example 2: Calculate [Tex]\bold{\int^2_1\int^4_3 (2x + 3y)~dy~dx}   [/Tex].

Solution:

[Tex]I = \int^2_1\int^4_3 (2x + 3y)~dy~dx  [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{2x^2}{2}+3xy]^{x=4}_{x=3}dy  [/Tex]

Putting the upper and lower limits for x, we get

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = \int^2_1[4^2+3(4)y-(3^2+3(3)y)]dy [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = \int^2_1(16+12y -9-9y)dy = \int^2_1(7+3y)dy [/Tex]

Now integrating w.r.t y, we get

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = [7y+ \frac{3y^2}{2}]^{y=2}_{y=1} [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = 7(2)+\frac{3(2^2)}{2}-(7(1)+\frac{3(1^2)}{2}) [/Tex]

⇒ I = 14 + 3 – 7 – 3/2 = 17/2

Example 3: Calculate [Tex]\bold{\int^1_0\int^4_2 x^2y~dy~dx}   [/Tex].

Solution:

[Tex]I = \int^1_0\int^4_2 x^2y~dy~dx  [/Tex]

[Tex]\Rightarrow I = \int^0_1[\frac{x^3y}{3}]^{x=4}_{x=2}dy  [/Tex]

Putting the upper and lower limits for x, we get

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = \int^1_0 (\frac{64y}{3}-\frac{8y}{3}) dy [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = \int^1_0(\frac{56y}{3})dy [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy \frac{56}{3}\int^1_0y~dy [/Tex]

Now integrating w.r.t y, we get

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = \frac{56}{3}[\frac{y^2}{2}]^{y=1}_{y=0} [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = \frac{56}{3}(\frac{1}{2}-0) [/Tex]

⇒ I = 56/6

Example 4: Calculate [Tex]\bold{\int^2_1\int^4_3 2xy~dy~dx}   [/Tex].

Solution:

Using [Tex]\int \int_R kf(x,y)dy~dx = k \int\int_R f(x,y)dy~dx [/Tex]

[Tex]I = \int^2_1\int^4_3 2xy~dy~dx  [/Tex]

[Tex]\Rightarrow I = 2\int^2_1\int^4_3 xy~dy~dx [/Tex]

[Tex]\Rightarrow I = \int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy = 2\int^2_1[\frac{x^2y}{2}]^{x=4}_{x=3}dy [/Tex]

Putting the upper and lower limits for x, we get

[Tex]\Rightarrow I= 2\int^2_1[\frac{4^2y}{2}-\frac{3^2y}{2}]dy [/Tex]

[Tex]\Rightarrow I= 2\int^2_1\frac{7y}{2}dy [/Tex]

Now integrating w.r.t y, we get

[Tex]\Rightarrow I= 2[\frac{7y^2}{4}]^{y=2}_{y=1} [/Tex]

[Tex]\Rightarrow I= 2[\frac{7(2^2)}{4}-\frac{7(1^2)}{4}] [/Tex]

⇒ I = 2(7 – 7/4) = 21/2

Example 5: Calculate [Tex]\bold{\int^5_0\int^6_0 x^3y^2~dy~dx}   [/Tex].

Solution:

Let [Tex]I = \int^5_0\int^6_0 x^3y^2~dy~dx [/Tex]

[Tex] \Rightarrow I= \int^5_0[\frac{x^4y^2}{4}]^{x=6}_{x=0}  [/Tex]

Putting the upper and lower limits for x, we get

[Tex]\Rightarrow I= \int^5_0[\frac{1296y^2}{4}-0]dy [/Tex]

[Tex]\Rightarrow I= \int^5_0\frac{324y^2}{2}dy [/Tex]

[Tex]\Rightarrow I= 162 \int^5_0 y^2dy [/Tex]

Now integrating w.r.t y, we get

[Tex]\Rightarrow I= 162[\frac{y^3}{3}]^{y=5}_{y=0} [/Tex]

[Tex]\Rightarrow I= 162[\frac{125}{3}-0] [/Tex]

⇒ I = 54(125) = 6750

What is the Meaning of Double Integral?

Double integral is the integration of a function with respect to two variables.

Can we Integrate the Double Integral using any of the Variables First?

Yes, the double integral can be integrated by using any of the variables first as the order of integration of variables do not matter in case of a double integral.

What are Some Applications of Double Integral?

Double integral is used to calculate the area and volume under one or more curves.

What is the Advantage of Double Integral over Single Integral?

A single integral can be used only to calculate the area under a 2D curve but a double integral also helps us to calculate volume of a 3D curve.