Double Integral

Double Integral is primarily used to integrate the area of a surface of a two-dimensional figure, such as rectangle, circle, square, triangle, quadrilateral, and pentagon.  ‘ ∫∫’ is known to be the mathematical symbol of double integral. Moreover, simple integration is the basis of double integral. Thus, it is crucial to understand the basics with simple integration examples before moving to a more complex double integral method. Also, double integrals assist you in computing the volume of a surface as well. It is a vital element of calculus. There are multiple types of integration in mathematics, simple integration, double integration, and triple integration. You need to understand the rules of every kind of integration to understand the other two types of integrals. 

The Geometrical Interpretation Of Double Integral

Double integral or double integration method can be defined as the process in which two variables x and y are involved, and you need to integrate with both of them. This process of integration is employed to assess the volume of a given function under a curve. Well, the geometrical interpretation of double integral can be easily defined as ‘3D volume under the surface.’ Here is the geometrical representation of an arbitrary surface when represented by a function z = f(x,y)

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(The image shows the geometrical representation of double integral)

You need to divide the said domains into small rectangles as per the image.

Where 

\[\iint_{D}^{}f(x,y)IA = \sum \triangle V\] Which is the double integral

                   = \[\sum F(x,y)\triangle A\] which implies under the graph

When you use double integral over the specific region or domain, you are actually adding different elements possessing a height of z and a base area of dA=dx.dy. Remember, you will have to add up these elements infinitely. This is the double integration formula.

The Main Difference Between Simple Integral and Double Integral

As mentioned above, it is imperative to understand the basic difference between the types of integrals to have a clear understanding of the concept. Whereas the simple integration helps you in finding the area under a curve, the double integration method assists you in seeking the volume under the surface. Now, this is known to be the fundamental difference between the two integrals. Technically, the simple integration of a positive function of a single variable represents the area of the specified region between the x-axis and the function. On the other hand, the double integral of two distinct variables constitutes the volume of a particular said region as between the surface as defined by the function as well as the plane containing the domain. 

Purpose Of Using Double Integrals

The basic purpose of the double integral in mathematics is to basically integrate three-dimensional functions. When in school, you would learn about a particular plane being described by x and y as well the lines and the curves that exist within these points. Double integrals are basically used for solving multiple single integrals that cannot be evaluated individually.

The Major Properties Of Double Integral

Different properties of the Double integral are parallel to those of single integrals:

• 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 

• 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

Solved Examples

Example 1

Make use of a simple sketch to set up the double integral for the mass of the planar region represented by R possessing a variable density of δ(x, y). 

Solution: The sketch would look like:

Image to be added soon 

Area = ΔA

You need to understand that since the said region is a planar region, its density would be measured in the mass/unit area. Now, the density over a small piece of the region would not vary much, and we assume it to be continuous. Thus, if we consider that Δm is the mass of the piece, we get:  

  Δm ≈ δ(x, y)ΔA,

Here, (x, y) is known as the point in the piece, while ΔA is the piece’s area. Now, if we make slices of the region into minute pieces and calculate the masses of the given pieces, then we need to remember that the region’s total mass would be the sum of all the masses of the pieces together.

If we assume ΔA ----> 0, then the approximation will improve, and you will get the sum as an integral.

M = δ(x, y) dA.

Now, if we cut the region into  n pieces and label them from 1 to n, and Δmi is known to be the mass of the ith piece, we get:

Δmi ≈ δ(xi,yi)ΔAi,

When we sum the masses of all these pieces, we get:

Mass of R =Δmi ≈ δ(xi,yi)ΔAi. i=1

Example 2: 

Solve: where f (x, y) = \[1\sqrt{x^{2}+y^{2}}\]

\[\sqrt{x^{2}+y^{}}\iint_{R}^{}f(x,y)dxdy\;where\; f(x,y)\]= 1 / \[\sqrt{x^{2}+y^{2}}\]

And R would be the region inside the circle of radius 1. The circle would be centered at (1,0)

Solution:

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 r = 2 cos θ

Since the equation of the circle in the polar coordinates would be r = 2 cos θ, we make use of the radial stripes having the limits of:

\[\int \int_{R}^{}f(x,y)dxdy\] = \[\int_{-\pi/2}^{\pi/2}\int_{0}^{}\int_{0}^{2 cos\theta}\frac{1}{r}rdrd\theta =\int_{-\pi/2}^{\pi/2}\int_{0}^{}\int_{0}^{2 cos\theta}drd\theta\]

(inner) r from 0 to 2cos θ; (outer) θ from −π/2 to π/2.

Thus, you get:

\[\int \int_{R}^{}f(x,y)dxdy\] = \[\int_{-\pi/2}^{\pi/2}\int_{0}^{}\;1\]

Where the inner integral is 2 cos θ.