Double integral is the extension of definite integrals. The double integral over the ordinates and abscissa is useful to find the volume of the solid generated by a curve about an axis, area bounded by the curve, mass of the solid bounded by the curve and further more to find the moment of inertia, centroid etc. The multiple integral can be calculated for the three dimensional solids. It is calculated using rectangular co-ordinate system, (x, y, z plane), spherical polar co-ordinates and cylindrical polar co-ordinates. In this section let us see the rules of evaluating the double integrals using rectangular co-ordinate system and polar co-ordinate system.

Area bounded by a curve and the co-ordinate plane: For any curve y = f (x) the area bounded by the curve with the x-axis between the lines x = a and x = b is given by
                                           $\int_{a}^{b}$ y dx = $\int_{a}^{b}$ f( x ) dx

Double Integral: The double integral $\int_{a}^{b}$ $\int_{c}^{d}$ f ( x, y ) . dx. dy gives area of the region bounded by the rectangle with sides x=a, x = b, y = c and y = d. It shows the area bounded by the region as shown below.
Double Integral Surface
Rules to evaluate double integral

Rule 1: Since the double integral should be evaluated with respect to x and also y, first we evaluate the inner integral, with respect to constant by retaining the variable y as constant.

Rule 2: Substitute the upper lower limits of x and simplify the expression by combining the like terms.

Rule 3: Evaluate the outer integral with respect to y.

Rule 4: Substitute the upper and lower limits of y evaluate the integral in numerical form.

Given a double integral with limits we use the well known polar form of substitution, x = r cos $\theta$ , y = r sin $\theta$.

This will satisfy the equation of circle, x2 + y2 = r2 , and tan $\theta$ = $\frac{y}{x}$.

When we substitute for x and y in polar form we substitute for dx. dy = r dr d$\theta$ which can be evaluated using the following jacobian
J = $\frac{\partial (x,y)}{\partial (r,\theta)}$

= $\begin{vmatrix}
\frac{\partial x}{\partial r} &\frac{\partial x}{\partial \theta} & \\
\frac{\partial y}{\partial r} &\frac{\partial y}{\partial \theta}
\end{vmatrix}$


= $\begin{vmatrix}
cos\theta & -rsin\theta\\
sin\theta & rcos\theta
\end{vmatrix}$

= r

Thus dx. dy = r . dr. d$\theta$
This method might be advantageous if the terms of the form x2 + y2 are involved in f (x, y)
and terms like $\sqrt{a^{2}-x^{2}}$ and $\sqrt{b^{2}-x^{2}}$ are involved in limits.
Double integrals is useful to find the area of the region bounded in a co-ordinate plane.

1. Area of the region f(x,y) = c is given by the formula Area = $\int$ $\int$ dx . dy
In polar form, the area is given by $\int$ $\int$ r dr. d$\theta$
2. When the curve is given by z = f( x, y)
the surface area of the solid is given by $\int$ $\int$ $\sqrt{\left ( \frac{\partial z}{\partial x} \right )^{2}+\left ( \frac{\partial z}{\partial y} \right )^{2}+1}$ . dx. dy

Apart from evaluating the area and the volume we can also compute mass, electric charge, center of mass and moment of inertia.

3. Mass of a Lamina whose density function is given by $\rho$ ( x, y ) can be calculated using the formula, mass ( m ) = $\iint_{D}^{}$ $\rho$ ( x, y ) dx. dy.
4. If an electric charge is distributed over a region D and the charge density (in units of charge per unit area) is given by $\sigma$ (x, y) at a point (x, y) in D,
then the total charge Q is = $\iint_{D}^{}$ $\sigma$ ( x, y ) dx. dy

5. If the density function of a lamina is given by $\rho$ (x, y), then the moment of the entire lamina about the x and y axis are denoted by Mx and My respectively.
The double integral formula is given by,

Mx = $\iint_{D}^{}$ y . $\rho$ ( x, y ) . dx. dy
and My = $\iint_{D}^{}$ x. $\rho$ ( x, y ) . dx. dy.

6. The center of mass of a lamina is given by the co-ordinates, ($\overline{x }$, $\overline{y }$).
When the lamina of density function $\rho$ (x, y) occupy the region D, then the co-ordinates are given by,
$\overline{x }$ = $\frac{M_{y}}{m}$ = $\frac{1}{m}$ $\iint_{D}^{}$ y. $\rho$ ( x, y ) . dx. dy

$\overline{y }$ = $\frac{M_{x}}{m}$ = $\frac{1}{m}$ $\iint_{D}^{}$ x. $\rho$ ( x, y ) . dx. dy

where m is given by m = = $\iint_{D}^{}$ $\rho$ (x, y) dx. dy.

7. The moment of inertia or the second moment of a lamina of mass m whose density function is given by $\rho$ (x, y) bounded by the region D can be calculated using the following integrals.
Moment of Inertia about the x-axis = Ix = $\iint_{D}^{}$ y2 . $\rho$ ( x, y ) dx dy

Moment of Inertia about the y-axis = Iy = $\iint_{D}^{}$ x2 . $\rho$ ( x, y ) dx dy

Moment of Inertia about the Origin = Polar Moment of Inertia
= $\iint_{D}^{}$ ( x2 + y2 ) . $\rho$ ( x, y ) dx dy

and Io = Ix + Iy


Solved Examples

Question 1: Evaluate $\int_{1}^{4}$ $\int_{0}^{\sqrt{4-x}}$ x y dy dx
Solution:
 
We have I = $\int_{1}^{4}$ $\int_{0}^{\sqrt{4-x}}$ x y dy dx

= $\int_{x=1}^{4}$ [$\int_{y=0}^{\sqrt{4-x}}$ x y dy] dx

= $\int_{x=1}^{4}$ x[$\frac{y^{2}}{2}$] $_{0}^{\sqrt{4-x}}$ dx [by evaluating inner integral assuming x as constant]

= $\int_{x=1}^{4}$ x [$\frac{1}{2}$ (4 - x)] dx

= $\frac{1}{2}$ $\int_{x=1}^{4}$ x (4 - x) dx

= $\frac{1}{2}$ $\int_{1}^{4}$ (4 x - x2) dx

= $\frac{1}{2}$ [4 $\frac{x^{2}}{2}$ - $\frac{x^{3}}{3}$]$_{1}^{4}$

= $\frac{1}{2}$ [(32 - $\frac{64}{3}$) - (2 - $\frac{1}{3}$)]

= $\frac{1}{2}$ $\frac{[32 - 5]}{3}$

= $\frac{9}{2}$
 

Question 2: If R is the region x2 + y2 $\ge$ 0, evaluate $\iint_{D}^{}$ (x + y) . dx . dy
Double Integral Example
Solution:
 
The given region R is bounded by the y-axis and the right side of the circle x2 + y2 = a2 as shown in the above figure.
We observe that the region R consists of two parts R1 and R2 .
In R1 , $\theta$ varies from 0 to $\pi$/2 and in R2 , $\theta$ varies from 3 $\pi$/2 to 2 $\pi$.
In both parts r varies from 0 to a.
Therefore,
$\iint_{R}^{}$ (x + y) . dx . dy = $\iint_{R_{1}}^{}$ (x + y) . dx . dy + $\iint_{R_{2}}^{}$ (x + y) . dx . dy

= $\int_{0}^{\pi/2}$ $\int_{0}^{a}$ r (cos $\theta$ + sin $\theta$) r dr . d$\theta$ + $\int_{3\pi/2}^{2\pi}$ $\int_{0}^{a}$ r (cos $\theta$ + sin $\theta$) r dr . d$\theta$

= $\int_{0}^{a}$ r2 . dr . [$\int_{0}^{\pi/2}$ (cos $\theta$ + sin $\theta$) d$\theta$ + $\int_{3\pi/2}^{2\pi}$ (cos $\theta$ + sin $\theta$) d$\theta$]

= $\frac{a^{3}}{3}$ [[sin $\theta$ - cos $\theta$] $_{0}^{\pi/2}$ + [ sin $\theta$ - cos $\theta$] $_{3\pi/2}^{2\pi}$]

= $\frac{a^{3}}{3}$ {(1 - 0) - (0 - 1)} + {(0 - 1) - (- 1 - 0)}

= $\frac{1}{3}$ a3 [2]

= $\frac{2}{3}$ a3
 


Practice Problems

Question 1: $\int_{0}^{1}$ $\int_{x}^{\sqrt{x}}$ xy dy dx
Question 2: Evaluate $\int$ $\int$ y dx dy over the region bounded by the first quadrant of the ellipse $\frac{x^{2}}{a^{2}}$ + $\frac{y^{2}}{b^{2}}$ = 1
Question 3: Change the integral $\int_{-a}^{a}$ $\int_{0}^{\sqrt{x^{2}+y^{2}}}$ dy dx into polars and hence evaluate the same.
Question 4: Find by double integration the area enclosed by the curve r = a (1 + cos $\theta$) between $\theta$ = 0 and $\theta$ = $\pi$.