Double Integral : A double integral is an integral of integral that is it’s integrand is an integral.
If we integrate a function \(z = f(x, y)\) with respect to variables \(x \) and \(y\) such that \( (x, y)\in D \subseteq R^2\) then such an integrating process is called a double integral.
Where \(D\) is domain of the function \(f(x, y)\).
It is denoted as \[\int\int_{D}f(x,y)dxdy\]
Double integral gives the volume under the surface \(z = f(x, y)\) such that \( (x, y)\in D\)
Example (1):
Evaluate \(\int_{y=2}^{3}\int_{x=0}^{3}(2+xy)dxdy\)
Solution:
Step 1: First evaluating inner integral
Since inner integral is with respect to \(x\). Hence
\[A(y) = \int_{x=0}^{3}(2+xy)dx = \left[2x+\frac{x^2y}{2}\right]_{x=0}^{3}\]
\[ =2*3+\frac{3^2y}{2}-0-0 \]
\[A(y) =6+\frac{9y}{2}\]
Step 2 Evaluating outer integral
Since outer integral is with respect to \(y\). Hence
\[\int_{y=2}^{3}A(y)dy =\int_{y=2}^{3}(6+\frac{9y}{2})dy\]
\[= \left[6y+\frac{9y^2}{4}\right]_{2}^{3}\]
\[=6*3+\frac{9*3^2}{4} -6*2-\frac{9*2^2}{4}\]
\[ =\frac{69}{4} \]
Therefore \(\int_{y=2}^{3}\int_{x=0}^{3}(2+xy)dxdy =\frac{69}{4}\)
Example (2):
Evaluate \(\int_{x=0}^{\frac{\pi}{2}}\int_{y=0}^{1}y cos(x)dydx\)
Solution:
Step 1: First evaluating inner integral
Here inner integral is with respect to \(y\). Hence
\[A(x) = \int_{y=0}^{1}(ycos(x))dy =cos(x) \left[\frac{y^2}{2}\right]_{y=0}^{1}\]
\[ =cos(x)(\frac{1^2}{2}-0 )\]
\[A(x) =\frac{cos(x)}{2}\]
Step 2 Evaluating outer integral
Now outer integral is with respect to \(x\). Hence
\[\int_{x=0}^{\frac{\pi}{2}}A(x)dx =\int_{x=0}^{\frac{\pi}{2}}(\frac{cos(x)}{2})dy\]
\[= \left[\frac{sin(x)}{2}\right]_{x=0}^{\frac{\pi}{2}}\]
\[=\frac{sin(\frac{\pi}{2})}{2}-\frac{sin(0)}{2}\]
\[ =\frac{1}{2}-0 =\frac{1}{2}\]
Therefore \(\int_{x=0}^{\frac{\pi}{2}}\int_{y=0}^{1}y cos(x)dydx =\frac{1}{2}\)
\section{Order of Integration}
Order of integration of the double integral can be understood as follows
\[\int\int_{D_1}f(x,y)dydx= \int_{x=a}^{b}\left(\int_{y=f(x)}^{g(x)}f(x,y)dy\right)dx\]
that is inner integral is with respect to \(y\) and outer is with respect to \(x\).
\[\int\int_{D_2}f(x,y)dydx= \int_{y=c}^{d}\left(\int_{x=f(y)}^{g(y)}f(x,y)dx\right)dy\]
Here inner integral is with respect to \(x\) and outer is with respect to \(y\).
This is called order of integration.
Change of Order of Integration
We can evaluate \(\int\int_{D_1}f(x,y)dydx= \int_{x=a}^{b}\left(\int_{y=f(x)}^{g(x)}f(x,y)dy\right)dx\) by evaluating
\(\int\int_{D_2}f(x,y)dydx= \int_{y=c}^{d}\left(\int_{x=f(y)}^{g(y)}f(x,y)dx\right)dy\) after determining the domain \(D_2\) from \(D_1\).
Note : Change of order integration of the double integral does not change it’s value .
Example (1):
Evaluate \(\int\limits_{x = 0}^1 {\int\limits_{y = x}^1 {{y^4}{e^{x{y^2}}}dydx} }\) by changing order of integration.
Solution:
Let \(I = \int\limits_{x = 0}^1 {\int\limits_{y = x}^1 {{y^4}{e^{x{y^2}}}dydx} }\)
Step 1: changing order of integration
Since inner integral is with respect to \(x\). Hence
For \(I\) we have domain \(D_1 = \{(x, y) : 0\le x \le 1, x\le y\le 1 \}\)
Since \( x =0 , x = 1, y = x \Rightarrow y =0, y = 1, x = y \)
It can shown in the following graphs
After changing order of integration, we have
\[\int\limits_{x = 0}^1 {\int\limits_{y = x}^1 {{y^4}{e^{x{y^2}}}dydx} } = \int\limits_{y = 0}^1 {\int\limits_{x= 0}^y {{y^4}{e^{x{y^2}}}dxdy} }\]
Step 2: Evaluation
\[I = \int_{y=0}^{1}\left(\int_{x=0}^{y}y^4e^{xy^2}dx\right)dy\]
\[ =\int_{y=0}^{1}y^4\left[\frac{e^{xy^2}}{y^2}\right]_{x=0}^{y}dy \]
\[ =\int_{y=0}^{1}y^4\left[\frac{e^{xy^2}}{y^2}\right]_{x=0}^{y}dy \]
\[ =\int_{y=0}^{1}y^4\left[\frac{e^{y^3}}{y^2}-\frac{1}{y^2}\right]dy \]
\[ =\int_{y=0}^{1}\left[{y^2}{e^{y^3}}-{y^2}\right]dy \]
\[=\left[\frac{e^{y^3}}{3}-\frac{y^3}{3}\right]_{y=0}^{1}\]
\[ =\frac{1}{3} (e-2) \]
Thus \(\int\limits_{x = 0}^1 {\int\limits_{y = x}^1 {{y^4}{e^{x{y^2}}}dydx} } =\frac{1}{3} (e-2) \)
Example (2):
Evaluate \(\int_{x=0}^{1}\int_{y=x}^{1}sin(y^2)dydx\) by changing order of integration.
Solution:
Let \(I = \int_{x=0}^{1}\int_{y=x}^{1}sin(y^2)dydx\)
\[ =\int_{x=0}^{1}\left(\int_{y=x}^{1}sin(y^2)dy\right)dx \]
For the given double integral \(I\) we have range \(D_1 = \{(x, y) : 0\le x \le 1, x\le y\le 1 \}\)
From \( x =0 , x = 1, y = x , y = 1 \)
\(\Rightarrow y =0, y = 1, x = 0, x = y \)
Therefore after changing order of integration, limits for integration can be written as set \(D_2 = \{(x, y) : 0\le x \le y, 0\le y\le 1 \}\)is displayed in the following figure
Now we can write \(\int_{x=0}^{1}\left(\int_{y=x}^{1}sin(y^2)dy\right)dx=\int_{y=0}^{1}\left(\int_{x=0}^{y}sin(y^2)dx\right)dy\)
\[ = \int_{y=0}^{1}sin(y^2)\left(\int_{x=0}^{y}dx\right)dy\]
\[ =\int_{y=0}^{1}sin(y^2)\left[x\right]_{x=0}^{y}dy \]
\[ =\int_{y=0}^{1}sin(y^2)\left[y\right]dy \]
\[ =-\frac{1}{2} \left[\cos \left(y^2\right)\right]_{y=0}^{1} \]
\[ =-\frac{1}{2}\left[\cos \left(1\right) -\cos \left(0\right)\right] \]
\[=-\frac{1}{2}\left[\cos \left(1\right) -1\right]\]
\[ =\frac{1-\cos \left(1\right)}{2} \]
Thus \(\int_{x=0}^{1}\int_{y=x}^{1}sin(y^2)dydx=\frac{1-\cos \left(1\right)}{2} \)
Example 3 : Evaluate the following double integral
\[I=\int_{x=0}^{2}{\int_{y=x}^{2}\left(2-xy\right)dxdy}\]
Solution :
Let
\[I=\int_{x=0}^{2}{\int_{y=x}^{2}\left(2-xy\right)dxdy}\]
\[=\int_{x=0}^{2}\left(\int_{y=x}^{2}\left(2-xy\right)dy\right)dx\]
Integrating inner integral
\[I=\int_{x=0}^{2}{\left(2y-x\frac{y^2}{2}\right)_x^2dy}\]
Evaluating at the y limits
\[I=\int_{x=0}^{2}\left(\left(2*2-x\frac{2^2}{2}\right)-\left(2x-x\frac{x^2}{2}\right)\right)dx\]
Or
\[=\int_{x=0}^{2}\left(4-4x+\frac{x^3}{2}\right)dx\]
Now integrating with respect to x
\[=\left(4x-2x^2+\frac{x^4}{8}\right)_0^2\]
And evaluating at the x limits
\[=8-8+2-0+0-0=2\]
Thus the value of the double integral is
\[\int_{x=0}^{2}{\int_{y=x}^{2}\left(2-xy\right)dxdy}=2\]