Iterated Integrals
Solved Problems
Example 3.
Calculate the iterated integral \[\int\limits_1^2 {\int\limits_0^y {x\sqrt {{y^2} + {x^2}} dxdy} }.\]
Solution.
Using the Fubini's theorem, we can write:
\[I = \int\limits_1^2 {\int\limits_0^y {x\sqrt {{y^2} + {x^2}} dxdy} } = \int\limits_1^2 {\left[ {\int\limits_0^y {x\sqrt {{y^2} + {x^2}} dx} } \right]dy} .\]
To calculate the inner integral, we make the substitution:
\[z = {y^2} + {x^2},\;\; \Rightarrow dz = 2xdx,\;\; \Rightarrow xdx = \frac{{dz}}{2}.\]
When \(x = 0,\) we have \(z = {y^2},\) and when \(x = y,\) we get \(z = 2{y^2}.\) Then
\[I = \int\limits_1^2 {\left[ {\int\limits_0^y {x\sqrt {{y^2} + {x^2}} dx} } \right]dy} = \int\limits_1^2 {\left[ {\int\limits_{{y^2}}^{2{y^2}} {\sqrt z \frac{{dz}}{2}} } \right]dy} = \int\limits_1^2 {\left[ {\left. {\left( {\frac{{{z^{\frac{3}{2}}}}}{3}} \right)} \right|_{{y^2}}^{2{y^2}}} \right]dy} = \frac{1}{3}\int\limits_1^2 {\left[ {{{\left( {2{y^2}} \right)}^{\frac{3}{2}}} - {{\left( {{y^2}} \right)}^{\frac{3}{2}}}} \right]dy} = \frac{1}{3}\int\limits_1^2 {\left[ {2\sqrt 2 {y^3} - {y^3}} \right]dy} = \frac{{2\sqrt 2 - 1}}{3}\int\limits_1^2 {{y^3}dy} = \frac{{2\sqrt 2 - 1}}{{12}}\left. {\left( {{y^4}} \right)} \right|_1^2 = \frac{{5\left( {2\sqrt 2 - 1} \right)}}{4}.\]
Example 4.
Evaluate the iterated integral \[\int\limits_0^1 {\int\limits_0^y {\ln \left( {{y^2} + 1} \right)dxdy} }.\]
Solution.
First we calculate the inner integral:
\[I = \int\limits_0^1 {\int\limits_0^y {\ln \left( {{y^2} + 1} \right)dxdy} } = \int\limits_0^1 {\left[ {\int\limits_0^y {\ln \left( {{y^2} + 1} \right)dx} } \right]dy} = \int\limits_0^1 {\left[ {\ln \left( {{y^2} + 1} \right)\left. x \right|_0^y} \right]dy} = \int\limits_0^1 {\ln \left( {{y^2} + 1} \right)ydy} .\]
Now we use integration by parts: \({\int} {udv} = uv - {\int} {vdu}.\) Let \(u = \ln \left( {{y^2} + 1} \right),\) \(dv = ydy.\) Then
\[du = \frac{d}{{dy}}\ln \left( {{y^2} + 1} \right) = \frac{{2ydy}}{{{y^2} + 1}},\;\; v = \int {ydy} = \frac{{{y^2}}}{2}.\]
Substituting this into integral, we have
\[I = \int\limits_0^1 {\ln \left( {{y^2} + 1} \right)ydy} = \left. {\left[ {\frac{{{y^2}}}{2}\ln \left( {{y^2} + 1} \right)} \right]} \right|_0^1 - \int\limits_0^1 {\frac{{{y^2}}}{2}\frac{{2ydy}}{{{y^2} + 1}}} = \left. {\left[ {\frac{{{y^2}}}{2}\ln \left( {{y^2} + 1} \right)} \right]} \right|_0^1 - \int\limits_0^1 {\frac{{{y^3}dy}}{{{y^2} + 1}}} = \frac{{\ln 2}}{2} - \int\limits_0^1 {\frac{{{y^3}dy}}{{{y^2} + 1}}} .\]
Finally we can calculate the outer integral:
\[\int\limits_0^1 {\frac{{{y^3}dy}}{{{y^2} + 1}}} = \int\limits_0^1 {\frac{{{y^3} + y - y}}{{{y^2} + 1}}dy} = \int\limits_0^1 {\left[ {y - \frac{y}{{{y^2} + 1}}} \right]dy} = \int\limits_0^1 {ydy} - \frac{1}{2}\int\limits_0^1 {\frac{{d\left( {{y^2} + 1} \right)}}{{{y^2} + 1}}} = \left. {\left[ {\frac{{{y^2}}}{2} - \frac{1}{2}\ln \left( {{y^2} + 1} \right)} \right]} \right|_0^1 = \frac{1}{2} - \frac{{\ln 2}}{2}.\]
The complete answer is
\[I = \frac{{\ln 2}}{2} - \frac{1}{2} + \frac{{\ln 2}}{2} = \ln 2 - \frac{1}{2}.\]
Example 5.
Express the integral as an integral with the order of integration reversed: \[I = \int\limits_0^5 {\int\limits_{\frac{x}{2}}^x {f\left( {x,y} \right)dydx} }.\]
Solution.
Here we have the region of type \(I\) (Figure \(3\)).
Figure 3. The region of integration is the triangle bounded by the lines \(y = \frac{x}{2}\) or \(x = 2y\) and \(y = x\) or \(x = y.\) The variable \(x\) runs through the interval \(0 \le x \le 5.\) By changing the order of integration we can write the initial integral as the sum of the two iterated integrals:
\[I = \int\limits_0^{2,5} {\int\limits_y^{2y} {f\left( {x,y} \right)dxdy} } + \int\limits_{2,5}^5 {\int\limits_y^5 {f\left( {x,y} \right)dxdy} } .\]
