Question

Reversing the order of integration: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals by reversing the order of integration.

${\int }_0^9{\int }_{\sqrt{y}}^3\frac{1}{1+x^3}dxdy$

Short answer

${\int }_0^{x^2}{\int }_0^3\frac{1}{1+x^3}dxdy=\frac{1}{3}\ln 28$

Step-by-step solution

Draw the region

The region determined by the limits of the given iterated integral is shown below,

Reversing the order of integration  

From the above diagram, reversing the order of integration.

${\int }_0^9{\int }_{\sqrt{y}}^3\frac{1}{1+x^3}dxdy\Rightarrow {\int }_0^{x^2}{\int }_0^3\frac{1}{1+x^3}dxdy$

Evaluate the integral

$$\begin{gathered} I={\int }_0^{x^2}{\int }_0^3\frac{1}{1+x^3}dxdy \\ I={\int }_0^{x^2}dy{\int }_0^3\frac{1}{1+x^3}dx \\ I={\int }_0^3\frac{x^2}{1+x^3}dx \end{gathered}$$

Substitute,

$$\begin{gathered} 1+x^3=t \\ 3x^{2}dx=dt \\ dx=\frac{1}{3x^2}dt \\ \mathrm{When}x=0,t=1+0=1 \\ \mathrm{When}x=3,t=1+3^3=28 \end{gathered}$$

Therefore,

$$\begin{gathered} I=\frac{1}{3}{\int }_1^{28}\frac{1}{t}dt \\ I=\frac{1}{3}{\left[\ln t\right]}_1^{28} \\ I=\frac{1}{3}\ln 28 \end{gathered}$$