Question

\(\int\limits_0^4 {\int\limits_{\sqrt x }^2 {\frac{1}{{{y^3} + 1}}dydx} } \).

Short answer

Thus, the value of integral is\(\frac{1}{3}\ln 3\).

Step-by-step solution

Given Data.

Consider the integral:

\(\int\limits_0^4 {\int\limits_{\sqrt x }^2 {\frac{1}{{{y^3} + 1}}dydx} } \)

From the given double integral, notice that the variable\(y\) varies from \(\sqrt x \) to \(2\) and the variable \(x\) varies from \(0\) to \(4\).

So, the region \(D\) can be written as:

\(D = \left\{ {\left( {x,y} \right)|\sqrt x \le y \le 2,0 \le x \le 4} \right\}\)

The sketch of the region \(D\) (Type I):

Reversing Order of Integration.

Rewrite the equation \(y = \sqrt x \)

\(y = \sqrt x \)

\({y^2} = x\)

\(x = {y^2}\)

The alternative description of \(D\) is

\(D = \left\{ {\left( {x,y} \right)|0 \le y \le 2,0 \le x \le {y^2}} \right\}\)

The region \(D\) (Type II):

Evaluation of Integral.

\(\begin{array}{c}\int\limits_0^4 {\int\limits_{\sqrt x }^2 {\frac{1}{{{y^3} + 1}}dydx} } = \int\limits_0^2 {\int\limits_0^{{y^2}} {\frac{1}{{{y^3} + 1}}dx} } dy\\ = \int\limits_0^2 {\frac{1}{{{y^3} + 1}}\left( x \right)_0^{{y^2}}dy} \\ = \int\limits_0^2 {\frac{{{y^2}}}{{{y^3} + 1}}dy} \end{array}\)

Put \({y^3} + 1 = t \Rightarrow 3{y^2}dy = dt\)

When \(y \to 0\) then \(t \to 1\)

When \(y \to 2\) then \(t \to 9\)

\(\begin{array}{c}\int\limits_0^4 {\int\limits_{\sqrt x }^2 {\frac{1}{{{y^3} + 1}}dydx} } = \frac{1}{3}\int\limits_1^9 {\frac{1}{t}dt} \\ = \frac{1}{3}\left( {\ln t} \right)_1^9\\ = \frac{1}{3}\left( {\ln 9 - \ln 1} \right)\\ = \frac{1}{3}\left( {\ln {3^2} - 0} \right)\\ = \frac{2}{3}\ln 3\end{array}\)

Therefore, the value of integral is\(\frac{1}{3}\ln 3\).