Question

To evaluate the value of iterated integral

\(\int_0^{\sqrt \pi } {\int_0^x {\int_0^{xz} {{x^2}} } } \sin ydydzdx\)

Short answer

The value of the given iterated integral is \(\frac{{{\pi ^2}}}{4} - 1\).

Step-by-step solution

The function and region of the given integral

The function is \(f(x,y,z) = {x^2}\sin y\).

The region is\(B = \{ (x,y,z)\mid 0 \le x \le \sqrt \pi ,0 \le y \le xz,0 \le z \le x\} \).

Integrate the given integral with respect to \(y\) and apply the limit as follows

The given integral is \(\int_0^{\sqrt \pi } {\int_0^x {\int_0^{xz} {{x^2}} } } \sin ydydzdx\).

\(\int_0^{\sqrt \pi } {\int_0^x {\int_0^{xz} {{x^2}} } } \sin ydydzdx = \int_0^{\sqrt \pi } {\int_0^x {\left( { - {x^2}\cos y} \right)} } dzdx\)

\( = \int_0^{\sqrt \pi } {\int_0^x {{x^2}} } (1 - \cos (xz))dzdx\)

Integrate the given integral with respect to \(z\) and apply the limit

On integrating the given integral with respect to \(z\) and applying the limit:

\(\begin{aligned}{c}\int_0^{\sqrt \pi } {\int_0^x {\int_0^{xz} {{x^2}} } } \sin ydydzdx = \int_0^{\sqrt \pi } {\left( {{x^2}\left( {z - \frac{1}{x}\sin (xz)} \right)} \right)_0^x} dx\\ = \int_0^{\sqrt \pi } {{x^2}} \left( {x - \frac{1}{x}\sin \left( {{x^2}} \right)} \right)dx\\ = \int_0^{\sqrt \pi } {\left( {{x^3} - x\sin \left( {{x^2}} \right)} \right)} dx\end{aligned}\)

By substituting \({x^2} = u,2xdx = du\), and \(x = 0 \Rightarrow u = 0,x = \sqrt \pi \Rightarrow u = \pi \) the above integral is rewritten as follows:

\(\begin{aligned}{c}\int_0^{\sqrt \pi } {\int_0^x {\int_0^{xz} {{x^2}} } } \sin ydydzdx = \int_0^{\sqrt \pi } {\left( {{x^2} - \sin \left( {{x^2}} \right)} \right)} (2xdx)\\ = \int_0^\pi {(u - \sin u)} du\end{aligned}\)

Integrate the given integral with respect to \(u\) and apply the limit

On integrating the given integral with respect to \(u\) and applying the limit:

\(\begin{aligned}{c}\int_0^{\sqrt \pi } {\int_0^x {\int_0^{xz} {{x^2}} } } \sin ydydzdx = \frac{1}{2}\left( {\frac{{{u^2}}}{2} + \cos u} \right)_0^\pi \\ = \frac{1}{2}\left( {\frac{{{\pi ^2}}}{2} - 2} \right)\\ = \frac{{{\pi ^2}}}{4} - 1\end{aligned}\)

Thus, the value of the given iterated integral is \(\underline {\frac{{{\pi ^2}}}{4} - 1} \).