Question

To evaluate the given integral.

\(\int_0^2 {\int_0^{{z^2}} {\int_0^{y - z} {(2x - y)} } } dxdydz\)

Short answer

The value of the given iterated integral is \(\frac{{16}}{{15}}\).

Step-by-step solution

The given function and region for the integral

The function is\(f(x,y,z) = 2x - y\).

The region is\(B = \left\{ {(x,y,z)\mid 0 \le x \le (y - z),0 \le y \le {z^2},0 \le z \le 2} \right\}\).

Integrate the given integral with respect to \(x\) and apply the limit of it

On integrating the given integral with respect to \(x\) and applying the limit of it:

\(\int_0^2 {\int_0^{{z^2}} {\int_0^{y - z} {(2x - y)} } } dxdydz = \int_0^2 {\int_0^{{z^2}} {\left( {\frac{{2{x^2}}}{2} - yx} \right)_0^{y - z}} } dydz\)

\( = \int_0^2 {\int_0^{{z^2}} {\left( {{{(y - z)}^2} - y(y - z)} \right)} } dydz\)

\( = \int_0^2 {\int_0^{{z^2}} {\left( {\left( {{z^2} - yz} \right)} \right)} } dydz\).

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

On integrating the given integral with respect to \(y\) and applying the limit of it:

\(\begin{aligned}{l}\int_0^2 {\int_0^{{z^2}} {\int_0^{y - z} {(2x - y)} } } dxdydz\\ = \int_0^2 {\left( {{z^2}y - \frac{{z{y^2}}}{2}} \right)_0^{{z^2}}} dz\\ = \int_0^2 {\left( {\left( {{z^2}\left( {{z^2}} \right) - \frac{{z{{\left( {{z^2}} \right)}^2}}}{2}} \right) - \left( {{z^2}(0) - \frac{{z{{(0)}^2}}}{2}} \right)} \right)} dz\\ = \int_0^2 {\left( {\left( {{z^4} - \frac{{{z^5}}}{2}} \right) - (0 - 0)} \right)} dz\\ = \int_0^2 {\left( {{z^4} - \frac{{{z^5}}}{2}} \right)} dz\end{aligned}\)

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 of it:

\(\begin{aligned}{l}\int_0^2 {\int_0^{{z^2}} {\int_0^{y - z} {(2x - y)} } } dxdydz = \left( {\left( {\frac{{{z^5}}}{5} - \frac{{{z^6}}}{{2(6)}}} \right)} \right)_0^2\\ = \left( {\left( {\frac{{{{(2)}^5}}}{5} - \frac{{{{(2)}^6}}}{{12}}} \right) - \left( {\frac{{{{(0)}^5}}}{5} - \frac{{{{(0)}^6}}}{{12}}} \right)} \right)\\ = \left( {\frac{{32}}{5} - \frac{{64}}{{12}}} \right) - (0 - 0)\\ = \frac{{32}}{5} - \frac{{16}}{3}\\ = \frac{{16}}{{15}}\end{aligned}\)

Thus, the value of the given iterated integral is \(\frac{{16}}{{15}}\).