Question

To evaluate the given integral .

\(\int_0^{\frac{\pi }{2}} {\int_0^y {\int_0^x {\cos } } } (x + y + z)dzdxdy\).

Short answer

The value of the given iterated integral is \(\frac{{ - 1}}{3}\).

Step-by-step solution

The function and region of the given integral

The function is \(f(x,y,z) = \cos (x + y + z)\).

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

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:

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

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

\(\begin{aligned}{l}\int_0^{\frac{\pi }{2}} {\int_0^y {\int_0^x {\cos } } } (x + y + z)dzdxdy = \int_0^{\frac{\pi }{2}} {\left( { - \frac{{\cos (2x + y)}}{2} + \cos (x + y)} \right)_0^y} dy\\ = \int_0^{\frac{\pi }{2}} {\left( {\left( { - \frac{{\cos (2y + y)}}{2} + \cos (y + y)} \right) - \left( { - \frac{{\cos (0 + y)}}{2} + \cos (0 + y)} \right)} \right)} dy\\ = \int_0^{\frac{\pi }{2}} {\left( {\left( { - \frac{{\cos (2y + y)}}{2} + \cos (y + y)} \right) - \left( { - \frac{{\cos (y)}}{2} + \cos (y)} \right)} \right)} dy\\ = \int_0^{\frac{\pi }{2}} {\left( { - \frac{{\cos (3y)}}{2} + \cos (2y) + \frac{{\cos (y)}}{2} - \cos (y)} \right)} dy\end{aligned}\)

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:

\(\begin{aligned}{l}\int_0^{\frac{\pi }{2}} {\int_0^y {\int_0^x {\cos } } } (x + y + z)dzdxdy = \int_0^{\frac{\pi }{2}} {\left( { - \frac{{\cos (3y)}}{2} + \cos (2y) + \frac{{\cos (y)}}{2} - \cos (y)} \right)} dy\\ = \left( { - \frac{{\sin (3y)}}{6} + \frac{{\sin (2y)}}{2} + \frac{{\sin y}}{2} - \sin y} \right)_0^{\frac{\pi }{2}}\\ = \left( {\frac{{ - \sin \left( {\frac{{3\pi }}{2}} \right)}}{6} + \frac{{\sin (\pi )}}{2} + \frac{{\sin \left( {\frac{\pi }{2}} \right)}}{2} - \sin \left( {\frac{\pi }{2}} \right)} \right) - \left\{ {\left( {\frac{{ - \sin \left( {\frac{{3(0)}}{2}} \right)}}{6} + \frac{{\sin (0)}}{2} + \frac{{\sin \left( {\frac{0}{2}} \right)}}{2} - \sin \left( {\frac{0}{2}} \right)} \right)} \right\}\end{aligned}\)On further simplification:

\(\begin{aligned}{l}\int_0^{\frac{\pi }{2}} {\int_0^y {\int_0^x {\cos } } } (x + y + z)dzdxdy = \left( {\frac{{ - \sin \left( {\pi + \frac{\pi }{2}} \right)}}{6} + \frac{0}{2} + \frac{1}{2} - 1} \right)\\ = \left( { - \frac{{ - \sin \left( {\frac{\pi }{2}} \right)}}{6} + 0 + \frac{1}{2} - 1} \right) - 0\\ = \frac{1}{6} + \frac{1}{2} - 1\\ = \frac{2}{3} - 1\\ = \frac{{ - 1}}{3}\end{aligned}\)

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