Question

Evaluate the iterated integral. $$ \int\limits_0^{2\pi}\int\limits_0^y\int\limits_0^x {\color{red}9} \cos(x+y+z)\,dz\,dx\,dy $$

Answer 1

$\displaystyle\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\int_{z=0}^{z=x}\cos(x+y+z)\,\mathrm dz\,\mathrm dx\,\mathrm dy=\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\sin(x+y+z)\bigg|_{z=0}^{z=x}\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\sin(2x+y)-\sin(x+y)\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}-\frac12\left(\cos(2x+y)-2\cos(x+y)\right)\bigg|_{x=0}^{x=y}\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}-\frac12\left((\cos3y-2\cos2y)-(\cos y-2\cos y)\right)\bigg|_{x=0}^{x=y}\,\mathrm dy$
$\displaystyle=-\frac12\int_{y=0}^{y=2\pi}(\cos3y-2\cos2y+\cos y)\,\mathrm dy$
$\displaystyle=-\frac12\left(\frac13\sin3y-\sin2y+\sin y\right)\bigg|_{y=0}^{y=2\pi}$
$=0$