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`