\section{Integration in Three Variables}

\begin{enumerate}[1.]
\item
{\it 
Let $R$ be the solid region in $\rr^3$ bounded by the planes
$x=0$, $y=0$, $z=0$, and $x+2y+z=6$.
Write 
$\displaystyle{\iiint_R  f(x,y,z)dV}$ as iterated integrals
where the order of integration is (a) $dzdydx$, (b) $dydxdz$.}
\medskip



\medskip\noindent{(a)} 
$$\int_0^6 \int_0^{3-x/2}\int_{0}^{6-x-2y} f(x,y,z)dzdydx$$



\medskip\noindent{(b)} 
$$\int_{0}^6\int_0^{6-z}\int_0^{3-\frac{x+z}{2} }  f(x,y,z)dydxdz$$

\bigskip



\item {\it Evaluate the integral
$\displaystyle \iiint_D y\, dV,$
 if $D$ is the tetrahedron with vertices $(0,0,0)$, $(4,2,0)$, $(0,2,0)$, and $(0,2,4)$.
}

\medskip

\begin{align*}
 \iiint_D y\, dV&
 = \int_0^2 dy \int_0^{2y} dx \int_0^{2y-x} ydz
 \\
 &= \int_0^2 dy \int_0^{2y} dx \left[ yz\right]_{z=0}^{2y-x} = \int_0^2 dy \int_0^{2y} (2y^2-yx) dx 
   \\
 &= \int_0^2 dy  \left[ 2y^2 x - \frac{yx^2}{2} \right]_{x=0}^{2y} = \int_0^2 (4y^3-2y^3) dy  
 \\
 &=\int_0^2 2y^3\, dy 
 = \left[ \frac{y^4}{2} \right]_{y=0}^{2} =8.
 \end{align*}
 


\medskip




\item{\it Let $D$ be the spatial region above the $xy$-plane, inside the cylinder $x^2+y^2=1$,
and inside the sphere $x^2+y^2+ z^2 = 4$. Evaluate the integral $\iiint_D z dV$. }

\medskip

We use cylindrical coordinates:
\begin{align*}
\iiint_D z dV
&= \int_0^1 \int_0^{2\pi}  \int_0^{\sqrt{4-r^2}}  zr\,dz\,d\theta\, dr
\\
&= \int_0^1 \int_0^{2\pi}  r\, \frac{z^2}{2} \Big|_0^{\sqrt{4-r^2}}  \,d\theta\,dr
\\
&=  \int_0^1\int_0^{2\pi}  r\, \frac{4-r^2}{2}  \,d\theta\, dr
\\
&=  \int_0^1 \pi (4r -r^3 )  \, dr
\\
&= \pi \Big( 2r^2 - \frac{r^4}{4}\Big) \Big|_0^1 = \pi\Big( 2-\frac{1}{4} \Big)
= \frac{7\pi}{4}.
\end{align*}


\item 
{\it Evaluate the integral
$\displaystyle \iiint_{R} x \, dV$,
if 
$R=\{(x,y,z):\ 0\leq z\leq y\leq x,\ x^2+y^2\leq 1\}$. }


\medskip

Let $D=\{(x,y):\ 0\leq y\leq x,\ x^2+y^2\leq 1\}$. Then 
\begin{align*}
 \iiint_{R} x \, dV&= \iint_D \int_0^y x\, dz \, dA 
= \iint_D xy\, dA 
\\
&= \int_0^{\pi/4} \int_0^1 (r\cos\theta)(r\sin\theta)r\, dr\,d\theta
\\
&=  \int_0^{\pi/4} \int_0^1 r^3\cos\theta\sin\theta \,dr\,d\theta
\\
&= \int_0^{\pi/4} \frac{r^3}{4}\Big|_{r=0}^1 \cos\theta\sin\theta \,d\theta
\\
&= \int_0^{\pi/4} \frac{ \cos\theta\sin\theta}{4} \,d\theta\\
&= \frac{\sin^2\theta}{8}\Big|_{\theta=0}^{\pi/4}
= \frac{(\sqrt{2}/2)^2}{8}=\frac{1}{16}.
\end{align*}

\end{enumerate}
