Example 3.9.1.

Evaluate \(\displaystyle {\iiint z \hspace {0.1cm} dV},\) where \(E\) is the tetrahedron bounded \(z = 0\hspace {0.1cm} ,\hspace {0.1cm} y = 0\hspace {0.1cm} , \hspace {0.1cm} x = 0\) and the plane \(x + y + z = 1\)

\begin {align*} \iiint \limits _E z\hspace {0.1cm}dV & = \int ^1_0 \int _0^{1-x}\int ^{1-x-y}_0 z \hspace {0.1cm}dz \hspace {0.1cm}dy\hspace {0.1cm} dx\\ & = \frac {1}{2} \int ^1_0 \int _0^{1-x}\big (1 - x- y\big )^2 \hspace {0.1cm}dy \hspace {0.1cm}dx\\\\ & = \frac {1}{24}\\ \end {align*}

A solid region \(E\) is of type 2 if it is of the form \[E = \big \{ (x,y,z)\hspace {0.1cm} \big | \hspace {0.1cm} (y,z)\in D\hspace {0.1cm}, \hspace {0.1cm} u_1(y,z) \leq x \leq u_2(y,z)\big \}\] and \(D\) is the projection of \(E\) on the \(yz\) plane, with surfaces \(\hspace {0.4cm} x = u_1(y,z)\hspace {0.2cm},\hspace {0.2cm} x = u_2(y,z)\).

\[\iiint \limits _E f(x,y,z)\hspace {0.1cm}dV = \iint \Bigg [ \int ^{u_2(y,z)}_{u_1(y,z)} f(x,y,z)\hspace {0.1cm}dx\Bigg ] \hspace {0.1cm} dA\]

A solid \(E\) is of type 3 if \[E = \big \{ (x,y,z)\hspace {0.1cm} \big | \hspace {0.1cm} (x,z)\in D\hspace {0.1cm}, \hspace {0.1cm} u_1(x,z) \leq x \leq u_2(x,z)\big \}\] then

\[\iiint \limits _E f(x,y,z)\hspace {0.1cm}dV = \iint \Bigg [ \int ^{u_2(x,z)}_{u_1(x,z)} f(x,y,z)\hspace {0.1cm}dy\Bigg ] \hspace {0.1cm} dA\]

Example 3.9.2.

Evaluate \(\displaystyle {\iiint \limits _E \sqrt {x^2 + z^2}\hspace {0.1cm}dV}\) where \(E\) is the region bounded by the paraboloid \( y = x^2 + z^2\) and the plane \(y =4\).

\(y = x^2 + z^2\implies z = \pm \sqrt {y - z^2}\)

\(E\) as type I region

\[E = \big \{ (x,y,z)\hspace {0.1cm} \big |\hspace {0.1cm} -2\leq x \leq 2 \hspace {0.1cm}, \hspace {0.1cm} x^2 \leq y \leq 4\hspace {0.1cm} , \hspace {0.1cm} -\sqrt {y - x^2} \leq z \leq \sqrt {y - z^2}\big \}\]

\[\iiint \sqrt {x^2 + z^2}\hspace {0.1cm}dV = \int ^2_{-2}\int ^4_{x^2}\int ^{\sqrt {y - x^2}}_{-\sqrt {y - x^2}}\sqrt {x^2 + z^2}\,dz\,dy\,dx\]

Alternatively, we take \(E\) as type 3 region.

\begin {align*} \iiint \sqrt {x^2 + z^2}\hspace {0.1cm}dV & = \iint \limits _{D_3} \Bigg [ \int ^4_{x^2 + z^2}\sqrt {x^2 + z^2}\hspace {0.1cm} dy\Bigg ]\hspace {0.1cm}dA\\\\ & = \iint \limits _{D_3}\big ( 4 -(x^2 + y^2)\big ) \hspace {0.1cm} \sqrt {x^2 + y^2}\hspace {0.1cm}dA\\ \end {align*}

Easier to convert to polar coordinates \(\hspace {0.3cm} x = r \cos \theta \hspace {0.5cm} z = r\sin \theta \)

\begin {align*} & = \int ^{2\pi }_0 \int ^2_0 (4-r^2)\hspace {0.1cm}(r)\hspace {0.1cm} \,r\,dr\,d\theta \\ & = \int ^{\pi }_0 d\theta \int ^2_0 \big (4r^2 - r^4\big )\, dr\\ & = 2\pi \cdot \Bigg (\frac {4r^3}{3}-\frac {r^5}{5}\Bigg |^2_0\\\\ & = \frac {128\pi }{15}\\\\ \end {align*}