The solid $W$ below is bounded by the circular paraboloid $$z = 2a\left( {1 - \frac{{{x^2} + {y^2}}}{{{{(3a)}^2}}}} \right)$$ and the $xy$ plane. At the point $(x,y,z) \in W$ its density is $$\delta (x,y,z) = {\delta _0} \cdot \frac{z}{{2a}}$$
- Calculate the volume $V$ , mass $m$ and average density $\overline \delta = \frac{m}{V}$ of $W$.
- Calculate the $z$-coordinate $\overline z $ of the center of mass of $W$. ($\overline x $ and $\overline y $ are ofcourse = 0 due to symmetry.)
Solution
The base of the paraboloid is a circle with a center coordinates at origins and radius $3a$. We will use cylindrical coordinates to simplify the calculations: \begin{equation} \notag \begin{split} V &= \int\limits_0^{2\pi } {\int\limits_0^{3a} {\int\limits_0^{2a\left( {1 - \frac{{{r^2}}}{{{{(3a)}^2}}}} \right)} {r\, dz \, dr \, d\theta = \int\limits_0^{2\pi } {\int\limits_0^{3a} {\left( {2ar - \frac{2}{{9a}}{r^3}} \right)\, dr \, d \theta } } } } } \\ &= \int\limits_0^{2\pi } {\frac{9}{2}} {a^3}\, d\theta = \frac{9}{2} \cdot {a^3} \cdot 2\pi = 9\pi {a^3} \\ \end{split} \end{equation} \begin{equation} \notag \begin{split} m &= \iiint\limits_D {\delta (x,y,z)\, dV = \int\limits_0^{2\pi } {\int\limits_0^{3a} {\int\limits_0^{2a\left( {1 - \frac{{{r^2}}}{{{{(3a)}^2}}}} \right)} {r \cdot {\delta _0} \cdot \frac{z}{{2a}}} } \, dz \, dr \, d\theta } } \\ &= \frac{{{\delta _0}}}{{2a}}\int\limits_0^{2\pi } {\int\limits_0^{3a} {r\int\limits_0^{2a\left( {1 - \frac{{{r^2}}}{{{{(3a)}^2}}}} \right)} z } \, dz \, dr \, d\theta } = \frac{{{\delta _0}}}{{2a}}\int\limits_0^{2\pi } {\int\limits_0^{3a} {r \cdot \left[ {\frac{1}{2}{z^2}} \right]_0^{2a\left( {1 - \frac{{{r^2}}}{{{{(3a)}^2}}}} \right)}\,} } dr \, d\theta \\ &= {\delta _0}a\int\limits_0^{2\pi } {\int\limits_0^{3a} {\left( {r - \frac{2}{{9{a^2}}}{r^3} + \frac{1}{{81{a^4}}}{r^5}} \right)\,} } dr \, d\theta \\ &= {\delta _0}a\int\limits_0^{2\pi } {\int\limits_0^{3a} {\left[ {\frac{1}{2}{r^2} - \frac{1}{{18{a^2}}}{r^4} + \frac{1}{{486{a^4}}}{r^6}} \right]_0^{3a}} } \, d\theta \\ &= \frac{3}{2}{a^3}{\delta _0}\int\limits_0^{2\pi } {d\theta } = \frac{3}{2}{a^3}{\delta _0} \cdot 2\pi = 3\pi {a^3}{\delta _0} \\ \end{split} \end{equation}
$$\overline \delta = \frac{m}{V} = \frac{{3\pi {a^3}{\delta _0}}}{{9\pi {a^3}}} = \frac{1}{3}{\delta _0}$$
$$\left( {\overline x ,\overline y ,\overline z } \right) = \left( {\frac{{{M_{yz}}}}{m},\frac{{{M_{xz}}}}{m},\frac{{{M_{xy}}}}{m}} \right)$$ where $${M_{yz}} = \iiint\limits_D {x\delta (x,y,z) dV},\,\,\,\,\,{M_{xz}} = \iiint\limits_D {y\delta (x,y,z) dV},\,\,\,\,\,{M_{xy}} = \iiint\limits_D {z\delta (x,y,z) dV}$$ \begin{equation} \notag \begin{split} {M_{xy}} &= \iiint\limits_D {z\delta (x,y,z)\, dV = \int\limits_0^{2\pi } {\int\limits_0^{3a} {\int\limits_0^{2a\left( {1 - \frac{{{r^2}}}{{{{(3a)}^2}}}} \right)} {r \cdot {\delta _0} \cdot \frac{z}{{2a}} \cdot z} } \, dz \, dr \, d\theta } } \\ &= \frac{{{\delta _0}}}{{2a}}\int\limits_0^{2\pi } {\int\limits_0^{3a} {r\int\limits_0^{2a\left( {1 - \frac{{{r^2}}}{{{{(3a)}^2}}}} \right)} {{z^2}} } \,dz \, dr \, d\theta } = \frac{{4{a^2}{\delta _0}}}{3}\int\limits_0^{2\pi } {\int\limits_0^{3a} {r \cdot \left[ {\frac{1}{3}{z^3}} \right]_0^{2a\left( {1 - \frac{{{r^2}}}{{{{(3a)}^2}}}} \right)}\,} } d r \, d\theta \\ &= \frac{{4{a^2}{\delta _0}}}{3}\int\limits_0^{2\pi } {\int\limits_0^{3a} {\left( {r - \frac{1}{{3{a^2}}}{r^3} + \frac{1}{{27{a^4}}}{r^5} - \frac{1}{{729{a^6}}}{r^7}} \right)\,} } dr \, d\theta \\ &= \frac{{4{a^2}{\delta _0}}}{3}\int\limits_0^{2\pi } {\int\limits_0^{3a} {\left[ {\frac{1}{2}{r^2} - \frac{1}{{12{a^2}}}{r^4} + \frac{1}{{162{a^4}}}{r^6} - \frac{1}{{5832{a^6}}}{r^8}} \right]_0^{3a}} } \, d\theta \\ &= \frac{3}{2}{a^4}{\delta _0}\int\limits_0^{2\pi } {d\theta } = \frac{3}{2}{a^4}{\delta _0} \cdot 2\pi = 3\pi {a^4}{\delta _0} \\ \end{split} \end{equation}
$$\overline z = \frac{{{M_{xy}}}}{m} = \frac{{3\pi {a^4}{\delta _0}}}{{3\pi {a^3}{\delta _0}}} = a$$
Therefore, the coordinates of the center of mass are $\left( {0,0,a} \right)$.
Comments:
I don't know the correct answer to this problem, so i would like to verify my answers.