Find the center mass of the solid bounded by planes $x+y+z=1,x=0,y=0$ and $z=0$, assuming a mass density of $$\rho(x,y,z) = 10 \sqrt{z}.$$
I could not set up the integral!
[Math] center mass of the solid
integrationmultivariable-calculus
Related Solutions
The mass of the solid is defined as
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV,$$
that is, the integral of body density at each point over the volume of the body. In this case we have $\rho \equiv B$ which is constant, therefore the mass will be a multiple of the body's volume:
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV = B \iiint\limits_{\mathcal{B}} \, dV.$$
This is a paraboloid and its volume can be found using the cylindrical coordinate substition. We find $z = 2(x^2+y^2) = 2r^2$ and the limits for $z$ will be $2r^2 \leq z \leq \sqrt{c/2}$. This was found equating $z=2r^2 = c$, therefore $r = \sqrt{c/2}$.
$$ \begin{align} \iiint\limits_{\mathcal{B}} \, dV & = \int_0^{2 \pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c r \, dz \, dr \, d \theta \\ & = 2 \pi \int_0^{\sqrt{c/2}} cr - 2r^3 \, dr \\ & = 2 \pi \left( \frac{cr^2}{2} - \frac{r^4}{2} \right) \Bigg\vert_0^{\sqrt{c/2}} \\ & = \pi \left( c \cdot \frac{c}{2} - \frac{c^2}{4} \right) \\ & = \frac{c^2 \pi}{4}. \end{align} $$
Hence $M = (Bc^2 \pi)/4$.
Coordinates for center of mass are defined as
$$ \begin{align} \overline{x} & = \frac{1}{M} \iiint x \rho \, dV \\ \overline{y} & = \frac{1}{M} \iiint y \rho \, dV \\ \overline{z} & = \frac{1}{M} \iiint z \rho \, dV. \end{align} $$
In physicist notation I've seen this written as
$$\mathbf{R} = \frac{1}{M} \iiint \rho \, \mathbf{r} \, dV.$$
It is interesting to note the following: the paraboloid is symmetric around $z$ axis. This means that the center of mass must be in the $z$ axis, for the $\overline{x}$ and $\overline{y}$ will cancel (if you don't believe this, write out the integral explicitly: you will have to integrate $\cos \theta$ and $\sin \theta$ over $[0,2 \pi]$, which is zero).
Therefore it is left for us to compute the $z$ coordinate. Leaving out the density out for a second (since it is uniform), we have
$$ \begin{align} \iiint\limits_{\mathcal{B}} z \, dV & = \int_0^{2\pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c z r \, dz \, dr \, d \theta \\ & = 2 \pi \int_0^{\sqrt{c/2}} \frac{r}{2} \left( c^2 - 4r^4 \right) \, dr \\ & = \pi \int_0^{\sqrt{c/2}} c^2 r - 4r^5 \, dr \\ & = \pi \left( \frac{(cr)^2}{2} - \frac{2r^6}{3} \right) \Bigg\vert_0^{\sqrt{c/2}} \\ & = \pi \left( \frac{c^2}{2} \cdot \frac{c}{2} - \frac{2}{3} \cdot \frac{c^3}{8} \right) \\ & = \pi \left( \frac{c^3}{4} - \frac{c^3}{12} \right) \\ & = \pi \left( \frac{c^3}{6} \right) \\ & = \frac{c^3 \pi}{6}. \end{align} $$
Finally
$$\overline{z} = \frac{B c^3 \pi}{6} \cdot \frac{4}{Bc^2 \pi} = \frac{2c}{3}.$$
Therefore
$$M = \frac{Bc^2 \pi}{4} \text{ and } \mathbf{R} = (\overline{x}, \overline{y}, \overline{z}) = \left( 0, 0, \frac{2c}{3} \right).$$
Hope this helps and best wishes.
The mass of the solid is $\int_W \rho \mathrm{d}V$ which in this case is equal to $\rho \int_W \mathrm{d}V$ as the solid has a uniform mass density. To find the mass consider a small rectangular element of sides $\mathrm{d}x$ and $\mathrm{d}y$ on the $xy$ plane. The volume of the cuboidal rod of solid with this element as the base is $z\mathrm{d}x\mathrm{d}y$. So the total volume of the solid will be $$\int_0^5\int_0^4(x+y+3) \mathrm{d}x \mathrm{d}y = 150$$ and therefore the mass = $1gm/cm^3\cdot150 cm^3 = 150 gm$
For the coordinates of center of mass the integrals are similar $$ \overline{x} = \frac{1}{m}\int_0^5\int_0^4\rho x(x+y+3) \mathrm{d}x \mathrm{d}y$$ $$ \overline{y} = \frac{1}{m}\int_0^5\int_0^4\rho y(x+y+3) \mathrm{d}x \mathrm{d}y$$ $$ \overline{z} = \frac{1}{m}\int_0^5\int_0^4\rho \frac{(x+y+3)^2}{2} \mathrm{d}x \mathrm{d}y$$
Best Answer
You can get bounds of integration by intersecting your various surfaces. For example: $x+y+z=1$ intersected with $z=0$ gives you $x+y+0=1$ so that $y=1-x$. Then intersect with $y=0$ and get $0=1-x$ so that $x=1$.
In the end, this region can be described as follows: $0 \leq z \leq 1-x-y$, $0 \leq y \leq 1-x$, $0 \leq x \leq 1$.
So to find the mass you'd need to integrate... $$\int_0^1 \int_0^{1-x} \int_0^{1-x-y} 10\sqrt{z}\,dz\,dy\,dx$$
Then the moment about the $yz$-plane is... $$\int_0^1 \int_0^{1-x} \int_0^{1-x-y} 10\sqrt{z} \cdot x\,dz\,dy\,dx$$