Using cylindrical coordinates to find center of mass of solid of uniform density given by
$$x^2 + y^2 \le \frac{1}{4},\quad x^2 + y^2 + (z-1)^2 \le 1,\quad z \le 1.$$
I don't know how to set up the integrals. Can someone please help?
[Math] Using cylindrical coordinates to find center of mass of solid of uniform density
calculuscylindrical coordinatesmultivariable-calculus
Related Solutions
To integrate this region in polar coordinates, it is advisable to break up the integral into two parts, as shown in the figures below:
The two parts of the integral are divided by the diagonal line through the upper right corner of the rectangle. Since the sides of the rectangle are $a$ and $b$, this diagonal line is at the angle $\arctan \frac ba.$
For $0 \leq \theta \leq \arctan \frac ba,$ you would integrate over $0 \leq r \leq a \sec\theta,$ and for $\arctan \frac ba \leq \theta \leq \frac\pi2,$ you would integrate over $0 \leq r \leq b \csc\theta.$
If you actually try this, I think you'll find that it is no easier than doing the integration in rectangular coordinates. It may even be worse.
An alternative approach, rather than combining $x^2+y^2$ into $r^2$, is to integrate the terms separately:
$$\begin{eqnarray} m &=& \int_0^a\int_0^b (1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b dy\,dx +\int_0^a\int_0^b x^2 \,dy\,dx +\int_0^a\int_0^b y^2 \,dy\,dx \end{eqnarray}$$ $$\begin{eqnarray} m\bar{x} &=& \int_0^a\int_0^b x(1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b x \,dy\,dx +\int_0^a\int_0^b x^3 \,dy\,dx +\int_0^a\int_0^b xy^2 \,dy\,dx \end{eqnarray}$$ $$\begin{eqnarray} m\bar{y} &=& \int_0^a\int_0^b y(1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b y \,dy\,dx +\int_0^a\int_0^b x^2y \,dy\,dx +\int_0^a\int_0^b y^3 \,dy\,dx \end{eqnarray}$$
Now you have nine integrals to solve, but they're all quite simple.
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.
Best Answer
Note that your solid is the intersection of a cylynder and a hemisphere. It is given by $$S:=\left\{(x,y,z)\in\mathbb{R}^3\::\; 1-\sqrt{1-(x^2+y^2)}\leq z\leq 1,\; x^2 + y^2 \leq \frac{1}{4}\right\}.$$ Moreover, by symmetry, the center of mass is along the $z$-axis.
Hence you need to evaluate the $z$-coordinate only: $$\bar z=\frac{\iiint_S z dxdydz}{\iiint_S dxdydz}$$ Use the cylindrical coordinates.
Can you take it from here?