Center of mass Double Integral?

Question

Can you help me with this problem?

Find the center of mass of a lamina whose region $R$ is given by the inequality: $$|x|+|y|\le 1,$$

and the density in the point $(x,y)$ is : $$\delta(x,y)=e^{x+y}.$$

The region $R$ is this one:

Is this the proper way to set up the integral for m:

$$\int_{-1}^{1}\int_{-x-1}^{x+1} \ e^{x+y} \ dy \ dx$$

Any help? Thanks

Answer 1

By definition the center of mass is situated in $$r_c = \frac{\int_{\Bbb R^2} r \rho(r)dr}{\int_{\Bbb R^2} \rho(r)dr},$$ therefore in your case you need to find the integrals

$$\int_{R} \exp(x+y)dxdy, \quad \int_{R} x\exp(x+y)dxdy, \quad \int_{R} y\exp(x+y)dxdy,$$ Where $R=\{(x,y)\in \Bbb R^2:|x|+|y|\le 1\}$.

Can you find these integrals?

Edit

The mass is found via $$M = \int_R \delta(x,y)dxdx = \int_{x=-1}^{x=1}\left( \int_{y=|x|-1}^{y=1-|x|}\exp(x+y)dy\right)dx$$

Answer 2

The integral should really look like

$$\frac{\displaystyle \iint_R dx \, dy \; (x,y) \delta}{\displaystyle \iint_R dx \, dy \delta} $$

The area in of the region $R$ is obviously $2$, but it is worth setting up the integral so you see that it is correct:

$$\begin{align}\iint_R dx \, dy &= \int_{-1}^0 dx \, \int_{-x-1}^{x+1} dy + \int_{0}^1 dx \, \int_{x-1}^{-x+1} dy \\ &= \int_{-1}^0 dx \, (2 x+2)- \int_0^1 dx \, (2 x-2) \\ &= [x^2+2 x]_{-1}^0 - [x^2-2 x]_0^1 \\ &= 1 - (-1) = 2 \end{align}$$

So, good. The denominator should now be clear:

$$\begin{align}\iint_R dx \, dy &= \int_{-1}^0 dx \, e^x \int_{-x-1}^{x+1} dy \,e^y + \int_{0}^1 dx \,e^x \int_{x-1}^{-x+1} dy \, e^y \\ &= \int_{-1}^0 \, dx \left ( e^{2 x+1} - e^{-1} \right ) + \int_0^1 dx \left (e - e^{2 x-1} \right ) \\ &= \frac12 \left (e-e^{-1} \right )+ e^{-1} + e - \frac12 \left ( e - e^{-1}\right )\\ &= e+e^{-1}\end{align} $$

The $x$ center of mass is

$$\begin{align}\iint_R dx \, dy x \delta &= \int_{-1}^0 dx \,x \, e^x \int_{-x-1}^{x+1} dy \,e^y + \int_{0}^1 dx\, x \,e^x \int_{x-1}^{-x+1} dy \, e^y \\ &=\int_{-1}^0 \, dx\, x \left ( e^{2 x+1} - e^{-1} \right ) + \int_0^1 dx \, x\left (e - e^{2 x-1} \right ) \end{align}$$

I think you can take it from here...