Calculate center of mass multiple integrals Can you help me with this problem?
Find the center of mass of a lamina whose region R is given by the inequality:

and the density in the point (x,y) is :

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
 A: The integral for total mass is $$\int_{xmin}^{xmax}\int_{ymin}^{ymax} \delta(x,y) \ dy \ dx$$ where the range $(xmin,xmax)$ is the whole range of the shape and the range $(ymin,ymax)$ is the range in $y$ at a given $x$.  Can you put the correct limits on?  Can you do the integral?
A: Draw an $s$-axis like so: $\nearrow$, and a $t$-axis like so: $\nwarrow$. Then the blue square appears in $(s,t)$-coordinates as
$$Q=\left\{(s,t)\>\biggm|\>|s|\leq{1\over\sqrt{2}}, \ |t|\leq{1\over\sqrt{2}}\right\}\ .$$
Due to symmetry the center of mass has $(s,t)$-coordinates $(\sigma,0)$, where it remains to determine $\sigma$. Since $x+y=\sqrt{2} s$ we have to solve
$$\sigma\int_Q e^{\sqrt{2}s}\ {\rm d}(s,t)=\int_Q s\>e^{\sqrt{2}s}\ {\rm d}(s,t)\ .$$
This immediately reduces to
$$\sigma\int_{-1/\sqrt{2}}^{1/\sqrt{2}}e^{\sqrt{2}s}\>ds=\int_{-1/\sqrt{2}}^{1/\sqrt{2}}s\> e^{\sqrt{2}s}\>ds\ .\tag{1}$$
Computing the integrals in $(1)$ leads to
$$\sigma={\sqrt{2}\over e^2-1}\ .$$
Therefore the $(x,y)$-coordinates of the centroid are given by
$${1\over e^2-1}\>(1,1)\ .$$
A: Rotate the entire thingy by angle $\frac{\pi}{4}$. Integrations will be
$$\int_{-1/\sqrt{2}}^{+1/\sqrt{2}}dx\int_{-1/\sqrt{2}}^{+1/\sqrt{2}}dy \;\;\delta'(x,y)$$
etc. where
$$\delta'(x,y) = \delta\left(\frac{x+y}{\sqrt{2}}, \frac{-x+y}{\sqrt{2}}\right)=e^{y\sqrt{2}}$$
