# Mass and center of mass of lamina in polar coordinates

I need some help with the following problem which is question number 15.5.4 in the seventh edition of Stewart Calculus. Here is the problem definition:

"Find the mass and center of mass of the lamina that occupies the region D and has the given density function $\rho$, where: $D={(x,y) | 0\le x \le a, 0 \le y \le b}$ and $\rho (x,y) =1+x^2+y^2$"

I did this in rectangular coordinates, but the work and answer are too complicated. I need help doing this in polar coordinates.

I see that $z=1+x^2 +y^2=1+r^2$, the graph of which is easy to visualize.

I need help getting started in converting the following into polar coordinates:

$m=\int\int_D \rho(x,y) dA =\int_0^a\int_0^b(1+x^2+y^2)dy dx$
$\bar{x}=\frac{1}{m}\int\int_Dx\rho(x,y)dA$
$\bar{y}=\frac{1}{m}\int\int_Dy\rho(x,y)dA$
Then solve for center of mass $(\bar{x},\bar{y})$

It would seem obvious that $m=\int\int_D \rho(x,y) dA =\int\int_D (1+r^2)r dr d\theta$, but the range of integration is what I do not understand. I tried using $0\le r\le \frac{b}{sin{\theta}}$ and $0\le \theta \le \arcsin{\frac{b}{r}}$ , but got an undefined result from my TI-89 calculator.

If someone can show me how to set up these integrals in polar coordinates, I think I could do the integration myself. However, I would hope to have someone check my answers to the integrals so that I make sure to geth the mass and center of mass correct.

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.

• Thank you and +1 for taking the time to answer this old question. – CodeMed May 5 '15 at 19:54