# 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.

-

## 1 Answer

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