Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.