# Thread: Finding the mass of a lamina?

1. ## Finding the mass of a lamina?

Here is the question...

Consider the lamina in the first quadrant of the XY plane bounded by the curve $\displaystyle x^2+y^2=R$. The density of the material is given by $\displaystyle \rho(x,y)=xy$.

a. Use rectangular coordinates to find the mass of this lamina.
b. Use polar coordinates to find the mass of this lamina.

The main thing I don't understand is how to find the limits of the area. I know I need to use $\displaystyle x^2+y^2=R$ and find its limits in the first quadrant, but I don't know what to do with the R. Can someone help me out with this? Thanks.

2. anyone?

3. In polar coordinates is...

$\displaystyle x=r\cdot \cos \theta$

$\displaystyle y=r\cdot \sin \theta$ (1)

... so that is...

$\displaystyle \rho(r,\theta)= r^{2}\cdot \sin \theta \cdot \cos \theta$ (2)

... and therefore ...

$\displaystyle M = \int_{0}^{R} \int_{0}^{\frac{\pi}{2}} r^{3}\cdot \sin \theta \cdot \cos \theta \cdot dr\cdot d\theta$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by Infernorage
Here is the question...

Consider the lamina in the first quadrant of the XY plane bounded by the curve $\displaystyle x^2+y^2=R$. The density of the material is given by $\displaystyle \rho(x,y)=xy$.

a. Use rectangular coordinates to find the mass of this lamina.
b. Use polar coordinates to find the mass of this lamina.

The main thing I don't understand is how to find the limits of the area. I know I need to use $\displaystyle x^2+y^2=R$ and find its limits in the first quadrant, but I don't know what to do with the R. Can someone help me out with this? Thanks.
First, dont' complain just because you don't get a response withing two hours! We have lives, you know.

Second, $\displaystyle x^2+ y^2= R^2$ is a circle with center at the origin and radius R. The first thing you should have done is draw at least a rough graph of that. The part in the first quadrant is a quarter of that circle. Now you have to decide whether the "inside integral" is with respect to x or y. It really doesn't matter here because of the symmetry but because it is more common, lets decide to do the "inside integral" with respect to y and the "outside" with respect to x.

Having decided that the "outside" integral will be with respect to x and so its limits must be constants, We see that to cover the entire circle, we must go from the the leftmost edge, x=0 to the rightmost point on the circle, (R, 0). The limits of integration on the "outside" integral will be from x= 0 to x= R.

Now, what are the limits on y for each x? Draw a vertical line on the quarter circle representing a specific value of x and you see that y goes from the x-axis, y= 0, to the circle. That circle is given by $\displaystyle x^2+ y^2= R^2$ so $\displaystyle y^2= R^2- x^2$ and $\displaystyle y= \pm\sqrt{R^2- x^2}$. Since we only want the part in the first quadrant, with $\displaystyle y\ge 0$, we only want $\displaystyle y= \sqrt{R^2- x^2}$.

Your integral is [tex]\int_{x= 0}^R\int_{y= 0}^{\sqrt{R^2-x^2}} xy dydx.

In polar coordinates, it is much easier because of the symmetry. Just as in rectangular coordinates, integrals over rectangles have constant limits of integeration, so here we have constant limits of integration. In order to cover the entire quarter circle, [/itex]\theta[/itex] must go from 0 to $\displaystyle \pi/2$. For each $\theta$, then, r goes from 0 out to R. The limits of integration of $\theta= 0$ to $\pi/2$, and r= 0 to R. Don't forget that the "differential of area" in polar coordinates is $\displaystyle r drd\theta$, not just $\displaystyle drd\theta$, and that $\displaystyle xy= (r cos(\theta))(r sin(\theta))= r^2 cos(\theta)sin(\theta)$.

5. Thanks for the info. There is only one thing I don't understand. HallsofIvy, you said that the equation $\displaystyle x^2+y^2= R^2$ is a circle with radius R, but the equation given was $\displaystyle x^2+ y^2= R$. Shouldn't the radius be $\displaystyle \sqrt{R}$ and the equation for y be $\displaystyle y= \sqrt{R- x^2}$? Sorry, just a little confused about that. Thanks.