Finding Moments for Elliptical Region

  • Thread starter Thread starter City88
  • Start date Start date
  • Tags Tags
    Moments
City88
Messages
13
Reaction score
0

Homework Statement



I'm not sure how to go about finding the following moments:

<br /> M_{x}= \int \int\ y dx dy
<br /> M_{y}= \int \int\ x dx dy

Where the region is bounded by the ellipse:
\frac{(x-2)^2}{16}} + \frac{(y-4)^2}{36}} = 1


Homework Equations


Listed above...

The Attempt at a Solution



I drew the ellipse and found the bounds to be
-2 \leq y \leq10
-2 \leq x \leq 6

Then I tried integrating with those bounds, but I can't seem to get the right answers. Any help/hints would be greatly appreciated.
 
Physics news on Phys.org
Those boundaries describe a rectangle, not an ellipse. Do you see why?
 
Ooohh yes, you're right.

So does this mean that I have to find the bounds of my integral as functions?
Would I have to find the equation of the ellipse in terms of y, and make that my bounds for the y-integral. and then keep the x integral as numbers?

like D = {(x,y)| a \leq x \leq b, g_{1}\leq y \leq g_{2}}
 
Yes that's what you would do. This problem may be easier in polar coordinates.
 
Actually I just found out that I might have to use polar coordinates.
But since the region is not a circle, but an ellipse, how would I be able to write my x and y in polar coordinate form?
Normally we have x=rcos(theta), and y=rsin(theta) for a circle.

would it be...
x=4rcos(theta) + 2
y=6rsin(theta) + 4
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top