# Finding Moments

1. Jul 3, 2008

### City88

1. The problem statement, all variables and given/known data

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

$$M_{x}= \int \int\$$ y dx dy
$$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

2. Relevant equations
Listed above...

3. 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.

2. Jul 3, 2008

### nicksauce

Those boundaries describe a rectangle, not an ellipse. Do you see why?

3. Jul 3, 2008

### City88

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)| $$a \leq x \leq b, g_{1}\leq y \leq g_{2}$$}

4. Jul 3, 2008

### nicksauce

Yes that's what you would do. This problem may be easier in polar coordinates.

5. Jul 5, 2008

### City88

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: Jul 5, 2008