Convert into polar coordinates

[City88]

Homework Statement

I'm trying to solve a double integral of a function which is bounded by the ellipse:
$$\frac{(x-2)^2}{16}} + \frac{(y-4)^2}{36}}$$ = 1
And I can't figure out how to write this in polar coordinate form, and also what my bounds for theta and radius would be.

Homework Equations

I know that for a circle, the polar coordinates are
x=rcos(theta) and y=rsin(theta)
theta going from 0 to 2pi, and r going from o to a(being the radius)

The Attempt at a Solution

Well I know that an ellipse is not a circle! so I can't apply this formula without doing some tweaking. Can anyone help with this "tweaking"?

I have started by thinking it would be something like..
x=rcos(theta) + 2
y=rsin(theta) + 4
Because the ellipse is not in the center, but I can't figure out what else to do...

 

[nicksauce]

Probably the easiest thing to do is to make a change of variables, say u = x-2, v = y-4.

 

[tiny-tim]

squashed circle

Hi City88!

(have a theta: θ and a pi: π )

I don't understand why you want to convert to polar coordinates.

Does this help? … An ellipse is a squashed circle, so you can represent it as x = a cosθ, y = b sinθ.

 

[City88]

Even after doing that, I'm still stuck..

x-2 = u = rcos(theta)
y-4 = v = rsin(theta)

Wouldn't I still have to add (more like multiply) something to the rcos(t)? The radius still isn't circular because the u² and v²'s are divided by 4² and 6².

 

[nicksauce]

[Deleted]

 

[City88]

Tiny-tim, my original question is
https://www.physicsforums.com/showthread.php?t=243360

I'm trying to solve the double integration, and can't figure out any other way to do it, well other than polar coordinates.

thanks for the θ and π!

 

[tiny-tim]

Tiny-tim, my original question is …

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

Hi City88!

You'd only convert to polar coordinates if your integrand was simpler in polar coordinates.

In this case, your integrand is either x or y which are obviously simpler in … x and y coordinates!

Just integrate along strips of thickness dx or dy.

 

[nicksauce]

Yes, earlier I wrote
"This problem may be easier in polar coordinates."... but I was using 'may' in the sense that I had not really thought about it.

 

[City88]

Thats what I first thought also...but then I realized the equation was just waaaay too messy for my liking. So then I figured polar coordinates may have to be used, since an ellipse sort of resembled a circle.

I just tried doing it again, but by keeping it in terms of x and y...and well it's messy!

for example, if I integrate with respect to x first, my bounds for that integral would be something like..

x goes from $$\pm$$ $$4\sqrt{1-\frac{(y-4)^{2}}{6^{2}}}+2$$

so is it just me, or does integrating over those bounds seem frightening? I feel as though there should be an easier way to do this. But I guess I'll just integrate this mess and try to make it work!

 

[tiny-tim]

messy bounds

… for example, if I integrate with respect to x first, my bounds for that integral would be something like..

x goes from $$\pm$$ $$4\sqrt{1-\frac{(y-4)^{2}}{6^{2}}}+2$$

so is it just me, or does integrating over those bounds seem frightening? I feel as though there should be an easier way to do this. But I guess I'll just integrate this mess and try to make it work!

Hi City88!

Messy bounds in the first integral don't matter, so long as the integrand is easy.

Of course, messy bounds in the first integral might make a messy integrand in the second integral …

so did it … ?