Question

How do you convert the ellipse 9x^2 + 16y^2 = 144 into polar coordinates?

Answer 1

x=r cos theta and y=r sin theta
You can plug those in and mess with it to see if you can get something nice.
There are also some versions used in astronomy. The include the definition of eccentricity... you would have to look those up to convert to those.

Answer 2

Yeah I have plugged those in for the equation. What do I do with the 144 though? Is that used in solving for r? 144r^2 = 9x^2 + 16y^2, which results in r = x/4 + y/3 ?

Answer 3

Once you do the substitution, the x's and y's should be gone. I don't think that you will find any solution that is elegant without going all the way to those astronomy versions.

Answer 4

I am doing a double integral with C being that ellipse, then using Green's Theorem to convert into a double integral. I'm just having trouble translating the given ellipse into proper bounds of integration

Answer 5

The standard form for the ellipse would actually be\[\frac{x ^{2}}{16}+\frac{y ^{2}}{9}=1\]

Answer 6

Oh, wow... would parametric be better? That's easier.

Answer 7

x=4cos(t) and y=3sin(t) over [0,2pi]

Answer 8

Ok I will give that try, just need to refresh myself on parametric stuff to find the bounds of integration using parametric equations

Answer 9

And I need to refresh on Green's to see what that takes... good luck.

Answer 10

Thanks for the help!

Answer 11

From your perspective, they can be seen as x and y components of vector valued functions. Substitute x values with 4 cos t, y values with 3 sin t, dx with -4 sin t, and dy with 3 cos t. Then integrate from 0 to 2pi... I think. :)

Answer 12

Maybe I should just be quiet... I'm really rusty. :)

Answer 13

What I ended up doing was using Green;s theorem to convert the equation in the original integral in M(x,y) and N(x,y). Then I did the partial derivative of M with respect to y and the partial derivative of N with respect to x

Answer 14

Both of those are 4x, so given Green;s theorem, that would be 4x - 4x

Answer 15

Resulting in an area of 0, because the integral of 4x-4x would be 0

Answer 16

Closed path and conservative vector field should result in zero. Did you test if it was conservative to start with?

Answer 17

Wait... yeah... that is the test... It's conservative!

Answer 18

Yeah, so when using Green's theorem I guess if it is conservative, then the area will be 0

Answer 19

Yes, because to use Green's requires a closed path.

Answer 20

Exactly. But for future reference, it's best to use parametric equations for ellipses, circles or lines and polar coordinates for other curves?

Answer 21

I would say so... parametric is almost the same as vector valued functions, so they relate well.

Answer 22

Alright awesome. Calc 3 is almost done, hallelujah!

Answer 23

Thanks again for all your help

Answer 24

Right on! You're welcome.