# Vector calculus for ellipse in polar coordinates

1. May 21, 2013

### dragonxhell

Hello =]

![Question][1]

I know that for a ellipse the parametric is x=a sin t , b= b cos t t:0 to 2pi (?)

for part a) I drew up the graph but not sure if it's right. the circle have 1 radius and for the ellipse I able to find the x= 2
(x-1)^2 =9
sqrt(x-1)^2=sqrt9
x-1=3
x=4

for y=2sqrt2
y^2=8
sqrty=sqrt8
y=sqrt4 sqrt 2
y= 2sqrt2 or 2.8

So the region should be the circle? since the ellipse like cover the whole circle?
Thank you very much for helping!
Cheers.

2. May 21, 2013

### CAF123

That would be valid if the ellipse was centered at the origin, which is not the case here. (but only a slight adjustment needs to be made).

There exists two solutions to each of these equations.

The region is not exactly a circle, but the unit circle is completely contained within the ellipse.

3. May 21, 2013

### dragonxhell

so for x it's 4 and -4 and for y its 2sqrt2 and -2sqrt 2 right?
i did a sketch of it, by any chance you know how to do this question? =(

Mod note: The image was way too large. Please scale the image down to about 900 x 600 pixels and edit this post to include it again, or add it in a new post in this thread.

Last edited by a moderator: May 21, 2013
4. May 21, 2013

### CAF123

No, for the first equation, you solve $(x-1)^2 = 9$ so $x-1 = \pm 3$ What are the two solutions?
Similarly for the other equation.

Your sketch shows the ellipse concentric with the unit circle, but this is incorrect since the ellipse has a centre (1,0). When you find the correct solutions to the above, it should fix things.

5. May 21, 2013

### LCKurtz

@dragonxhell: To get started on this problem you need to substitute the polar coordinate values for $x$ and $y$ into the equation of the ellipse and solve it for $r$. That should lead you to the answer for part b. Then you will use the form$$\int_0^{2\pi} \int_{r_{inner}}^{r_{outer}} \frac 1 {r^3}r\,drd\theta$$

6. May 21, 2013

### dragonxhell

o so the centre is (1,0) and the equation you show me I have already obtained
But don't you move the 1 over? To just simply get x by it self? So it should be 4 and -3 (-4+1?)

7. May 21, 2013

### dragonxhell

Is this right? Can somebody double check this for me? It seem right to me...

(x−1)²/9 + y²/8 = 1
8(x−1)² + 9y² = 72
8x² − 16x + 8 + 9y² = 72
9(x²+y²) − x² − 16x + 8 = 72
9(x²+y²) = x² + 16x + 64
9(x²+y²) = (x + 8)²
9r² = (r cosθ + 8)²3r = r cosθ + 88 + r cosθ = 3r
a = 8, b = 3
c)R is region outside circle: r = 1*and inside ellipse: r = 8/(3−cosθ)
Limits:*1 ≤ r ≤ 8/(3−cosθ)
0 ≤ θ ≤ 2πƒ
= 1/(x²+y²)^(3/2)
= 1/(r²)^(3/2)
= 1/r³∫∫ ƒ dA = ∫ [0 to 2π] ∫ [1 to 8/(3−cosθ)] 1/r³ * r dr dθR. . . . .
= ∫ [0 to 2π] ∫ [1 to 8/(3−cosθ)] 1/r² dr dθ. . . . .
= ∫ [0 to 2π] (−1/r) | [1 to 8/(3−cosθ)] dθ. . . . .
= ∫ [0 to 2π] (−(3−cosθ)/8 − (−1)) dθ. . . . . = ∫ [0 to 2π] 1/8 (cosθ + 5) dθ. . . . .
= 1/8 (sinθ + 5θ) | [0 to 2π]*. . . . .
= 1/8 [(0 + 10π) − (0 + 0)]. . . . .
= 5π/4

Also for part a) the R region it is the region between the ellipse to the circle right?

8. May 21, 2013

### LCKurtz

Yes. Very nice.

9. May 22, 2013

### dragonxhell

Can you please check this question for me because my answer is different to my friends and they say its wrong...
Given the equation: G= (x^3 -3xy^2)I +(y^3-3x^2y)j +2k is conservative find the potential function for the fields that are conservative
fx=x^3-3xy^2
fy=y^3-3x^2y
fz=z

integral (x^3-3xy^2)dx
=x^4/4-3x^2y^2/2 +h(y, z)
fy or f'y= -3x^2 y +hy(y, z)
fy=y^3-3x^2y= -3x^2y+hy(y, z)
hy(y,z)=y^3

Integral y^3 dy = y^4/4 +gz
fz=0 because no z term ^
Integral 0dz =c
F= x^4/4-3x^2y^2/2 +y^4/4+c <-potential function

10. May 22, 2013

### LCKurtz

There is no F anywhere. Don't you mean grad f = G?

Is $f_z = z$ or $f_z = 2$? Look at G.
Poorly written. At this point you have$$f = \frac {x^4} 4 - \frac {3x^2y^2} 2 +\frac{y^4} 4 + h(z)$$
Now you need to take $f_z$ and set it equal to the third term, whether it is a $2$ or a $z$. You don't get $f_z=0$ either way.

11. May 22, 2013

### dragonxhell

sorry I type it all up on my phone and when I submit it it's so messy....
so you are saying
my integral y^4/4 must equal to fz which is z