Line integral over vector field of a shifted ellipse

[TheBoro76]

This is part of a larger question, but this is the part I am having difficulty with. I have had an attempt, but am not sure where I am making a mistake. Any help would be very, very appreciated.
1. Homework Statement
Let C2 be the part of an ellipse with centre at (4,0), horizontal semi-axis a=5, and vertical semi-axis b=3, from (0,-9/5) to (0,9/5) (i.e. anti-clockwise)

calculate:

$$\int_{C2} \mathbf v \cdot d\mathbf r$$
where $\mathbf v = \frac{1}{2}(-y\mathbf i + x \mathbf j)$

Hint: use t:-t0 ->t0 as limits when parametrising #C_2# and explain why cos(t0)=-4/5 and sin(t0)=3/5

Homework Equations

For the integral:
$$I=\int_{C2} \mathbf v \cdot d\mathbf r = \int_a^b v_1dx+v_2dy$$

The Attempt at a Solution

I first parametrised the ellipse, by considering the ellipse as a stretched and shifted unit circle. Using this, I get
x=5cos(t)+4
y=3sin(t)

when substituting in t0 from the question, I get x=0 and y=9/5, as expected. I also get the expected values of x=9 and y=0 when t=0.

differentiating these I get:
dx=-5sin(t) dt
dy=3cos(t) dt

and from $\mathbf v$ I get:
$v_1=-y$ and $v_2=x$

so substituting these values into the equation for the integral, I get:
$$I=\frac{1}{2}\int_{-t_0}^{t_0}-ydx+xdy$$
$$\rightarrow I=\frac{1}{2}\int_{-t_0}^{t_0}-3sin(t)*-5sin(t)+(5cos(t)+4)(3cos(t)dt$$
$$I=\frac{1}{2} \int_{-t_0}^{t_0}15(sin^2(t)+cos^2(t))+12cos(t) dt=\frac{1}{2}\int_{-t_0}^{t_0}15+12cos(t) dt$$
$$=\frac{1}{2}(15t+12sin(t))|_{-t_0}^{t_0}=\frac{1}{2}(30(t_0)+24sin(t_0))$$
Using the values of t0, from the question, I get:
$$=15*sin^{-1}(\frac{3}{5})+\frac{36}{5} \approx 16.85$$

But, using Greens theorem, this is an expression for the area of this part of the ellipse (when considering the closed integral of C=C1 and C2 and C1 is y=0). I know the area of an ellipse is $A=\pi ab$. For the ellipse in the question, the area would be about $15\pi\approx47.12$. Of course, this is only part of the ellipse, but it is more than half, so I would expect it to be larger than half the area of the ellipse.

Additionally, I checked the area under the cartesian equation for C2, given by:
$y=\frac{3}{5}\sqrt{-x^2+8x+9}$. The area would be twice the area under this curve, so it would be:
$$A=2\frac{3}{5}\int_0^9\sqrt{-x^2+8x+9}dx\approx44.67$$, which is closer to what I would expect.

I don't know what the actual answer is, but I'm pretty sure I'm wrong. Any advice or hints would be much appreciated

Thanks in advance!

 

[andrewkirk]

Perhaps it's to do with how you use Green's theorem.
To use it to calculate area don't you need $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$? In this case, with $P(x,y)=-y$ and $Q(x,y)=+x$, what is $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}$?

 

[TheBoro76]

Hi thanks for the reply andrewkirk. My understanding is that you do need $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}=1$. But because there is a factor of 1/2 in the $\mathbf v$, P(x,y)=-y/2 and Q(x,y)=x/2 so $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{1}{2}-\,-\frac{1}{2}=1$ ?

 

[SteamKing]

This is part of a larger question, but this is the part I am having difficulty with. I have had an attempt, but am not sure where I am making a mistake. Any help would be very, very appreciated.
1. Homework Statement
Let C2 be the part of an ellipse with centre at (4,0), horizontal semi-axis a=5, and vertical semi-axis b=3, from (0,-9/5) to (0,9/5) (i.e. anti-clockwise)

I think if you make a quick sketch of the ellipse described by C2 and plot the end points on that sketch, you will find that the first point is in the third quadrant w.r.t. the center of the ellipse and the second is in the second quadrant. Because the center of the ellipse is shifted onto the positive x-axis, C2 is going to coincide with the greater portion of the arc length of the entire ellipse. Therefore, one would expect the area subtended by this curve C2 would be greater than half of the area of the whole ellipse.

Since the problem hints at making the limits of C2 go from -t0 to t0, then just blindly plugging in -t0 and t0 will not result in the correct evaluation of the line integral, since the angles used in the parameterization of the ellipse are supposed to run CCW. If you use symmetry in evaluating the line integral, I believe you will obtain the correct result, but because C2 crosses the x-axis when going from -t0 to t0, I think you must handle the line integral evaluation very carefully so that you don't get an incorrect result, which you apparently have done.

 

[andrewkirk]

The problem might be that you are choosing 0.6435 as arcsin of 3/5. The angle t0 needs to be between pi/2 and pi, because of the quadrants the start and end points are in (as SK points out), so one should choose the arcsin instead as pi-0.6435. If you use that instead, I think you'll get the right result.

 

[TheBoro76]

Thanks for your responses I tried both of your suggestions. but couldn't quite get them to work (I probably made a mistake somewhere). However, if I used the value t0=arccos(-4/5), I got the answer I expected, So I'm going with that.
Thanks again!