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**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

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!!