Question

Line integrals. Still looking for help.

Evaluate integral F.dr anticlockwise around the ellipse 4x^2+y^2=4 if F(x)=(e^(x)siny+3y)i +(e^(x)cosy+2x-2y)j

very stuck :(

Answer 1

f(x,y) = 4x^2+y^2=4

\( \nabla f(x,y)\) would be normal component

Answer 2

Im sorry i dont really understand. Im trying to approach it with Greens theorem.

Answer 3

I think we need the parametric equation of ellipse

Answer 4

it should be x=cost t and y=2sint for 0<t<2pi

Answer 5

x = acos(t)
y = bsin(t)

dr = idx + jdy

take dot product ... then use green theorem evaluate the integral!!

Answer 6

okay thanks, but what are the limits for the integral- i know the answer in the back of the book indicates the coordinates have been converted to the polar coordinate system but i have a lot of trouble figuring out the limits for r...

Answer 7

also above where you say to take the dot product- of what sorry?

Answer 8

the limits for x is -a to a
for b it's -b to b
use the data to calculate limit for t

Answer 9

dr and F are both vectors ... you gotta take dot product

Answer 10

Oops ... sorry ... if you change all x's and y's to t's in F ... you don't need Green theorem ... you will just be able to calculate it directly ...

sorry gotta go ... :(
have exam in an hour

Answer 11

okay, thanks for the help. Good luck

Answer 12

Anyone else? if i try to find F(r(t)) i get horrible expressions like e^(cost)sin(2sint).....etc

Answer 13

I get the right answer if i plug the horrible expression in my calculator and let it integrate for me, but its not ideal as the answer is in decimal form even though it equals -2pi. and plus you have to show working for exams.

Answer 14

ok so our path C is the curve 4x^2+y^2=4 ---> x^2+y^2/4=1. now we know by Green's theorem that:
\[\int\limits \int\limits_{D}N_{x}-M_{y}dA\]
where M is the first component and N is the second component of F(x)
ok so let's first find N_x-M_y...
N_x=e^xcos(y)+2
M_y=e^xcos(y)+3
N_x-M_y=-1
Let's now look at our region D bounded by the curve C
now our limits of integration are:
-1<=x<=1
-4sqrt(1-x^2)<=y<=4sqrt(1-x^2)
and so we'll have:
\[\int\limits_{-1}^{1}\int\limits_{-2\sqrt{1-x^2}}^{2\sqrt{1-x^2}}-1dydx=\int\limits_{-1}^{1}-4\sqrt{1-x^2}dx=-4(\pi/2)=-2\pi\]
notice that I took out the -4 since that's just a constant and I'm left to integrating sqrt(1-x^2) from -1 to 1. know notice that we are also just finding the area of a semicircle with radius one. so that is just pi/2 then multiply that to -4 to get -2pi

Answer 15

and that's why Green's theorem makes life easier... sometimes :))

Answer 16

anonymoustwo44- thank you so much! very helpful answer!! :)