Question

verify green's theorem in the plane for integral (3x^2-8y^2)dx + (4y-6xy)dy, where C is the boundry of the region defined by x=0, y=0 and x + y=1

Answer 1

\[\int\limits_cPdx+Qdy=\int\int\limits_D\left(\frac{\partial Q}{dx}-\frac{\partial P}{dy}\right)dA\]so first we might do well to draw D

Answer 2

so it looks like our limits are pretty easy to find. Think you can tell me the bounds of our double integral?

Answer 3

@kay-jay please try to stick around for a moment after posting so we can interact on this question

Answer 4

am here

Answer 5

cool, so do you think you can find the bounds of our double integral now that I have drawn out the region D ?

Answer 6

yes but not so sure about them

Answer 7

well what does x vary from?

Answer 8

from what to what...?

Answer 9

0 to 1

Answer 10

and how about y, what is it bound by?

Answer 11

y=x

Answer 12

not quite

Answer 13

x+y=1
so solving for y gives...?

Answer 14

the question is 15 marks how about we start with line ine integration

Answer 15

y=1-x

Answer 16

well Green's theorem is the easier part I think
doing this without greens theorem will require me to look at my notes...

Answer 17

yes, so what are the bounds on y ?

Answer 18

y= -1 m not so sure

Answer 19

calculating partal integrals would be easy...but coming to the intervals
x interval would be from 0 to 1,and
y integral would be from 0 to 1-x ...(as x+y=1 is given)

Answer 20

exactly^\[0\le x\le1\]\[0\le y\le1-x\]so those will be our bounds on the integral
now you just need to do the\[\left(\frac{\partial Q}{dx}-\frac{\partial P}{dy}\right)\]part, which should be pretty easy by this time in the course

Answer 21

then what to do now

Answer 22

please help me... please

Answer 23

what did you get for the expression above\[\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\]???

Answer 24

i just don't get it

Answer 25

oh yah let me work it out

Answer 26

exectly what is the answer suppose to be

Answer 27

I don't know for sure because I am having difficulty with the line integral method, but by Green's Theorem I got \(\large -\frac72\)

Answer 28

ohky please do it for me and i will compare it with my solution

Answer 29

I'm not going to type out the whole thing, but I will walk you through it.
just do the partial derivative part\[\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\]and tell me what you get
\that is a good start

Answer 30

i got - \[\left(\begin{matrix}11 \\ 3\end{matrix}\right)\]

Answer 31

derivative of N=-6y and the derivative of M=-16y

Answer 32

ok, so one minus the other is....?

Answer 33

ohk let me try it again

Answer 34

\[N_x-M_y=?\]

Answer 35

your partials are right, but you are just not doing the algebra correctly :P

Answer 36

eix... practise makes perfect i wl kep trying

Answer 37

N-M=10y

Answer 38

there we go :)

Answer 39

I had the wrong answer though too, let me redo it

Answer 40

now I get -5/3

Answer 41

i got \[\left(\begin{matrix}5 \\ 3\end{matrix}\right)\]

Answer 42

oh yeah, dang negative sign :P
you're right

Answer 43

by the way, to write fractions write \frac53

Answer 44

oh... thak you but it is 15 marks

Answer 45

and put it into the equation editor, or enclose it in brackets if you know latex

Answer 46

ohk thanks

Answer 47

I'm fairly confident we have done the green's theorem part right, and that 5/3 is the right answer, but the line integral is giving me a headache

Answer 48

and thats the part i must gain marks in cause green's theorem is only 5 marks

Answer 49

I'm working on it, but I am unsure how to paramaterize t
I am thinking everything from 0 to 1
let me play with it for a bit...

Answer 50

please do

Answer 51

so I'm thinking to parameterize the curves we do\[C_1:x=t,y=0~~~~~~~~~~~0\le t\le1\]\[C_2:x=t,y=1-t~~~~~0\le t\le1\]\[C_3:x=0,y=t~~~~~~~~~~~~0\le t\le1\]