Question

Verify my answer!

Green's Theorem anyone ?

Answer 1

Evaluate by Green’s Theorem

∫c [(3x^2-8y^2 )dx+(4y-6xy)dy]

where C is the boundary of the region bounded= by y=√x,y=x^2

Answer 2

Verify formula too, and let me know if i have done any mistake in any step...

Answer 3

isnt it dp/dy-dq/dx instead of -dp/dy+dq/dx, does that not make a difference

Answer 4

reference ?

Answer 5

http://en.wikipedia.org/wiki/Green's_theorem

Answer 6

nvm u are right i forgot that for the cross product its DEL X vector not Vec X del first

Answer 7

\[\oint _C F.dr = \iint_R curl(F) dA\]

Answer 8

ok looks good!

Answer 9

hey hartnn the doc is not opening for me...

Answer 10

if i copy paste here, it'll look ugly

Answer 11

Evaluate by Green’s Theorem
∫_c▒〖[(3x^2-8y^2 )dx+(4y-6xy)dy] where C is the boundary of the region bounded= by y=√x,y=x^2 〗
Solution: Module 4
F.dr= (3x^2-8y^2 )dx+(4y-6xy)dy….(1)
Let P=3x^2-8y^2 and Q=4y-6xy
∴∂P/∂y=-16y and ∂Q/∂x= -6y
By Green’s Theorem,∫_c▒〖Pdx+Qdy=∬_R▒〖(∂Q/∂x-∂P/∂y)dx dy〗〗
∫_c▒〖(3x^2-8y^2 )dx+(4y-6xy)dy=∬_R▒〖(-6y+16y)dx dy…(2)〗〗
From 1 and 2,∫_c▒〖F.dr= ∬_R▒〖10y dx dy〗〗
Since we need to integrate w.r.t x first,
y=√x gives x=y^2
y=x^2 gives x=+√y
Also, √x=x^2, gives x=0,1
when x=0,y=0, when x=1,y=1
So, the limits for y are from 0 to 1.


∫_c▒〖F.dr= 〗 10 ∫_(y=0)^(y=1 )▒〖y [∫_(x=y^2)^(x=√y)▒〖dx 〗] 〗 dy=10 ∫_(y=0)^(y=1)▒y [x]_(y^2)^√y dy
= 10 ∫_(y=0)^(y=1)▒y[√y-y^2]dy
=10 ∫_0^1▒〖(y〗^(3/2) -y^3)dy
=10 [y^(5/2)/(5/2)-y^4/4]_0^1=10 [2/5-1/4]=4-5/2
=3/2

Answer 12

basically i got the answer as 3/2 :P

Answer 13

3/2 is the final answer ?

Answer 14

what software are you using for thaose pictures

Answer 15

i got it as 3/2, not sure whether its correct

for that pic?? i used google.com :P

Answer 16

\(\huge ∬_R10y dx dy\)

one of the middle step

R is region bounded by
y= sqrt x
y=x^2

Answer 17

another middle step
\(10 \int \limits_0^1 y[√y-y^2]dy \)

Answer 18

firstly, whether F.dr = 10y is correct or not...

Answer 19

http://www.wolframalpha.com/input/?i=%5Cint+%5Climits_0%5E1+%5Cint+%5Climits_%28x%5E2%29%5E%28sqrt%28x%29%29+10y+dydx

Answer 20

3/2 is right ^

Answer 21

awesome!
so steps before setting up that integral are correct or not...

Answer 22

everything looks good to me in that matrix code lol

Answer 23

Thanks :)

Answer 24

my msoffice got messed up need a reinstall, np :)

Answer 25

you should save it on google docs

Answer 26

everyone has google accounts these days

Answer 27

tried that also :
https://drive.google.com/file/d/0B7ss_Y8M_VEgdWJVMXdQOWpBYlk/edit?usp=sharing

Answer 28

it wont show the equations or any other advanced formatting >.<

Answer 29

that said
"You need permission"

Answer 30

sorry, try now ... it wil work

Answer 31

haha! can't see any equations :P

Answer 32

nvm,
needed confirmation on this part only after wolf confirmed that integral = 1.5

F.dr= (3x^2-8y^2 )dx+(4y-6xy)dy….(1)
Let P=3x^2-8y^2 and Q=4y-6xy
∴∂P/∂y=-16y and ∂Q/∂x= -6y

Answer 33

\(\vec{F}.d\vec{r} = Mdx + Ndy\)


\(Curl(\vec{F}) = Nx - My\)

Answer 34

\(Curl(\vec{F}) = 10y\) is perfectly right ! and yes after that its a trivial double integral

Answer 35

thanks again :)
back to helping :)

Answer 36

np :)