Question

Evaluate the line integral directly
\[\oint_c xydx+x^2dy\]

Answer 1

What is c?

Answer 2

from 0 to 1

Answer 3

(0,0) to (1,0)?

Answer 4

Sorry it's (0,0) (0,1) (3,0) (3,1)

Answer 5

\[\int_0^1 \int_0^3 xydx+x^2 dy?\]

Answer 6

you need to parametrize your path

Answer 7

what do you mean?

Answer 8

have you ever done any of these problems before?

Answer 9

No, I have evaluated the integral using greens theorem, but I am asked to evaluate it directly. How is that different from greens theorem?

Answer 10

http://www.math.dartmouth.edu/~m43s12/notes/class6.pdf

Answer 11

do I use \[\int_c f(x,y)ds\]
?

Answer 12

from (0,0) to (0,1) the integral is zero.
x=0 dx =0
y=t dy =dt
The quantity to be integrated is zero. So the integral is zero.

Answer 13

From (3,0) to (3,1)
x=3 , dx =0
y=t dy =dt
\[
xydx+x^2dy= 9 dt\\
\int_0^1 9 dt =9
\]

Answer 14

Is your c the boundary of the rectangle?

Answer 15

I believe so

Answer 16

From (0,0) to (3,0)
y=0 dy=0
x=t dx =dt
xydx+x^2dy=0
So the integral is zero

Answer 17

From (3,1) to (0,1)
y=1 dy =0
x=3 dx =0
xydx+x^2dy=0
The integral is zero

Answer 18

but using greens theorem I get 4.5

Answer 19

I also got 4.5 using Green's Theorem. Let me double check.

Answer 20

Here is my mistake
From (3,1) to (0,1)
y=1 dy =0
x=t dx =dt
xydx+x^2dy=t dt
The integral
\[
\int_3^0 t dt=-\frac 92
\]

0 + 9-9/2 + 0 =9/2

Answer 21

Are you still there?

Answer 22

Yep sorry I had to go for a little bit, but I'm back. Oh that makes sense now.

Answer 23

Thank you so much!

Answer 24

yw