Question

Can someone check if this is correct? I need to calculate the curvilinear integral...

Answer 1

This is what i did:
\[\int\limits_{0}^{4} \sqrt{2x}\sqrt{1+ \frac{ 1 }{ 2x }} dx\]
And after calculating it i got 26/3

Answer 2

in the assignment i have that the y^2 = 2x
so do i just use sqrt(2x) as y and calculate its derivative or do i calculate the derivative for 2x only?

Answer 3

Also i have two more assignments with the similar problem and i am not sure where to start with these:

Answer 4

I've done the z2 one, by putting the limits 0 and 1 in the integral and replacing dy with 2xdx and y with x^2
\[\int\limits_{0}^{1} 2xx^2dx+x^22xdx = \int\limits_{0}^{1}4x^3dx=1\]

Answer 5

for the first one,
did you use this:
\[ds= \sqrt{1 + (y')^2}dx\]
?

Answer 6

yes that is the \[\sqrt{1+\frac{ 1 }{ 2x }}\]

Answer 7

i am just not sure if my y'^2 is correct

Answer 8

yep, that part looks good,
how did you figure out the limits?

Answer 9

i drew the picture and x goes from 0 to 4

Answer 10

yep, that's right, its simply 0 to 4

Answer 11

so the first one is good?
how about the second one?
where do i start?

Answer 12

the graph looks like

Answer 13

a is grater than 0 so just the top side then?

Answer 14

top right

Answer 15

i think so...
what do you think @phi

Answer 16

i think thats the way to go...
substitute y=a-x
the limits would be from x=0 to x=a
then do the same thing with ds as earlier...

Answer 17

I think the path is the diamond. For example if you are at (0, -a) (lower point)
we get 0 + a = a which shows that point is on the path.

I am thinking we do 4 times the answer for one leg (but I'm only 90% sure of that )

Answer 18

yep...
on second thought i agree, the path has to be the entire diamond

Answer 19

so does that mean i have 4 y equations?

Answer 20

ok, all the paths have the same (y')^2 = 1

Answer 21

if y= a-1 isn't y' = -1?

Answer 22

y'=-1 => (y')^2 =1

Answer 23

yes, but then you square it

Answer 24

i didn't see the square

Answer 25

if y= a-1

not sure where you got that. the four paths are
y= -x + a
y = x - a
y= -x -a
y= x + a

I did 2 of the legs and got the same answer, and I think all 4 will be the same

Answer 26

so is it then \[\int\limits_{0}^{a} x(a-x)*\sqrt{1+1}\]

Answer 27

i wrote it wrong it was y= a-x

Answer 28

yes, with dx
that is the upper right leg.

Answer 29

and i do the same for all 4?

Answer 30

you could.

Answer 31

they are all 0

Answer 32

although thinking about it, we have to be careful of the limits as we traverse the path in one direction. And does it matter if clockwise versus counter clockwise ?

Answer 33

they aren't 0

Answer 34

i got :