Calculate Line Integrals for Vector Field F on C1 and C2

[hils0005]

Homework Statement

Calculate \intF dr if C = C1 + C2 where C1 is the line segment from P1(-1,pi,-1) to P2(0,0,0) and C2 is the line segment from P2(0,0,0) to P3(2,0,4)

vectorF= yz i + (xz - e^(z)siny) j + (e^(z)cosy + xy) k

The Attempt at a Solution

Im having problems setting up the C1 integral:
I have it parametrized as x=t, y=-pi t z=t, as t goes 0 to 1

r(t)= t i - pit j + t k

r'(t)= i - pi j + k

F= -pi t^2 i + (t^2-e^(t)sin(pit)) j + (e^(t)cos(pit) + pit^2) k

F dot r'(t) = -pi t^2 + pi e^(t)sin(pit) + e^tcos(pit)

I would then take integral dt as t goes from 0 to 1

I think I have answered C2 correctly, I'm not sure if my parameters are correct

 

[Defennder]

Big time-saving hint: What is curl F ?

 

[hils0005]

dF2/dx - dF1/dy =0 z-z=0
dF3/dy - dF2/dz =0 (-e^zsiny + x) + (-x + e^zsiny) = 0
dF1/dz - dF3/dx = 0 y-y=0

Curl test?? how does that help?

 

[Defennder]

So what does that tell you about F?

 

[hils0005]

that it is path independent or conservative?

 

[gabbagabbahey]

that it is path independent or conservative?

Yes, which also tells you that you can write F as the gradient of some scalar \varphi...So if you can find such a scalar, you can easily compute the path integral using the fundamental theorem of gradients.

However, the way your original question is worded leads e to believe that they want you to evaluate the path integral along that specific path, so I would use the gradient method only as a check.

 

[gabbagabbahey]

C1 is the line segment from P1(-1,pi,-1) to P2(0,0,0)...

Im having problems setting up the C1 integral:
I have it parametrized as x=t, y=-pi t z=t, as t goes 0 to 1

r(t)= t i - pit j + t k

That doesn't seem right; r(0)=(0,0,0) which is not your starting point, and r(1)=(1,-\pi,1) which is neither your starting point or your finish point!

 

[hils0005]

x=t-1
y= pi t + t
z=t-1

t goes from -2 to -1,
would that work?

 

[gabbagabbahey]

x=t-1
y= pi t + t
z=t-1

t goes from -2 to -1,
would that work?

Nope; now your starting point is r(-2)=(-3,-2pi-2,-3) and your finish point is just as incorrect.

Let's think this through logically instead of stabbing blindly in the dark...You know that c1 is the direct line from P1 to P2; so you can write down a general parameterized line for r: \mathbf{r}(t)=(at+b,ct+d,ft+g)...say you let t go from 0 to 1; what must a,b,c,d,f and g be for \mathbf{r}(0) to be P1 and \mathbf{r}(1) to be P2?

 

[hils0005]

x=t-1
y=pit^2
z=t-1

 

[gabbagabbahey]

x and z look good, but y isn't even linear in t!...how did you come up with that? Is y(1) really 0?

 

[hils0005]

its late...what else can i say
I don't know what to use for t going from 0 to 1
y=pi t + pi
y(0)=pi
y(1)=2pi

y=pit - pi
y(0)=-pi
y(1)=pi t - pi=0

 

[gabbagabbahey]

well now y is linear, and y(1)=0 , but y(0)=-pi not +pi

Perhaps you should get some rest and leave this until morning?

 

[hils0005]

Great idea, thanks for your help

 

[kathrynag]

Homework Statement

Calculate \intF dr if C = C1 + C2 where C1 is the line segment from P1(-1,pi,-1) to P2(0,0,0) and C2 is the line segment from P2(0,0,0) to P3(2,0,4)

vectorF= yz i + (xz - e^(z)siny) j + (e^(z)cosy + xy) k

The Attempt at a Solution

Im having problems setting up the C1 integral:
I have it parametrized as x=t, y=-pi t z=t, as t goes 0 to 1

r(t)= t i - pit j + t k

r'(t)= i - pi j + k

F= -pi t^2 i + (t^2-e^(t)sin(pit)) j + (e^(t)cos(pit) + pit^2) k

F dot r'(t) = -pi t^2 + pi e^(t)sin(pit) + e^tcos(pit)

I would then take integral dt as t goes from 0 to 1

I think I have answered C2 correctly, I'm not sure if my parameters are correct

Here's a big hint for finding x, y and z,
P1(-1,pi,-1) to P2(0,0,0)
So, we have (-1, pi, -1) + some t
(-1, pi,-1)+(1,a, b)t
So x= -1+t
I got this by having our original point and taking 0 from our second point and and -1 from the first point.
So, we have -1+[0-(-1)]t=-1+t
Follow this in the same manner for finding y and z.

 

[Defennder]

Yes, which also tells you that you can write F as the gradient of some scalar \varphi...So if you can find such a scalar, you can easily compute the path integral using the fundamental theorem of gradients.

However, the way your original question is worded leads e to believe that they want you to evaluate the path integral along that specific path, so I would use the gradient method only as a check.

Actually my intention wasn't to find the scalar potential which itself could be rather tedious, but to show that one could evaluate the line integral from P1 to P3 directly bypassing P2 to save time.