Evaluate the Following Line Integral Part 3

[bugatti79]

Homework Statement

\displaystyle \int x^2dx+y^2dy+z^2dz where C is the line segment from(0,0,0) to (1,2,-1) and (1,2,-1) to (3,2,0)

Homework Equations

\displaystyle\int_c \vec F(t) d \vec r(t)= (x^2 i+ y^2 j+z^2 k)d \vec r(t)

where d \vec r(t) for C_1=ti+2tj-tk and d \vec r(t) for C_1=(1+2t)i+2j+(-1+t)k

Therefore \displaystyle \int_{C1} ((1+2t)^2 i+4^2 j+(-1+t)^2 k)(2i+0j+k)

Is this correct so far?

 

[bugatti79]

hAlso, How do I determine the limits.

Note I have used LaTex to make it prettier. Thanks

 

[I like Serena]

Hi bugatti79!

Pretty! I like it!

Apparently you have split the curve C into 2 subcurves: C1 and C2.
At least... you named the 2nd subcurve C1. Is that a typo?

In your choice for those 2 curves, you selected equations such that at t=0 you're in the first point, and at t=1 you're in the 2nd point.
But these are equations for r(t) and not for dr(t).
To find dr(t) you still need to take the derivative.

The complete curve would consist of the sum of the integral over C1 and the integral over C2.
With your choice for the equations, both integrals would have limits 0 and 1.

 

[bugatti79]

Hi bugatti79!

Pretty! I like it!

Apparently you have split the curve C into 2 subcurves: C1 and C2.
At least... you named the 2nd subcurve C1. Is that a typo?

In your choice for those 2 curves, you selected equations such that at t=0 you're in the first point, and at t=1 you're in the 2nd point.
But these are equations for r(t) and not for dr(t).
To find dr(t) you still need to take the derivative.

The complete curve would consist of the sum of the integral over C1 and the integral over C2.
With your choice for the equations, both integrals would have limits 0 and 1.

Hi,

Yes, that's a typo.

Ah, that's making sense now regarding the limits...I will re attempt tomorrow, its bedtime! Thanks

 

[bugatti79]

Homework Statement

\displaystyle \int x^2dx+y^2dy+z^2dz where C is the line segment from(0,0,0) to (1,2,-1) and (1,2,-1) to (3,2,0)

Homework Equations

\displaystyle\int_c \vec F(t) d \vec r(t)= (x^2 i+ y^2 j+z^2 k)d \vec r(t)

where \vec r(t) for C_1=ti+2tj-tk and \vec r(t) for C_2=(1+2t)i+2j+(-1+t)k

Therefore \displaystyle \int_{C1} ((1+2t)^2 i+4^2 j+(-1+t)^2 k)(2i+0j+k)

Is this correct so far?

Hi,
The equations I have written for C1 and C2 are for r(t) yes...the above should be correct now...?

 

[I like Serena]

Hi,
The equations I have written for C1 and C2 are for r(t) yes...the above should be correct now...?

Almost. You corrected the typo on the definition of C2.

But your integral is for C2 and not for C1.
And you need a result for C which is C1 combined with C2.

 

[bugatti79]

Almost. You corrected the typo on the definition of C2.

But your integral is for C2 and not for C1.
And you need a result for C which is C1 combined with C2.

Yes, that's another typo.

\displaystyle \int_{C2} ((1+2t)^2 i+4^2 j+(-1+t)^2 k)(2i+0j+k)

Ok, and since the 2 brackets is a dot product the basis i,j and k will go to 1 upon multiplcation.

Similar for C1. Evalute the 2 integrals and then add. Should get a number because F.dr is as scalar. Correct?

Thanks

 

[I like Serena]

Yes, that's another typo.

\displaystyle \int_{C2} ((1+2t)^2 i+4^2 j+(-1+t)^2 k)(2i+0j+k)

Ok, and since the 2 brackets is a dot product the basis i,j and k will go to 1 upon multiplcation.

Similar for C1. Evalute the 2 integrals and then add. Should get a number because F.dr is as scalar. Correct?

Thanks

Yes. :)

Btw, you integral should also include "dt".