Line integral of vector field from Apostol calculus

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Homework Help Overview

The discussion revolves around calculating the work done by various force fields along specified paths in three-dimensional space, as presented in problems from Apostol's Calculus volume 2. The problems involve vector fields and line integrals, with specific paths defined by geometric shapes such as squares and triangles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss different parameterizations for the paths involved in the problems, questioning the correctness of their approaches and the resulting calculations of work done.
  • Some participants attempt to apply Stokes' theorem to simplify the calculations for certain problems, while others express confusion about how to properly parametrize the curves.
  • There is a focus on verifying the results against textbook answers, with some participants noting discrepancies in their calculations.

Discussion Status

The discussion is ongoing, with participants exploring various parameterizations and mathematical approaches. Some have provided hints and suggestions for using theorems like Stokes' theorem, while others are still grappling with the initial setup of the problems. There is no explicit consensus on the correct methods or solutions yet.

Contextual Notes

Participants are working under the constraints of homework assignments, which may limit the information they can share or the methods they can use. There are also indications of confusion regarding the parameterization of curves and the application of vector calculus theorems.

Richardbryant
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Homework Statement



Here are the three problems that i couldn't solve from the book Calculus volume 2 by apostol

10.9 Exercise

2. Find the amount of work done by the force f(x,y)=(x^2-y^2)i+2xyj in moving a particle (in a counter clockwise direction) once around the square bounded by the coordinate axes and the lines x=a and y=a, a>0

4) A force field f in 3-space is give by the formula f(x,y,z)=yzi+xzj+x(y+1)k. Calculate the work done by f in moving a particle once around the triangle with verticles (0,0,0),(1,1,1),(-1,1,-1) in that order

6) Calculate the work done by the force field f(x,y,z)=(y-z)i+(z-x)j+(x-y)k along the curve of intersection of the sphere x^2+y^2+z^2=4 and the plane z=ytanb where 0<b<pi/2. The path is tranversed in a direction that appears counterclockwise when viewed from high above th xy-plane.

Homework Equations



3. The Attempt at a Solution [/B]

2)(0,0)-->(a,0) r(t)=ati r'(t)=ai
(a,0)-->(a,a) r(t)=atj r'(t)=aj
(a,a)-->(0,a) r(t)=-ati r'(t)=-ai
(0,a)-->(0,0) r(t)=-atj r'(t)=-aj

∫ƒ.dr= ∫(a^2 t^2)i⋅ai dt (0-->1)+ ∫(-a^2t^2)i⋅aj dt + ∫(a^2t^2)i -ai⋅dt (-1-->0) + ∫(-a^2t^2)i⋅-aj dt
=2a^3/3
yet the correct answer is 2a^3 by the textbook

4)(0,0,0)-->(1,1,1) r(t)=ti+tj+tk r'(t)=i+j+k
(1,1,1)-->(-1,1,-1) r(t)=(1-2t)i+(1-2t)k r'(t)=-2i+-2k
∫ƒ.dr= ∫t^2i+t^2j+(t^2+t)k⋅(i+j+k) dt (0-->1) + ∫(1-2t)^2i +(1-2t)k ⋅(-2i-2k) dt (0-->1)
=t^3+(5/2)t^2-2t
=3/2
yet the correct answer is 0 by the textbook

6)i stuck in parametrize the equation
 
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Richardbryant said:

Homework Statement



Here are the three problems that i couldn't solve from the book Calculus volume 2 by apostol

10.9 Exercise

2. Find the amount of work done by the force f(x,y)=(x^2-y^2)i+2xyj in moving a particle (in a counter clockwise direction) once around the square bounded by the coordinate axes and the lines x=a and y=a, a>0

4) A force field f in 3-space is give by the formula f(x,y,z)=yzi+xzj+x(y+1)k. Calculate the work done by f in moving a particle once around the triangle with verticles (0,0,0),(1,1,1),(-1,1,-1) in that order

6) Calculate the work done by the force field f(x,y,z)=(y-z)i+(z-x)j+(x-y)k along the curve of intersection of the sphere x^2+y^2+z^2=4 and the plane z=ytanb where 0<b<pi/2. The path is tranversed in a direction that appears counterclockwise when viewed from high above th xy-plane.

Homework Equations



3. The Attempt at a Solution [/B]

2)(0,0)-->(a,0) r(t)=ati r'(t)=ai
(a,0)-->(a,a) r(t)=atj r'(t)=aj
(a,a)-->(0,a) r(t)=-ati r'(t)=-ai
(0,a)-->(0,0) r(t)=-atj r'(t)=-aj

∫ƒ.dr= ∫(a^2 t^2)i⋅ai dt (0-->1)+ ∫(-a^2t^2)i⋅aj dt + ∫(a^2t^2)i -ai⋅dt (-1-->0) + ∫(-a^2t^2)i⋅-aj dt
=2a^3/3
yet the correct answer is 2a^3 by the textbook

4)(0,0,0)-->(1,1,1) r(t)=ti+tj+tk r'(t)=i+j+k
(1,1,1)-->(-1,1,-1) r(t)=(1-2t)i+(1-2t)k r'(t)=-2i+-2k
∫ƒ.dr= ∫t^2i+t^2j+(t^2+t)k⋅(i+j+k) dt (0-->1) + ∫(1-2t)^2i +(1-2t)k ⋅(-2i-2k) dt (0-->1)
=t^3+(5/2)t^2-2t
=3/2
yet the correct answer is 0 by the textbook

6)i stuck in parametrize the equation

(2) Using the simpler parametrizations ##t = x## or ##t = y## on the segments, we have:
$$\begin{array}{l} W_{(0,0) \to (a,0)} = \int_0^a x^2 \, dx\\
W_{(a,0) \to (a,a)} = \int_0^a 2 a y \, dy\\
W_{(a,a) \to (0,a)} = \int_0^a (-) [(a-x)^2 - a^2] \, dx \\
W_{(0,a) \to (0,0)} = 0
\end{array}
$$
I get the book's value.

(4) ##\vec{F } = \vec{F_1} + \vec{F_2}##, where ##\vec{F_1} = \langle yz, zx ,xy\rangle## and ##\vec{F_2} = \langle 0,0 ,x \rangle##.
Note that ##\vec{F_1} = \vec{\nabla} V##, where ##V = xyz##, so the contribution of ##\vec{F_1}## to the total work is ##\int_C \vec{\nabla} V \cdot \,d \vec{r} = 0##, by Stokes theorem. Thus, we need only look at ##\int_C \langle 0,0, x \rangle \cdot \langle dx,dy,dz \rangle##, which is a lot easier than the original problem. Of course, this simplification does not help much if you have not yet seen Stokes theorem.

(6) What type of ##xy## curve do you see if you view it from above the ##xy## plane? How would you parametrize such a curve?
 
Last edited:
Richardbryant said:

Homework Statement



Here are the three problems that i couldn't solve from the book Calculus volume 2 by apostol

6) Calculate the work done by the force field f(x,y,z)=(y-z)i+(z-x)j+(x-y)k along the curve of intersection of the sphere x^2+y^2+z^2=4 and the plane z=ytanb where 0<b<pi/2. The path is tranversed in a direction that appears counterclockwise when viewed from high above th xy-plane.

6)i stuck in parametrize the equation

Another hint for ##6##: If you calculate it you will find ##\nabla \times \vec f = \langle -2,-2,-2\rangle ##. You might try using Stokes' theorem that ##\int_C \vec f \cdot d\vec r =\iint_S \nabla \times \vec f \cdot d\vec S## where ##S## is the plane circular cross-section.
 
Last edited:
Richardbryant said:
(0,0)-->(a,0) r(t)=ati r'(t)=ai
(a,0)-->(a,a) r(t)=atj r'(t)=aj
The parameterization ##r(t) = at \hat{j}## doesn't move you from (a,0) to (a,a). It goes from (0,0) to (0,a).
 

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