# Line integral of vector field from Apostol calculus

1. Jun 3, 2017

### Richardbryant

1. The problem statement, all variables and given/known data

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.
2. Relevant equations

3. The attempt at a solution

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. Jun 3, 2017

### Ray Vickson

(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: Jun 3, 2017
3. Jun 3, 2017

### LCKurtz

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: Jun 3, 2017
4. Jun 3, 2017

### vela

Staff Emeritus
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).