Line integral of vector field from Apostol calculus

Similarly for the other two segments. You need to think about how to get each segment to go from one end point to the other.
  • #1
Richardbryant
24
0

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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  • #2
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?
 
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  • #3
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:
  • #4
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).
 

1. What is the definition of a line integral of a vector field?

The line integral of a vector field is a mathematical concept that represents the total sum of a vector field's values along a given curve or path.

2. How do you calculate the line integral of a vector field?

The line integral of a vector field is calculated by first parametrizing the given curve or path, then integrating the dot product of the vector field and the parametrized curve with respect to the parameter.

3. What is the significance of the line integral of a vector field in real-world applications?

The line integral of a vector field has various applications in physics and engineering, such as determining work done by a force along a curved path or calculating the flow of a fluid through a given surface.

4. Can you provide an example of calculating a line integral of a vector field?

For example, if we have a vector field F(x,y) = (x, y), and a curve C defined by x = t and y = t^2, with t ranging from 0 to 1, the line integral of F along C would be calculated as: ∫F(x,y)⋅dr = ∫(t, t^2)⋅(1, 2t)dt = ∫(t+2t^3)dt = 5/4.

5. How is the line integral of a vector field related to the fundamental theorem of calculus?

The fundamental theorem of calculus states that the integral of a function can be calculated by evaluating its antiderivative at the limits of integration. The line integral of a vector field is a generalization of this concept, where the antiderivative is replaced by a vector field and the function is replaced by a curve. It can be seen as a higher-dimensional version of the fundamental theorem of calculus.

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