4306.sagews

Calculate the line integral of $\int_{C^+} \,(-ydx+xdy) \,$ for each of the curves $C^+$ from (0,0) to (1,1)

1. Problem 1: $C: y=x^2$    YouTube Video

2. Problem 2: $C: y=\sqrt{x}$

3. Problem 3: $C: y=\sin ({{\pi x} \over 2})$

Cool graphs with explanations are at the bottom of each problem!

Other Sage pages for Line Integral of Work Type:    SOLVER  

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Formula Sheet  $\int_{C^+} P(x,y,z)\, dx+Q(x,y,z) \,dy+R(x,y,z) \,dz = \int_C \vec F \cdot d \vec s = \int_{t_1}^{t_2}\, \lt P(t), Q(t), R(t) \gt \cdot \lt \dot{x}, \dot{y},\dot{z} \gt \, dt$

where C is the curve parameterized by $\vec s(t)= \lt x(t),\, y(t),\, z(t) \gt$ with given orientation $t \in [t1,t2]$ and $\vec F= \lt P,Q,R \gt$.

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SOLUTION TO PROBLEMS 

Check type of line integral: This line integral is of work type (Type II) since the integrand is a vector function (it has components: P=-y and Q=x) - watch the video.

Preparation for SOLVER: We must parameterize the curve C with a vector function $\vec s(t)$ in 1 parameter t, find the interval and check that the order of this interval puts the orientation of the curve as given.

Work particular to the given problems: All curves are 2d and given explicitly with interval: $x \in [0,1 ]$. We simply let x=t, y=f(t) and z=0 with interval: $t\in [0,1 ]$

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Problem 1: Calculate the line integral of work type $\int_C \,(-ydx+xdy)$ where $C: y=x^2$ from (0,0) to (1,1).

Step 0: The program declares our variables. We are given variables (x,y,z). We need t as parameter.

var ('t'); var('x y z')

t (x, y, z)

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Step 1: We define $C(t)=\vec s(t)$, the interval of t and $\vec F$. Changes with problem; you must input your parametrization of the curve, your interval for t (check orientation) and your vector function F(x,y,z).

s=vector((t,t^2,0))
t1=0; t2=1
F=vector((-y,x,0))

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Step 2: The program finds the vector function with the partial derivatives: $d \vec s(t)=\lt {\dot{x}},\, {\dot{y}},\, {\dot{z}} \gt$

ds=diff(s,t)
view(ds)

\newcommand{\Bold}[1]{\mathbf{#1}}\left(1,\,2 \, t,\,0\right)

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Step 3: The program defines a standard sage function for "change of variables" for a 1-variable parametrization (curve).

The program changes the variables of $\vec F$ from a vector in x, y, z to a vector of t using the components of s(t).

def changevar(f, eqn, newvar1):
    return f.substitute(eqn)

F=changevar(F,x==s[0],t)
F=changevar(F,y==s[1],t)
F=changevar(F,z==s[2],t)
view(F)

\newcommand{\Bold}[1]{\mathbf{#1}}\left(-t^{2},\,t,\,0\right)

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Step 4: Then program finds the dot product of the vectors from Step2 and Step3.

integrand=F.dot_product(ds)
view(integrand)

\newcommand{\Bold}[1]{\mathbf{#1}}t^{2}

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Step 6: The program computes the integral (work).

result=integral(integrand,(t,t1,t2))
view(result)

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{3}

n(result)

\newcommand{\Bold}[1]{\mathbf{#1}}0.333333333333333

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So the work done by $\vec F = \lt -y, \, x \gt$ along the line curve $y=x^2$ from (0,0) to (1,1) is: $\int_C \,(-ydx+xdy) =1/3 \approx 0.33$  

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Graph 1: We graph the vector field $\vec F$ and the curve $y=x^2$  from (0,0) to (1,1).

• Remember the orientation of the curve is from (0,0) to (1,1).
• With respect to this orientation, the vector field  $\vec F$ is "mildly positive" with respect to the curve, i.e. they are pointing slightly in the direction of the orientation.
• So some "work" is being done along the line. 

xmin=-2; xmax=2; ymin=-2; ymax=2; zmin=-1; zmax=1
VF=plot_vector_field(vector((-y,x)), (x,xmin,xmax),(y,ymin,ymax), color='red')
C=parametric_plot(vector((t,t^2)),(t,0,1))
show(VF+C)

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Problem 2: Calculate the line integral of work type $\int_C \,(-ydx+xdy) =0\,$ where $C: y=\sqrt{x}$

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Step 0: The program declares our variables. We are given variables (x,y,z). We need t as parameter.

var ('t'); var('x y z')

\newcommand{\Bold}[1]{\mathbf{#1}}t

\newcommand{\Bold}[1]{\mathbf{#1}}\left(x, y, z\right)

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Step 1: We define $C(t)=\vec s(t)$, the interval of t and $\vec F$. Changes with problem; you must input your parametrization of the curve, your interval for t (check orientation) and your vector function F(x,y,z).

s2=vector((t,sqrt(t),0))
t1=0; t2=1
F2=vector((-y,x,0))

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Step 2: The program finds the vector function with the partial derivatives: $d \vec s(t)=\lt {\dot{x}},\, {\dot{y}},\, {\dot{z}} \gt$

ds2=diff(s2,t)
view(ds2)

\newcommand{\Bold}[1]{\mathbf{#1}}\left(1,\,\frac{1}{2 \, \sqrt{t}},\,0\right)

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Step 3: The program defines a standard sage function for "change of variables" for a 1-variable parametrization (curve).

The program changes the variables of $\vec F$ from a vector in x, y, z to a vector of t using the components of s(t).

def changevar(h, eqn, newvar1):
    return h.substitute(eqn)

F2=changevar(F2,x==s2[0],t)
F2=changevar(F2,y==s2[1],t)
F2=changevar(F2,z==s2[2],t)
view(F2)

\newcommand{\Bold}[1]{\mathbf{#1}}\left(-\sqrt{t},\,t,\,0\right)

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Step 4: Then program finds the dot product of the vectors from Step2 and Step3.

integrand2=F2.dot_product(ds2)
view(integrand2)

\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{2} \, \sqrt{t}

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Step 6: The program computes the integral (work).

result2=integral(integrand2,(t,t1,t2))
view(result2)

\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{3}

n(result2)

\newcommand{\Bold}[1]{\mathbf{#1}}-0.333333333333333

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So the work done by $\vec F = \lt -y, \, x \gt$ along the line curve $y=\sqrt{x}$ from (0,0) to (1,1) is: $\int_C \,(-ydx+xdy) = -1/3 \approx -0.33$  

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Graph 2: We graph the vector field $\vec F$ and the curve $y=\sqrt{x}$  from (0,0) to (1,1).

• Remember the orientation of the curve is from (0,0) to (1,1).
• With respect to this orientation, the vector field  $\vec F$ is "mildly negative" That is, the arrows point "backwards". 
• So some "work" is being lost along the line. 

xmin=-2; xmax=2; ymin=-2; ymax=2; zmin=-1; zmax=1
VF2=plot_vector_field(vector((-y,x)), (x,xmin,xmax),(y,ymin,ymax), color='red')
C2=parametric_plot(vector((t,sqrt(t))),(t,0,1))
show(VF2+C2)

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Problem 3: Calculate the line integral of work type $\int_C \,(-ydx+xdy) =0\,$ where $C: \,\, y=sin(\frac{\pi x}{2})$

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Step 0: The program declares our variables. We are given variables (x,y,z). We need t as parameter.

var ('t'); var('x y z')

\newcommand{\Bold}[1]{\mathbf{#1}}t

\newcommand{\Bold}[1]{\mathbf{#1}}\left(x, y, z\right)

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Step 1: We define $C(t)=\vec s(t)$, the interval of t and $\vec F$. Changes with problem; you must input your parametrization of the curve, your interval for t (check orientation) and your vector function F(x,y,z).

s3=vector((t,sin(pi*t/2),0))
t1=0; t2=1
F3=vector((-y,x,0))

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Step 2: The program finds the vector function with the partial derivatives: $d \vec s(t)=\lt {\dot{x}},\, {\dot{y}},\, {\dot{z}} \gt$

ds3=diff(s3,t)
view(ds3)

\newcommand{\Bold}[1]{\mathbf{#1}}\left(1,\,\frac{1}{2} \, \pi \cos\left(\frac{1}{2} \, \pi t\right),\,0\right)

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Step 3: The program defines a standard sage function for "change of variables" for a 1-variable parametrization (curve).

The program changes the variables of $\vec F$ from a vector in x, y, z to a vector of t using the components of s(t).

def changevar(h, eqn, newvar1):
    return h.substitute(eqn)

F3=changevar(F3,x==s3[0],t)
F3=changevar(F3,y==s3[1],t)
F3=changevar(F3,z==s3[2],t)
view(F3)

\newcommand{\Bold}[1]{\mathbf{#1}}\left(-\sin\left(\frac{1}{2} \, \pi t\right),\,t,\,0\right)

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Step 4: Then program finds the dot product of the vectors from Step2 and Step3.

integrand3=F3.dot_product(ds3)
view(integrand3)

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{2} \, \pi t \cos\left(\frac{1}{2} \, \pi t\right) - \sin\left(\frac{1}{2} \, \pi t\right)

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Step 6: The program computes the integral (work).

result3=integral(integrand3,(t,t1,t2))
view(result3)

\newcommand{\Bold}[1]{\mathbf{#1}}\frac{\pi - 4}{\pi}

n(result3)

\newcommand{\Bold}[1]{\mathbf{#1}}-0.273239544735163

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So the work done by $\vec F = \lt -y, \, x \gt$ along the line curve $y=\sin ({{\pi x} \over 2})$ from (0,0) to (1,1) is: $\int_C \,(-ydx+xdy) = \frac{\pi-4}{\pi} \approx -0.27$  

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Graph 1: We graph the vector field $\vec F$ and the curve $y=\sin ({{\pi x} \over 2})$  from (0,0) to (1,1).

• Remember the orientation of the curve is from (0,0) to (1,1).
• With respect to this orientation, it is hard to tell exactly the direction of the vector field  $\vec F$. 
• Our calculations show us that some work is lost.

xmin=-2; xmax=2; ymin=-2; ymax=2; zmin=-1; zmax=1
VF3=plot_vector_field(vector((-y,x)), (x,xmin,xmax),(y,ymin,ymax), color='red')
C3=parametric_plot(vector((t,sin(pi*t/2))),(t,0,1))
show(VF3+C3)