︠4aa36d05-fce6-40fe-9180-a8384d2c6b20i︠ %html

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

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

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 $\vec F= \lt P,Q,R \gt$.

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: $t\in [0,1 ]$

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.

︡bba83a67-9795-45b5-aa68-b20b4f1f42c4︡{"html": "

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

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 $\\vec F= \\lt P,Q,R \\gt$.

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: $t\\in [0,1 ]$

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.

"}︡ ︠a052acd7-4fdd-46fc-90ff-1a232da7b753︠ var ('t'); var('x y z') ︡129e2ccb-a2e4-415f-babe-612671ad797d︡{"stdout": "t\n(x, y, z)"}︡ ︠cca50f66-695e-4a83-9a6b-79347f2fa573i︠ %html

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).

︡35129994-c3de-4cbb-bf05-436211621d4b︡{"html": "

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).

"}︡ ︠462f5e7b-ca3b-4451-bdc9-b0e1c48f5633︠ s=vector((t,t^2,0)) t1=0; t2=1 F=vector((-y,x,0)) ︡0b56db24-8edd-42dc-91e5-1aa3f72ab4ec︡︡ ︠efc9cd45-5c86-4bec-b74c-559f88743b04i︠ %html

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

︡d6d13d36-a773-43fb-8e21-45518a29996d︡{"html": "

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

"}︡ ︠e94961a1-8def-4e16-a17f-c2cb0e2ba772︠ ds=diff(s,t) view(ds) ︡6094e07a-99d1-48e7-ad81-21882d12921f︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(1,\\,2 \\, t,\\,0\\right)"}︡ ︠b206efd6-673b-4b46-bb2e-827f02625b61i︠ %html

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).

︡c3cba7a7-ea2c-40c3-a1dd-eeaf45c76ddc︡{"html": "

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).

"}︡ ︠fc6634c3-998d-47df-ad1c-5264420d3d19︠ def changevar(f, eqn, newvar1): return f.substitute(eqn) ︡14db6185-57d5-400b-b211-2c830bdc0ba9︡︡ ︠9f6c0781-c6c9-4800-986b-89789261707a︠ F=changevar(F,x==s[0],t) F=changevar(F,y==s[1],t) F=changevar(F,z==s[2],t) view(F) ︡01821975-a7f7-411f-a655-5cdac9cb5563︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(-t^{2},\\,t,\\,0\\right)"}︡ ︠62cc7552-7a69-4c8b-b66f-a7f5cdd52732i︠ %html

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

︡fa72e9e6-3a54-4dc1-a750-85e373e693ec︡{"html": "

\r\n

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

"}︡ ︠27ac0e3e-7dc8-4bfa-b2bc-dd2528122692︠ integrand=F.dot_product(ds) view(integrand) ︡d7b4506a-d9c8-4ff7-8ea9-959627584ffa︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}t^{2}"}︡ ︠69c5400a-62a4-4fc6-a98a-4f256c558661i︠ %html

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

︡e8caeefa-8168-42bf-b18c-23dc9a0c1d32︡{"html": "

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

"}︡ ︠c9d4d240-4b6c-450e-9856-5bc42a9f4a59︠ result=integral(integrand,(t,t1,t2)) view(result) ︡e03f58e8-e8de-40f1-b76d-e3a3138cf7ab︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{1}{3}"}︡ ︠3790f11c-68e3-4afc-b8ea-656f1608060c︠ n(result) ︡66c21418-123c-4fd6-9491-96bb0bff4da4︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}0.333333333333333"}︡ ︠56728856-9c94-4a49-9b42-5a94e98971a3i︠ %html

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: $\int_C \,(-ydx+xdy) =1/3 \approx 0.33$

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.

︡7252d2f7-9088-44c3-bedb-924906265cdd︡{"html": "

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: $\\int_C \\,(-ydx+xdy) =1/3 \\approx 0.33$

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.

"}︡ ︠42e96f94-eec4-4a94-8da4-bb92fd323a9d︠ 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) ︡7f6e7a71-10b5-443a-8757-06a466cf6999︡{"html": "

"}︡ ︠9e34c263-4634-4c33-8b79-9366fd8e0495i︠ %html

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

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

︡665fa4bc-cec5-49a7-82c1-fb08f3b82f8c︡{"html": "

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

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

"}︡ ︠3e922d2f-e4eb-471e-9a75-a156ce365c41︠ var ('t'); var('x y z') ︡1f618959-88ce-46bb-bfa9-f2685026755f︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}t"}︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(x, y, z\\right)"}︡ ︠7439e287-3c0e-4449-bfd9-ecf40e118fd7i︠ %html

︡f434cfdc-121e-4715-beb4-4f40c0d19852︡{"html": "

"}︡ ︠8b4ddb55-8d90-4cc0-a536-e0f4a7205fed︠ s2=vector((t,sqrt(t),0)) t1=0; t2=1 F2=vector((-y,x,0)) ︡ca2d74d4-e616-42d5-8cf1-0cadccf6a466︡︡ ︠a98e1d85-c013-4354-8e26-df43679ce241i︠ %html

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

︡cdca81a6-8ced-4a57-b2af-76caea3f3f4c︡{"html": "

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

"}︡ ︠0d24fdf1-563a-4172-846a-85fa24cc7182︠ ds2=diff(s2,t) view(ds2) ︡a2346a2f-91c1-445d-ae52-4ffae2496fbe︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(1,\\,\\frac{1}{2 \\, \\sqrt{t}},\\,0\\right)"}︡ ︠fe2c7edd-38e4-4697-9ac9-130fe8eb6485i︠ %html

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).

︡e53dedac-344a-44f6-90ec-c9e088ec5784︡{"html": "

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).

"}︡ ︠4c2466e2-3311-4453-ba86-01164cd2f8bd︠ def changevar(h, eqn, newvar1): return h.substitute(eqn) ︡c352b9f9-d693-4551-a38b-9a92af3f434d︡︡ ︠f208a9fd-9c04-49ca-ae72-b773b45bca4e︠ F2=changevar(F2,x==s2[0],t) F2=changevar(F2,y==s2[1],t) F2=changevar(F2,z==s2[2],t) view(F2) ︡d59a0599-9366-44d2-87a1-b3a3165246f3︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(-\\sqrt{t},\\,t,\\,0\\right)"}︡ ︠89a352a2-098e-429a-861d-ebeede781d27i︠ %html

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

︡565c514d-598c-40d2-975f-6f070344332e︡{"html": "

\r\n

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

"}︡ ︠9be258d4-cb42-425a-bdbb-0c9613fc7896︠ integrand2=F2.dot_product(ds2) view(integrand2) ︡76a7836d-57f9-4a69-9607-951bfda9ad7f︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{2} \\, \\sqrt{t}"}︡ ︠49dd7feb-af42-45cb-882b-3ed5f8624e30i︠ %html

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

︡12527d60-c72c-4d72-b5df-3f12003d08f2︡{"html": "

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

"}︡ ︠1873d5f6-5335-4180-96de-2186eeaf0a37︠ result2=integral(integrand2,(t,t1,t2)) view(result2) ︡39045168-38b4-4c4d-be71-cfbf03f5fb33︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}-\\frac{1}{3}"}︡ ︠d1bb629a-33ea-4546-a785-b632fba0e484︠ n(result2) ︡4de2d66d-e151-421c-b585-2d0c664e4d73︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}-0.333333333333333"}︡ ︠5f101c26-dc1d-4dee-8af1-992a5c4a015ei︠ %html

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: $\int_C \,(-ydx+xdy) = -1/3 \approx -0.33$

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.

︡8ed0b057-f6e5-4a36-a516-c1fd8a87894f︡{"html": "

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: $\\int_C \\,(-ydx+xdy) = -1/3 \\approx -0.33$

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.

"}︡ ︠81de53c5-44fc-4ae0-8e8a-ed11115ad8dd︠ 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) ︡f1dde260-8a44-4e2e-b524-e5c86e0d65cc︡{"html": "

"}︡ ︠131b81b1-93c5-44a2-aca4-2a3050f90ea4i︠ %html

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

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

︡cd443f56-714e-4667-9a97-1a38d44f51a0︡{"html": "

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

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

"}︡ ︠48ccae07-53b9-4095-9597-9357ddfa2c1d︠ var ('t'); var('x y z') ︡d6712ad2-41e1-43e3-8a7a-f51688a5c541︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}t"}︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(x, y, z\\right)"}︡ ︠d17ab36a-2e16-4e78-ac43-589b51fb669ai︠ %html

︡7ef79968-bc85-49d1-ac4a-e5ccdd6ed4c5︡{"html": "

"}︡ ︠8ddc81e0-f924-4590-ac05-2cdc56fdbaaf︠ s3=vector((t,sin(pi*t/2),0)) t1=0; t2=1 F3=vector((-y,x,0)) ︡1aaffb38-488d-4b8b-8c66-ba071c7989da︡︡ ︠7b1f1356-6139-4acf-8d18-dc833dec0d5fi︠ %html

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

︡0b0b4cad-e1a3-4214-ae5d-31627b0e9d75︡{"html": "

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

"}︡ ︠9383c62f-1897-42d9-9871-0cd2f24407b0︠ ds3=diff(s3,t) view(ds3) ︡69137c27-bed4-42e4-8552-3fa028e590bd︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(1,\\,\\frac{1}{2} \\, \\pi \\cos\\left(\\frac{1}{2} \\, \\pi t\\right),\\,0\\right)"}︡ ︠c7ca400d-fec1-4faf-846a-ae6835805ddci︠ %html

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).

︡0e695bd9-52e0-4b33-8fcd-aa37102a314a︡{"html": "

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).

"}︡ ︠2a44f2db-c6ed-42bd-9101-700a65d767a4︠ def changevar(h, eqn, newvar1): return h.substitute(eqn) ︡a1d01004-f3b6-4ed7-9cd1-6e371089b84a︡︡ ︠a43c7b94-63c1-4dda-8454-6b6ef9e77418︠ F3=changevar(F3,x==s3[0],t) F3=changevar(F3,y==s3[1],t) F3=changevar(F3,z==s3[2],t) view(F3) ︡4aa68bf9-16fb-4a3c-a4c0-2cdd2025b39c︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(-\\sin\\left(\\frac{1}{2} \\, \\pi t\\right),\\,t,\\,0\\right)"}︡ ︠1c094741-1d82-44b0-b312-7f47bbc22b44i︠ %html

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

︡d36572d7-fc6e-4f56-87f5-0c6daa5d6740︡{"html": "

\r\n

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

"}︡ ︠8d357e8b-868b-4fc6-a50c-7ef211586cb7︠ integrand3=F3.dot_product(ds3) view(integrand3) ︡14df000f-41bc-4337-a48f-bc2077083c0b︡{"html": "\\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)"}︡ ︠a2356e11-8714-4e7f-8c37-4d2f3f0034c9i︠ %html

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

︡e86f7c3b-419a-4b78-84d1-95062b47b325︡{"html": "

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

"}︡ ︠c4cdddc4-ad5a-4c7f-9af0-ca95537145dc︠ result3=integral(integrand3,(t,t1,t2)) view(result3) ︡a2189e09-2a80-44e5-a03f-ef335207b89f︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\frac{\\pi - 4}{\\pi}"}︡ ︠7f9c3d9b-d68e-4ae0-8563-b96a7fff289e︠ n(result3) ︡a0ddf704-9f46-41cd-b6a6-a8a949911a42︡{"html": "\\newcommand{\\Bold}[1]{\\mathbf{#1}}-0.273239544735163"}︡ ︠adf64db0-2444-4bd6-bd7e-3283926b8f14i︠ %html

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: $\int_C \,(-ydx+xdy) = \frac{\pi-4}{\pi} \approx -0.27$

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.

︡1e8e156e-2c8b-4b3c-8472-1b3041b955b3︡{"html": "

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: $\\int_C \\,(-ydx+xdy) = \\frac{\\pi-4}{\\pi} \\approx -0.27$

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.

"}︡ ︠6f618b12-0184-49f2-8a62-fe4124c9002d︠ 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) ︡8ca56324-99b0-4f6d-b9d5-c8b40c7865fe︡{"html": "

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