1
$\begingroup$

When a==a,

Clear["Global`*"]; c[t_] := {a*Cos[t], a*Sin[t]}; vf = {y^2, 0}; Integrate[vf.Dt[{x, y}, t] /. Thread[{x, y} -> c[t]], {t, 0, Pi},Assumptions-> a>0] (* Integrate returns unevaluated. *)

When a==2, I can get the result.

Clear["Global`*"]; c[t_] := {2*Cos[t], 2*Sin[t]}; vf = {y^2, 0}; Integrate[vf.Dt[{x, y}, t] /. Thread[{x, y} -> c[t]], {t, 0, Pi}] (*-(32/3)*)
Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$
  • Mathematica not easy to handle differential form. Here we use traditional form.
c[t_] = {a*Cos[t], a*Sin[t]}; vf = {y^2, 0}; Integrate[c'[t] . vf /. Thread[{x, y} -> c[t]], {t, 0, Pi}]

-((4 a^3)/3)

  • Another way is set Constants -> {a} in Dt.
c[t_] := {a*Cos[t], a*Sin[t]}; vf = {y^2, 0}; Integrate[ vf . Dt[{x, y}, t, Constants -> {a}] /. Thread[{x, y} -> c[t]], {t, 0, Pi}, Assumptions -> a > 0]

-((4 a^3)/3)

Improve this answer
$\endgroup$
1
  • $\begingroup$ The answer is great. Thanks a lot!! @cvgmt $\endgroup$ Commented Mar 4, 2022 at 6:11

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.