Length of implicit curve: Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Improper integrals in curve lengthDoes the formula for arc length hold for other coordinate systems?Arc length of $xsin x$Why is the formula $ Arc length = int r times dtheta $ not correct?Length of curve: cannot solve integralArc length of a polar curveThe length of a parametric curveArea inside curve in polar coordinates.Compute spiral length from parametric curve.Shortest curve on sphere

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Length of implicit curve:



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Improper integrals in curve lengthDoes the formula for arc length hold for other coordinate systems?Arc length of $xsin x$Why is the formula $ Arc length = int r times dtheta $ not correct?Length of curve: cannot solve integralArc length of a polar curveThe length of a parametric curveArea inside curve in polar coordinates.Compute spiral length from parametric curve.Shortest curve on sphere










0












$begingroup$


How to find the length of the curve $(x^2+y^2)^3 = 4(x^2 + xy +y^2)^2$?
I have found an explicit polar form $r = 2-cos(2theta)$ by setting $x = rcos(theta)$ and $y = rsin(theta)$. Now I can have a parametrization in $theta$ but I can also use the derived formula for the length of a curve in explicit polar form: $$intsqrtr(theta)^2+r'(theta)^2mathrm dtheta$$
However this doesn't seem to work, I can't reduce the square root to anything simpler.










share|cite|improve this question









$endgroup$











  • $begingroup$
    Are you sure there is a nice answer? It seems that neither Maple nor WolframAlpha can evaluate the integral algebraically.
    $endgroup$
    – Milten
    Mar 27 at 20:43















0












$begingroup$


How to find the length of the curve $(x^2+y^2)^3 = 4(x^2 + xy +y^2)^2$?
I have found an explicit polar form $r = 2-cos(2theta)$ by setting $x = rcos(theta)$ and $y = rsin(theta)$. Now I can have a parametrization in $theta$ but I can also use the derived formula for the length of a curve in explicit polar form: $$intsqrtr(theta)^2+r'(theta)^2mathrm dtheta$$
However this doesn't seem to work, I can't reduce the square root to anything simpler.










share|cite|improve this question









$endgroup$











  • $begingroup$
    Are you sure there is a nice answer? It seems that neither Maple nor WolframAlpha can evaluate the integral algebraically.
    $endgroup$
    – Milten
    Mar 27 at 20:43













0












0








0





$begingroup$


How to find the length of the curve $(x^2+y^2)^3 = 4(x^2 + xy +y^2)^2$?
I have found an explicit polar form $r = 2-cos(2theta)$ by setting $x = rcos(theta)$ and $y = rsin(theta)$. Now I can have a parametrization in $theta$ but I can also use the derived formula for the length of a curve in explicit polar form: $$intsqrtr(theta)^2+r'(theta)^2mathrm dtheta$$
However this doesn't seem to work, I can't reduce the square root to anything simpler.










share|cite|improve this question









$endgroup$




How to find the length of the curve $(x^2+y^2)^3 = 4(x^2 + xy +y^2)^2$?
I have found an explicit polar form $r = 2-cos(2theta)$ by setting $x = rcos(theta)$ and $y = rsin(theta)$. Now I can have a parametrization in $theta$ but I can also use the derived formula for the length of a curve in explicit polar form: $$intsqrtr(theta)^2+r'(theta)^2mathrm dtheta$$
However this doesn't seem to work, I can't reduce the square root to anything simpler.







integration






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 27 at 19:57









delivosadelivosa

1066




1066











  • $begingroup$
    Are you sure there is a nice answer? It seems that neither Maple nor WolframAlpha can evaluate the integral algebraically.
    $endgroup$
    – Milten
    Mar 27 at 20:43
















  • $begingroup$
    Are you sure there is a nice answer? It seems that neither Maple nor WolframAlpha can evaluate the integral algebraically.
    $endgroup$
    – Milten
    Mar 27 at 20:43















$begingroup$
Are you sure there is a nice answer? It seems that neither Maple nor WolframAlpha can evaluate the integral algebraically.
$endgroup$
– Milten
Mar 27 at 20:43




$begingroup$
Are you sure there is a nice answer? It seems that neither Maple nor WolframAlpha can evaluate the integral algebraically.
$endgroup$
– Milten
Mar 27 at 20:43










1 Answer
1






active

oldest

votes


















0












$begingroup$

First of all, I think that you made a small mistake since, if I am correct,
$$r(theta)=2+sin(2theta)$$
Now, using



$$L=intsqrtr(theta)^2+r'(theta)^2, dtheta$$ and simplifying, you end with the monster
$$L=frac1sqrt2int sqrt8 sin (2 t)+3 cos (4 t)+13, dtheta$$ which,as Milten commented, is far away to be pleasant. So, numerical integration seems to be the way to go.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Okay thanks, i'll try numeric integration.
    $endgroup$
    – delivosa
    Mar 28 at 6:46











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

First of all, I think that you made a small mistake since, if I am correct,
$$r(theta)=2+sin(2theta)$$
Now, using



$$L=intsqrtr(theta)^2+r'(theta)^2, dtheta$$ and simplifying, you end with the monster
$$L=frac1sqrt2int sqrt8 sin (2 t)+3 cos (4 t)+13, dtheta$$ which,as Milten commented, is far away to be pleasant. So, numerical integration seems to be the way to go.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Okay thanks, i'll try numeric integration.
    $endgroup$
    – delivosa
    Mar 28 at 6:46















0












$begingroup$

First of all, I think that you made a small mistake since, if I am correct,
$$r(theta)=2+sin(2theta)$$
Now, using



$$L=intsqrtr(theta)^2+r'(theta)^2, dtheta$$ and simplifying, you end with the monster
$$L=frac1sqrt2int sqrt8 sin (2 t)+3 cos (4 t)+13, dtheta$$ which,as Milten commented, is far away to be pleasant. So, numerical integration seems to be the way to go.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Okay thanks, i'll try numeric integration.
    $endgroup$
    – delivosa
    Mar 28 at 6:46













0












0








0





$begingroup$

First of all, I think that you made a small mistake since, if I am correct,
$$r(theta)=2+sin(2theta)$$
Now, using



$$L=intsqrtr(theta)^2+r'(theta)^2, dtheta$$ and simplifying, you end with the monster
$$L=frac1sqrt2int sqrt8 sin (2 t)+3 cos (4 t)+13, dtheta$$ which,as Milten commented, is far away to be pleasant. So, numerical integration seems to be the way to go.






share|cite|improve this answer









$endgroup$



First of all, I think that you made a small mistake since, if I am correct,
$$r(theta)=2+sin(2theta)$$
Now, using



$$L=intsqrtr(theta)^2+r'(theta)^2, dtheta$$ and simplifying, you end with the monster
$$L=frac1sqrt2int sqrt8 sin (2 t)+3 cos (4 t)+13, dtheta$$ which,as Milten commented, is far away to be pleasant. So, numerical integration seems to be the way to go.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 28 at 4:18









Claude LeiboviciClaude Leibovici

126k1158135




126k1158135











  • $begingroup$
    Okay thanks, i'll try numeric integration.
    $endgroup$
    – delivosa
    Mar 28 at 6:46
















  • $begingroup$
    Okay thanks, i'll try numeric integration.
    $endgroup$
    – delivosa
    Mar 28 at 6:46















$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
Mar 28 at 6:46




$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
Mar 28 at 6:46

















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