Length of implicit curve: The Next CEO of Stack OverflowImproper 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:
The Next CEO of Stack OverflowImproper 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
$begingroup$
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.
integration
asked Mar 27 at 19:57
delivosadelivosa
1066
$endgroup$
add a comment |
$begingroup$
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.
integration
asked Mar 27 at 19:57
delivosadelivosa
1066
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
integration
asked Mar 27 at 19:57
delivosadelivosa
1066
$endgroup$
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
integration
asked Mar 27 at 19:57
delivosadelivosa
1066
asked Mar 27 at 19:57
delivosadelivosa
1066
asked Mar 27 at 19:57
delivosadelivosa
1066
asked Mar 27 at 19:57
delivosadelivosa
1066
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Mar 28 at 4:18
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
2 days ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Mar 28 at 4:18
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
2 days ago
add a comment |
$begingroup$
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.
answered Mar 28 at 4:18
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
2 days ago
add a comment |
$begingroup$
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.
answered Mar 28 at 4:18
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
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.
answered Mar 28 at 4:18
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 28 at 4:18
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 28 at 4:18
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 28 at 4:18
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
2 days ago
add a comment |
$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
2 days ago
$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
2 days ago
$begingroup$
Okay thanks, i'll try numeric integration.
$endgroup$
– delivosa
2 days ago
add a comment |
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$begingroup$
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