How to evaluate this line integral over a plane curve?
$begingroup$
How do I integrate $f$ over the given curve?
$$f(x,y)= frac{x^3}{y};quad C: y=frac{x^2}{2} quad text{for}; 0 leq x leq 2.$$
I can't figure this out... can anyone show me how to solve it?
The answer is supposed to be $displaystyle frac{10sqrt{5}-2}{3}$.
multivariable-calculus
edited Nov 25 '12 at 23:08
amWhy
192k28225439
$endgroup$
add a comment |
$begingroup$
How do I integrate $f$ over the given curve?
$$f(x,y)= frac{x^3}{y};quad C: y=frac{x^2}{2} quad text{for}; 0 leq x leq 2.$$
I can't figure this out... can anyone show me how to solve it?
The answer is supposed to be $displaystyle frac{10sqrt{5}-2}{3}$.
multivariable-calculus
edited Nov 25 '12 at 23:08
amWhy
192k28225439
$endgroup$
add a comment |
$begingroup$
How do I integrate $f$ over the given curve?
$$f(x,y)= frac{x^3}{y};quad C: y=frac{x^2}{2} quad text{for}; 0 leq x leq 2.$$
I can't figure this out... can anyone show me how to solve it?
The answer is supposed to be $displaystyle frac{10sqrt{5}-2}{3}$.
multivariable-calculus
edited Nov 25 '12 at 23:08
amWhy
192k28225439
$endgroup$
How do I integrate $f$ over the given curve?
$$f(x,y)= frac{x^3}{y};quad C: y=frac{x^2}{2} quad text{for}; 0 leq x leq 2.$$
I can't figure this out... can anyone show me how to solve it?
The answer is supposed to be $displaystyle frac{10sqrt{5}-2}{3}$.
multivariable-calculus
multivariable-calculus
edited Nov 25 '12 at 23:08
amWhy
192k28225439
edited Nov 25 '12 at 23:08
amWhy
192k28225439
edited Nov 25 '12 at 23:08
amWhy
192k28225439
edited Nov 25 '12 at 23:08
amWhy
192k28225439
edited Nov 25 '12 at 23:08
amWhy
192k28225439
192k28225439
asked Nov 25 '12 at 21:22
user39794
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1 Answer
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$begingroup$
Parametrize the curve $C$ by $gamma : [0,2] rightarrow mathbb{R}^2$ with $ gamma(t) = left( t, frac{t^2}{2} right).$ Then,
$$ int_{gamma} f(x,y) mathrm{ds} = int_0^2 f(gamma(t)) ||dot{gamma}(t)|| dt = int_0^2 frac{t^3}{frac{t^2}{2}} ||(1, t)|| dt = int_0^2 2t sqrt{1 + t^2} dt.$$
Solve this integral by substitution.
answered Nov 25 '12 at 21:36
levaplevap
47k23273
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Parametrize the curve $C$ by $gamma : [0,2] rightarrow mathbb{R}^2$ with $ gamma(t) = left( t, frac{t^2}{2} right).$ Then,
$$ int_{gamma} f(x,y) mathrm{ds} = int_0^2 f(gamma(t)) ||dot{gamma}(t)|| dt = int_0^2 frac{t^3}{frac{t^2}{2}} ||(1, t)|| dt = int_0^2 2t sqrt{1 + t^2} dt.$$
Solve this integral by substitution.
answered Nov 25 '12 at 21:36
levaplevap
47k23273
$endgroup$
add a comment |
$begingroup$
Parametrize the curve $C$ by $gamma : [0,2] rightarrow mathbb{R}^2$ with $ gamma(t) = left( t, frac{t^2}{2} right).$ Then,
$$ int_{gamma} f(x,y) mathrm{ds} = int_0^2 f(gamma(t)) ||dot{gamma}(t)|| dt = int_0^2 frac{t^3}{frac{t^2}{2}} ||(1, t)|| dt = int_0^2 2t sqrt{1 + t^2} dt.$$
Solve this integral by substitution.
answered Nov 25 '12 at 21:36
levaplevap
47k23273
$endgroup$
add a comment |
$begingroup$
Parametrize the curve $C$ by $gamma : [0,2] rightarrow mathbb{R}^2$ with $ gamma(t) = left( t, frac{t^2}{2} right).$ Then,
$$ int_{gamma} f(x,y) mathrm{ds} = int_0^2 f(gamma(t)) ||dot{gamma}(t)|| dt = int_0^2 frac{t^3}{frac{t^2}{2}} ||(1, t)|| dt = int_0^2 2t sqrt{1 + t^2} dt.$$
Solve this integral by substitution.
answered Nov 25 '12 at 21:36
levaplevap
47k23273
$endgroup$
Parametrize the curve $C$ by $gamma : [0,2] rightarrow mathbb{R}^2$ with $ gamma(t) = left( t, frac{t^2}{2} right).$ Then,
$$ int_{gamma} f(x,y) mathrm{ds} = int_0^2 f(gamma(t)) ||dot{gamma}(t)|| dt = int_0^2 frac{t^3}{frac{t^2}{2}} ||(1, t)|| dt = int_0^2 2t sqrt{1 + t^2} dt.$$
Solve this integral by substitution.
answered Nov 25 '12 at 21:36
levaplevap
47k23273
edited Dec 1 '18 at 14:38
answered Nov 25 '12 at 21:36
levaplevap
47k23273
answered Nov 25 '12 at 21:36
levaplevap
47k23273
answered Nov 25 '12 at 21:36
levaplevap
47k23273
47k23273
add a comment |
add a comment |
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