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Computing the volume inside a lemniscate.

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0

$begingroup$

Compute the volume of the lemniscate given as the graph of $x^4=x^2-y^2$.

So I thought the best way to solve this would be to take the parametrized curves, $gamma_1(t)=(t,sqrtt^2-t^4)$ and the curve $gamma_2(t)=(t,0)$ from $-1leq tleq 1$ and $dotgamma_1=(1,frac2t-4t^3)2sqrtt^2-t^4)$

Then using this integral for finding the area inside a curve $gamma_1$ I get $int_C xdy=int_-1^1 tfrac2t-4t^3)2sqrtt^2-t^4dt$. But this integral seems pretty difficult to compute. So I'm not sure if there is some change of variables that would make sense here, or if I'm not computing the correct curve.

multivariable-calculus

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asked Mar 20 at 19:47

AColoredReptileAColoredReptile

401210

$endgroup$

add a comment | 

0

$begingroup$

Compute the volume of the lemniscate given as the graph of $x^4=x^2-y^2$.

So I thought the best way to solve this would be to take the parametrized curves, $gamma_1(t)=(t,sqrtt^2-t^4)$ and the curve $gamma_2(t)=(t,0)$ from $-1leq tleq 1$ and $dotgamma_1=(1,frac2t-4t^3)2sqrtt^2-t^4)$

Then using this integral for finding the area inside a curve $gamma_1$ I get $int_C xdy=int_-1^1 tfrac2t-4t^3)2sqrtt^2-t^4dt$. But this integral seems pretty difficult to compute. So I'm not sure if there is some change of variables that would make sense here, or if I'm not computing the correct curve.

multivariable-calculus

share|cite|improve this question

asked Mar 20 at 19:47

AColoredReptileAColoredReptile

401210

$endgroup$

add a comment | 

$begingroup$

Compute the volume of the lemniscate given as the graph of $x^4=x^2-y^2$.

So I thought the best way to solve this would be to take the parametrized curves, $gamma_1(t)=(t,sqrtt^2-t^4)$ and the curve $gamma_2(t)=(t,0)$ from $-1leq tleq 1$ and $dotgamma_1=(1,frac2t-4t^3)2sqrtt^2-t^4)$

Then using this integral for finding the area inside a curve $gamma_1$ I get $int_C xdy=int_-1^1 tfrac2t-4t^3)2sqrtt^2-t^4dt$. But this integral seems pretty difficult to compute. So I'm not sure if there is some change of variables that would make sense here, or if I'm not computing the correct curve.

multivariable-calculus

share|cite|improve this question

asked Mar 20 at 19:47

AColoredReptileAColoredReptile

401210

$endgroup$

share|cite|improve this question

asked Mar 20 at 19:47

AColoredReptileAColoredReptile

401210

share|cite|improve this question

asked Mar 20 at 19:47

AColoredReptileAColoredReptile

401210

asked Mar 20 at 19:47

AColoredReptileAColoredReptile

401210

asked Mar 20 at 19:47

AColoredReptileAColoredReptile

401210

1 Answer
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1

$begingroup$

Be careful, because $int_-1^1 fract^2-2t^4sqrtt^2-t^4 dt < 0$ (you can assume $t>0$ because the integrand is even, so $int_-1^1dots = 2 int_0^1dots$ and the calculation is straightforward with variables changes $t = sqrtz$ and $z = 1-s$). Also $max y = 1/sqrt2$, but $4int_0^1/sqrt 2 fract^2-2t^4sqrtt^2-t^4 dt$ is not the right answer: can you see why?

An easy way to calculate the area inside $x^4 = x^2 - y^2$ is the following: because the domain of integration is symmetric respect the $x$- and $y$-axis, the total area is $4 int_0^1 sqrtx^2-x^4 dx = 4/3$ (you can calculate this integral with the same variable changes!).

share|cite|improve this answer

edited Mar 20 at 23:13

answered Mar 20 at 21:59

dcolazindcolazin

4395

$endgroup$

add a comment | 

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1

$begingroup$

Be careful, because $int_-1^1 fract^2-2t^4sqrtt^2-t^4 dt < 0$ (you can assume $t>0$ because the integrand is even, so $int_-1^1dots = 2 int_0^1dots$ and the calculation is straightforward with variables changes $t = sqrtz$ and $z = 1-s$). Also $max y = 1/sqrt2$, but $4int_0^1/sqrt 2 fract^2-2t^4sqrtt^2-t^4 dt$ is not the right answer: can you see why?

An easy way to calculate the area inside $x^4 = x^2 - y^2$ is the following: because the domain of integration is symmetric respect the $x$- and $y$-axis, the total area is $4 int_0^1 sqrtx^2-x^4 dx = 4/3$ (you can calculate this integral with the same variable changes!).

share|cite|improve this answer

edited Mar 20 at 23:13

answered Mar 20 at 21:59

dcolazindcolazin

4395

$endgroup$

add a comment | 

1

$begingroup$

Be careful, because $int_-1^1 fract^2-2t^4sqrtt^2-t^4 dt < 0$ (you can assume $t>0$ because the integrand is even, so $int_-1^1dots = 2 int_0^1dots$ and the calculation is straightforward with variables changes $t = sqrtz$ and $z = 1-s$). Also $max y = 1/sqrt2$, but $4int_0^1/sqrt 2 fract^2-2t^4sqrtt^2-t^4 dt$ is not the right answer: can you see why?

An easy way to calculate the area inside $x^4 = x^2 - y^2$ is the following: because the domain of integration is symmetric respect the $x$- and $y$-axis, the total area is $4 int_0^1 sqrtx^2-x^4 dx = 4/3$ (you can calculate this integral with the same variable changes!).

share|cite|improve this answer

edited Mar 20 at 23:13

answered Mar 20 at 21:59

dcolazindcolazin

4395

$endgroup$

add a comment | 

$begingroup$

Be careful, because $int_-1^1 fract^2-2t^4sqrtt^2-t^4 dt < 0$ (you can assume $t>0$ because the integrand is even, so $int_-1^1dots = 2 int_0^1dots$ and the calculation is straightforward with variables changes $t = sqrtz$ and $z = 1-s$). Also $max y = 1/sqrt2$, but $4int_0^1/sqrt 2 fract^2-2t^4sqrtt^2-t^4 dt$ is not the right answer: can you see why?

An easy way to calculate the area inside $x^4 = x^2 - y^2$ is the following: because the domain of integration is symmetric respect the $x$- and $y$-axis, the total area is $4 int_0^1 sqrtx^2-x^4 dx = 4/3$ (you can calculate this integral with the same variable changes!).

share|cite|improve this answer

edited Mar 20 at 23:13

answered Mar 20 at 21:59

dcolazindcolazin

4395

$endgroup$

share|cite|improve this answer

edited Mar 20 at 23:13

answered Mar 20 at 21:59

dcolazindcolazin

4395

answered Mar 20 at 21:59

dcolazindcolazin

4395

answered Mar 20 at 21:59

dcolazindcolazin

4395