Mixing
Length of an Astroid
$begingroup$
I am having trouble with this question:
Calculate the length of the astroid of $x^{frac23}+y^{frac23}=1$. s = ?
I approached it by doing the following:
- setting $x^{frac23}=1$ because then I can find $x=sqrt{1}=1$
- Then I set $$s = 4int_0^1 sqrt{1 + left(frac{dy}{dx}right)^2}dx$$
- Then I did implicit differentiation to get
$$frac23 x^{-frac13} + frac23 y^{-frac13}left(frac{dy}{dx}right) = 0,\
frac{dy}{dx} = -left(frac{y}{x}right)^{frac13}.$$
I stopped there...
I appreciate the help. Thank you!
calculus
edited Apr 12 '14 at 22:42
user84413
22.9k11848
asked Apr 12 '14 at 21:48
Vanessa VitielloVanessa Vitiello
116110
$endgroup$
add a comment |
$begingroup$
I am having trouble with this question:
Calculate the length of the astroid of $x^{frac23}+y^{frac23}=1$. s = ?
I approached it by doing the following:
- setting $x^{frac23}=1$ because then I can find $x=sqrt{1}=1$
- Then I set $$s = 4int_0^1 sqrt{1 + left(frac{dy}{dx}right)^2}dx$$
- Then I did implicit differentiation to get
$$frac23 x^{-frac13} + frac23 y^{-frac13}left(frac{dy}{dx}right) = 0,\
frac{dy}{dx} = -left(frac{y}{x}right)^{frac13}.$$
I stopped there...
I appreciate the help. Thank you!
calculus
edited Apr 12 '14 at 22:42
user84413
22.9k11848
asked Apr 12 '14 at 21:48
Vanessa VitielloVanessa Vitiello
116110
$endgroup$
$begingroup$
Can you write $y$ as a function of $x$, i.e., $y=f(x)$?
$endgroup$
– David H
Apr 12 '14 at 22:14
add a comment |
$begingroup$
I am having trouble with this question:
Calculate the length of the astroid of $x^{frac23}+y^{frac23}=1$. s = ?
I approached it by doing the following:
- setting $x^{frac23}=1$ because then I can find $x=sqrt{1}=1$
- Then I set $$s = 4int_0^1 sqrt{1 + left(frac{dy}{dx}right)^2}dx$$
- Then I did implicit differentiation to get
$$frac23 x^{-frac13} + frac23 y^{-frac13}left(frac{dy}{dx}right) = 0,\
frac{dy}{dx} = -left(frac{y}{x}right)^{frac13}.$$
I stopped there...
I appreciate the help. Thank you!
calculus
edited Apr 12 '14 at 22:42
user84413
22.9k11848
asked Apr 12 '14 at 21:48
Vanessa VitielloVanessa Vitiello
116110
$endgroup$
I am having trouble with this question:
Calculate the length of the astroid of $x^{frac23}+y^{frac23}=1$. s = ?
I approached it by doing the following:
- setting $x^{frac23}=1$ because then I can find $x=sqrt{1}=1$
- Then I set $$s = 4int_0^1 sqrt{1 + left(frac{dy}{dx}right)^2}dx$$
- Then I did implicit differentiation to get
$$frac23 x^{-frac13} + frac23 y^{-frac13}left(frac{dy}{dx}right) = 0,\
frac{dy}{dx} = -left(frac{y}{x}right)^{frac13}.$$
I stopped there...
I appreciate the help. Thank you!
calculus
calculus
edited Apr 12 '14 at 22:42
user84413
22.9k11848
asked Apr 12 '14 at 21:48
Vanessa VitielloVanessa Vitiello
116110
edited Apr 12 '14 at 22:42
user84413
22.9k11848
asked Apr 12 '14 at 21:48
Vanessa VitielloVanessa Vitiello
116110
edited Apr 12 '14 at 22:42
user84413
22.9k11848
edited Apr 12 '14 at 22:42
user84413
22.9k11848
edited Apr 12 '14 at 22:42
user84413
22.9k11848
22.9k11848
asked Apr 12 '14 at 21:48
Vanessa VitielloVanessa Vitiello
116110
asked Apr 12 '14 at 21:48
Vanessa VitielloVanessa Vitiello
116110
asked Apr 12 '14 at 21:48
Vanessa VitielloVanessa Vitiello
116110
116110
$begingroup$
Can you write $y$ as a function of $x$, i.e., $y=f(x)$?
$endgroup$
– David H
Apr 12 '14 at 22:14
add a comment |
$begingroup$
Can you write $y$ as a function of $x$, i.e., $y=f(x)$?
$endgroup$
– David H
Apr 12 '14 at 22:14
$begingroup$
Can you write $y$ as a function of $x$, i.e., $y=f(x)$?
$endgroup$
– David H
Apr 12 '14 at 22:14
$begingroup$
Can you write $y$ as a function of $x$, i.e., $y=f(x)$?
$endgroup$
– David H
Apr 12 '14 at 22:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$ds^2=dx^2+dy^2$$
$$ds=sqrt{dx^2+dy^2}=sqrt{1+left(frac{dy}{dx}right)^2}dx$$
$$s=int sqrt{1+left(frac{dy}{dx}right)^2}dx$$
You had the right approach, but the problem was with the implicit differentiation. Since you have $y$ in there, it becomes a problem when integrating.
Instead, isolate $y$ from the original and take the derivative of that:
$$x^{frac23}+y^{frac23}=1$$
$$y^{frac23}=1-x^{frac23}$$
$$y=left(1-x^{frac23}right)^frac32$$
Now when you find $frac{dy}{dx}$, it will only be in terms of $x$. So take the derivative of that:
$$frac{dy}{dx}=frac32 left(1-x^{frac23}right)^frac12 left(-frac23x^{-frac13}right)$$
Get rid of the fractions:
$$frac{dy}{dx}=left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)$$
Now plug it in:
$$s=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)^2}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right) left(x^{-frac23}right)right)}dx=4int_0^1 sqrt{1+left(x^{-frac23}-1right)}dx=4int_0^1 sqrt{x^{-frac23}}dx=4int_0^1 x^{-frac13}dx$$
So easy to integrate now!
$$s=4int_0^1 x^{-frac13}dx=4left(frac32x^frac23Biggr|_0^1right)=4left(frac32right)=therefore 6$$
answered Apr 12 '14 at 22:25
ShaharShahar
3,1742917
$endgroup$
add a comment |
$begingroup$
A potentially easier way to do this is to parametrize the astroid by taking advantage of the trig identity $cos^2(theta)+sin^2(theta) = 1$.
Take $x = cos^3(t)$ and $y = sin^3(t)$ where $0leq t leq 2pi$.
By symmetry, we can simply find the arclength of $1/4$th the astroid and multiply by $4$ at the end. The bounds on our integration will be $0 leq t leq pi/2$.
Now simply use the formula for arclength:
$$int_0^{pi/2} sqrt{left( frac{dx}{dt}right)^2 + left( frac{dy}{dt} right)^2} dt$$
answered Apr 12 '14 at 22:38
Kaj HansenKaj Hansen
27.5k43779
$endgroup$
$begingroup$
Thank you very much! This was very helpful!
$endgroup$
– Vanessa Vitiello
May 8 '14 at 22:40
add a comment |
$begingroup$
You can also continue using your formula for s.
$$s=4int_0^1sqrt{1+left(frac{dy}{dx}right)^2}dx$$
When you fill in $frac{dy}{dx} = -frac{y^frac{1}{3}}{x^frac{1}{3}}$ in $s$, you get
$$s=4int_0^1 sqrt{1+left(frac{y^frac{1}{3}}{x^frac{1}{3}}right)^2}dx$$
$$s=4int_0^1 sqrt{frac{x^frac{2}{3}+y^frac{2}{3}}{x^frac{2}{3}}}dx$$
$$s=4int_0^1 frac{1}{x^frac{1}{3}}sqrt{x^frac{2}{3}+{y^frac{2}{3}}}dx$$
But the original formula says that $x^frac{2}{3}+y^frac{2}{3}=1$, so
$$s=4int_0^1 frac{1}{x^frac{1}{3}}dx=4int_0^1 x^frac{-1}{3}dx=4left(frac{3}{2}right)=6$$
answered Jan 24 at 15:48
kilian vounckxkilian vounckx
1
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$begingroup$
Great answer and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 24 at 15:55
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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oldest
votes
$begingroup$
$$ds^2=dx^2+dy^2$$
$$ds=sqrt{dx^2+dy^2}=sqrt{1+left(frac{dy}{dx}right)^2}dx$$
$$s=int sqrt{1+left(frac{dy}{dx}right)^2}dx$$
You had the right approach, but the problem was with the implicit differentiation. Since you have $y$ in there, it becomes a problem when integrating.
Instead, isolate $y$ from the original and take the derivative of that:
$$x^{frac23}+y^{frac23}=1$$
$$y^{frac23}=1-x^{frac23}$$
$$y=left(1-x^{frac23}right)^frac32$$
Now when you find $frac{dy}{dx}$, it will only be in terms of $x$. So take the derivative of that:
$$frac{dy}{dx}=frac32 left(1-x^{frac23}right)^frac12 left(-frac23x^{-frac13}right)$$
Get rid of the fractions:
$$frac{dy}{dx}=left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)$$
Now plug it in:
$$s=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)^2}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right) left(x^{-frac23}right)right)}dx=4int_0^1 sqrt{1+left(x^{-frac23}-1right)}dx=4int_0^1 sqrt{x^{-frac23}}dx=4int_0^1 x^{-frac13}dx$$
So easy to integrate now!
$$s=4int_0^1 x^{-frac13}dx=4left(frac32x^frac23Biggr|_0^1right)=4left(frac32right)=therefore 6$$
answered Apr 12 '14 at 22:25
ShaharShahar
3,1742917
$endgroup$
add a comment |
$begingroup$
$$ds^2=dx^2+dy^2$$
$$ds=sqrt{dx^2+dy^2}=sqrt{1+left(frac{dy}{dx}right)^2}dx$$
$$s=int sqrt{1+left(frac{dy}{dx}right)^2}dx$$
You had the right approach, but the problem was with the implicit differentiation. Since you have $y$ in there, it becomes a problem when integrating.
Instead, isolate $y$ from the original and take the derivative of that:
$$x^{frac23}+y^{frac23}=1$$
$$y^{frac23}=1-x^{frac23}$$
$$y=left(1-x^{frac23}right)^frac32$$
Now when you find $frac{dy}{dx}$, it will only be in terms of $x$. So take the derivative of that:
$$frac{dy}{dx}=frac32 left(1-x^{frac23}right)^frac12 left(-frac23x^{-frac13}right)$$
Get rid of the fractions:
$$frac{dy}{dx}=left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)$$
Now plug it in:
$$s=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)^2}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right) left(x^{-frac23}right)right)}dx=4int_0^1 sqrt{1+left(x^{-frac23}-1right)}dx=4int_0^1 sqrt{x^{-frac23}}dx=4int_0^1 x^{-frac13}dx$$
So easy to integrate now!
$$s=4int_0^1 x^{-frac13}dx=4left(frac32x^frac23Biggr|_0^1right)=4left(frac32right)=therefore 6$$
answered Apr 12 '14 at 22:25
ShaharShahar
3,1742917
$endgroup$
add a comment |
$begingroup$
$$ds^2=dx^2+dy^2$$
$$ds=sqrt{dx^2+dy^2}=sqrt{1+left(frac{dy}{dx}right)^2}dx$$
$$s=int sqrt{1+left(frac{dy}{dx}right)^2}dx$$
You had the right approach, but the problem was with the implicit differentiation. Since you have $y$ in there, it becomes a problem when integrating.
Instead, isolate $y$ from the original and take the derivative of that:
$$x^{frac23}+y^{frac23}=1$$
$$y^{frac23}=1-x^{frac23}$$
$$y=left(1-x^{frac23}right)^frac32$$
Now when you find $frac{dy}{dx}$, it will only be in terms of $x$. So take the derivative of that:
$$frac{dy}{dx}=frac32 left(1-x^{frac23}right)^frac12 left(-frac23x^{-frac13}right)$$
Get rid of the fractions:
$$frac{dy}{dx}=left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)$$
Now plug it in:
$$s=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)^2}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right) left(x^{-frac23}right)right)}dx=4int_0^1 sqrt{1+left(x^{-frac23}-1right)}dx=4int_0^1 sqrt{x^{-frac23}}dx=4int_0^1 x^{-frac13}dx$$
So easy to integrate now!
$$s=4int_0^1 x^{-frac13}dx=4left(frac32x^frac23Biggr|_0^1right)=4left(frac32right)=therefore 6$$
answered Apr 12 '14 at 22:25
ShaharShahar
3,1742917
$endgroup$
$$ds^2=dx^2+dy^2$$
$$ds=sqrt{dx^2+dy^2}=sqrt{1+left(frac{dy}{dx}right)^2}dx$$
$$s=int sqrt{1+left(frac{dy}{dx}right)^2}dx$$
You had the right approach, but the problem was with the implicit differentiation. Since you have $y$ in there, it becomes a problem when integrating.
Instead, isolate $y$ from the original and take the derivative of that:
$$x^{frac23}+y^{frac23}=1$$
$$y^{frac23}=1-x^{frac23}$$
$$y=left(1-x^{frac23}right)^frac32$$
Now when you find $frac{dy}{dx}$, it will only be in terms of $x$. So take the derivative of that:
$$frac{dy}{dx}=frac32 left(1-x^{frac23}right)^frac12 left(-frac23x^{-frac13}right)$$
Get rid of the fractions:
$$frac{dy}{dx}=left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)$$
Now plug it in:
$$s=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)^2}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)left(1-x^{frac23}right)^frac12 left(-x^{-frac13}right)right)}dx=4int_0^1 sqrt{1+left(left(1-x^{frac23}right) left(x^{-frac23}right)right)}dx=4int_0^1 sqrt{1+left(x^{-frac23}-1right)}dx=4int_0^1 sqrt{x^{-frac23}}dx=4int_0^1 x^{-frac13}dx$$
So easy to integrate now!
$$s=4int_0^1 x^{-frac13}dx=4left(frac32x^frac23Biggr|_0^1right)=4left(frac32right)=therefore 6$$
answered Apr 12 '14 at 22:25
ShaharShahar
3,1742917
answered Apr 12 '14 at 22:25
ShaharShahar
3,1742917
answered Apr 12 '14 at 22:25
ShaharShahar
3,1742917
answered Apr 12 '14 at 22:25
ShaharShahar
3,1742917
3,1742917
add a comment |
add a comment |
$begingroup$
A potentially easier way to do this is to parametrize the astroid by taking advantage of the trig identity $cos^2(theta)+sin^2(theta) = 1$.
Take $x = cos^3(t)$ and $y = sin^3(t)$ where $0leq t leq 2pi$.
By symmetry, we can simply find the arclength of $1/4$th the astroid and multiply by $4$ at the end. The bounds on our integration will be $0 leq t leq pi/2$.
Now simply use the formula for arclength:
$$int_0^{pi/2} sqrt{left( frac{dx}{dt}right)^2 + left( frac{dy}{dt} right)^2} dt$$
answered Apr 12 '14 at 22:38
Kaj HansenKaj Hansen
27.5k43779
$endgroup$
$begingroup$
Thank you very much! This was very helpful!
$endgroup$
– Vanessa Vitiello
May 8 '14 at 22:40
add a comment |
$begingroup$
A potentially easier way to do this is to parametrize the astroid by taking advantage of the trig identity $cos^2(theta)+sin^2(theta) = 1$.
Take $x = cos^3(t)$ and $y = sin^3(t)$ where $0leq t leq 2pi$.
By symmetry, we can simply find the arclength of $1/4$th the astroid and multiply by $4$ at the end. The bounds on our integration will be $0 leq t leq pi/2$.
Now simply use the formula for arclength:
$$int_0^{pi/2} sqrt{left( frac{dx}{dt}right)^2 + left( frac{dy}{dt} right)^2} dt$$
answered Apr 12 '14 at 22:38
Kaj HansenKaj Hansen
27.5k43779
$endgroup$
$begingroup$
Thank you very much! This was very helpful!
$endgroup$
– Vanessa Vitiello
May 8 '14 at 22:40
add a comment |
$begingroup$
A potentially easier way to do this is to parametrize the astroid by taking advantage of the trig identity $cos^2(theta)+sin^2(theta) = 1$.
Take $x = cos^3(t)$ and $y = sin^3(t)$ where $0leq t leq 2pi$.
By symmetry, we can simply find the arclength of $1/4$th the astroid and multiply by $4$ at the end. The bounds on our integration will be $0 leq t leq pi/2$.
Now simply use the formula for arclength:
$$int_0^{pi/2} sqrt{left( frac{dx}{dt}right)^2 + left( frac{dy}{dt} right)^2} dt$$
answered Apr 12 '14 at 22:38
Kaj HansenKaj Hansen
27.5k43779
$endgroup$
A potentially easier way to do this is to parametrize the astroid by taking advantage of the trig identity $cos^2(theta)+sin^2(theta) = 1$.
Take $x = cos^3(t)$ and $y = sin^3(t)$ where $0leq t leq 2pi$.
By symmetry, we can simply find the arclength of $1/4$th the astroid and multiply by $4$ at the end. The bounds on our integration will be $0 leq t leq pi/2$.
Now simply use the formula for arclength:
$$int_0^{pi/2} sqrt{left( frac{dx}{dt}right)^2 + left( frac{dy}{dt} right)^2} dt$$
answered Apr 12 '14 at 22:38
Kaj HansenKaj Hansen
27.5k43779
edited Aug 9 '16 at 23:42
answered Apr 12 '14 at 22:38
Kaj HansenKaj Hansen
27.5k43779
answered Apr 12 '14 at 22:38
Kaj HansenKaj Hansen
27.5k43779
answered Apr 12 '14 at 22:38
Kaj HansenKaj Hansen
27.5k43779
27.5k43779
$begingroup$
Thank you very much! This was very helpful!
$endgroup$
– Vanessa Vitiello
May 8 '14 at 22:40
add a comment |
$begingroup$
Thank you very much! This was very helpful!
$endgroup$
– Vanessa Vitiello
May 8 '14 at 22:40
$begingroup$
Thank you very much! This was very helpful!
$endgroup$
– Vanessa Vitiello
May 8 '14 at 22:40
$begingroup$
Thank you very much! This was very helpful!
$endgroup$
– Vanessa Vitiello
May 8 '14 at 22:40
add a comment |
$begingroup$
You can also continue using your formula for s.
$$s=4int_0^1sqrt{1+left(frac{dy}{dx}right)^2}dx$$
When you fill in $frac{dy}{dx} = -frac{y^frac{1}{3}}{x^frac{1}{3}}$ in $s$, you get
$$s=4int_0^1 sqrt{1+left(frac{y^frac{1}{3}}{x^frac{1}{3}}right)^2}dx$$
$$s=4int_0^1 sqrt{frac{x^frac{2}{3}+y^frac{2}{3}}{x^frac{2}{3}}}dx$$
$$s=4int_0^1 frac{1}{x^frac{1}{3}}sqrt{x^frac{2}{3}+{y^frac{2}{3}}}dx$$
But the original formula says that $x^frac{2}{3}+y^frac{2}{3}=1$, so
$$s=4int_0^1 frac{1}{x^frac{1}{3}}dx=4int_0^1 x^frac{-1}{3}dx=4left(frac{3}{2}right)=6$$
answered Jan 24 at 15:48
kilian vounckxkilian vounckx
1
$endgroup$
$begingroup$
Great answer and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 24 at 15:55
add a comment |
$begingroup$
You can also continue using your formula for s.
$$s=4int_0^1sqrt{1+left(frac{dy}{dx}right)^2}dx$$
When you fill in $frac{dy}{dx} = -frac{y^frac{1}{3}}{x^frac{1}{3}}$ in $s$, you get
$$s=4int_0^1 sqrt{1+left(frac{y^frac{1}{3}}{x^frac{1}{3}}right)^2}dx$$
$$s=4int_0^1 sqrt{frac{x^frac{2}{3}+y^frac{2}{3}}{x^frac{2}{3}}}dx$$
$$s=4int_0^1 frac{1}{x^frac{1}{3}}sqrt{x^frac{2}{3}+{y^frac{2}{3}}}dx$$
But the original formula says that $x^frac{2}{3}+y^frac{2}{3}=1$, so
$$s=4int_0^1 frac{1}{x^frac{1}{3}}dx=4int_0^1 x^frac{-1}{3}dx=4left(frac{3}{2}right)=6$$
answered Jan 24 at 15:48
kilian vounckxkilian vounckx
1
$endgroup$
$begingroup$
Great answer and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 24 at 15:55
add a comment |
$begingroup$
You can also continue using your formula for s.
$$s=4int_0^1sqrt{1+left(frac{dy}{dx}right)^2}dx$$
When you fill in $frac{dy}{dx} = -frac{y^frac{1}{3}}{x^frac{1}{3}}$ in $s$, you get
$$s=4int_0^1 sqrt{1+left(frac{y^frac{1}{3}}{x^frac{1}{3}}right)^2}dx$$
$$s=4int_0^1 sqrt{frac{x^frac{2}{3}+y^frac{2}{3}}{x^frac{2}{3}}}dx$$
$$s=4int_0^1 frac{1}{x^frac{1}{3}}sqrt{x^frac{2}{3}+{y^frac{2}{3}}}dx$$
But the original formula says that $x^frac{2}{3}+y^frac{2}{3}=1$, so
$$s=4int_0^1 frac{1}{x^frac{1}{3}}dx=4int_0^1 x^frac{-1}{3}dx=4left(frac{3}{2}right)=6$$
answered Jan 24 at 15:48
kilian vounckxkilian vounckx
1
$endgroup$
You can also continue using your formula for s.
$$s=4int_0^1sqrt{1+left(frac{dy}{dx}right)^2}dx$$
When you fill in $frac{dy}{dx} = -frac{y^frac{1}{3}}{x^frac{1}{3}}$ in $s$, you get
$$s=4int_0^1 sqrt{1+left(frac{y^frac{1}{3}}{x^frac{1}{3}}right)^2}dx$$
$$s=4int_0^1 sqrt{frac{x^frac{2}{3}+y^frac{2}{3}}{x^frac{2}{3}}}dx$$
$$s=4int_0^1 frac{1}{x^frac{1}{3}}sqrt{x^frac{2}{3}+{y^frac{2}{3}}}dx$$
But the original formula says that $x^frac{2}{3}+y^frac{2}{3}=1$, so
$$s=4int_0^1 frac{1}{x^frac{1}{3}}dx=4int_0^1 x^frac{-1}{3}dx=4left(frac{3}{2}right)=6$$
answered Jan 24 at 15:48
kilian vounckxkilian vounckx
1
answered Jan 24 at 15:48
kilian vounckxkilian vounckx
1
answered Jan 24 at 15:48
kilian vounckxkilian vounckx
1
answered Jan 24 at 15:48
kilian vounckxkilian vounckx
1
1
$begingroup$
Great answer and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
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– dantopa
Jan 24 at 15:55
add a comment |
$begingroup$
Great answer and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 24 at 15:55
$begingroup$
Great answer and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 24 at 15:55
$begingroup$
Great answer and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 24 at 15:55
add a comment |
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Can you write $y$ as a function of $x$, i.e., $y=f(x)$?
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– David H
Apr 12 '14 at 22:14