Calculate the area of $z=frac{x^2}{2}+frac{y^2}{2};$ that is enclosed by $x^2+frac{y^2}{4}=1$












1












$begingroup$


The exercise is the text in the title. I'm studying surface integrals. To start I thougt to make a change to cartesian coordinates with $z$ as a function of $x$ and $y$, that is, $z=frac{x^2+y^2}{2}$ . With this I make the following parameterization: $O(x,y)=[x/2,y,z=frac{x^2+y^2}{2}]$ Then I calculated the norm of the normal vector of $O(x,y)$ that is $(-x,frac{-y}{2}, 1/2)$.
Is it okay what I did? How did I continue the exercise?






calculus integration multivariable-calculus surfaces surface-integrals







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$endgroup$












  • $begingroup$
    I don't understand your approach. Moreover are you sure that you have written all the formulas correctly? Have you a final resul for this exercise?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 11:18












  • $begingroup$
    @RobertZ What I did was parameterize trivially. My parameterization in principle was $L (x, y)=(x,y,(x^+y^)/2)$ but since the module of the vector product of the partial derivatives oficina L(x,y) did not allow me to use using the change in polar corrdinates x^2+(y^2)/2=r^2 in the integral of surface area I decided to use O(x,y) that does allow me to make this change to polar
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:42
















1












$begingroup$


The exercise is the text in the title. I'm studying surface integrals. To start I thougt to make a change to cartesian coordinates with $z$ as a function of $x$ and $y$, that is, $z=frac{x^2+y^2}{2}$ . With this I make the following parameterization: $O(x,y)=[x/2,y,z=frac{x^2+y^2}{2}]$ Then I calculated the norm of the normal vector of $O(x,y)$ that is $(-x,frac{-y}{2}, 1/2)$.
Is it okay what I did? How did I continue the exercise?






calculus integration multivariable-calculus surfaces surface-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't understand your approach. Moreover are you sure that you have written all the formulas correctly? Have you a final resul for this exercise?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 11:18












  • $begingroup$
    @RobertZ What I did was parameterize trivially. My parameterization in principle was $L (x, y)=(x,y,(x^+y^)/2)$ but since the module of the vector product of the partial derivatives oficina L(x,y) did not allow me to use using the change in polar corrdinates x^2+(y^2)/2=r^2 in the integral of surface area I decided to use O(x,y) that does allow me to make this change to polar
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:42














1












1








1





$begingroup$


The exercise is the text in the title. I'm studying surface integrals. To start I thougt to make a change to cartesian coordinates with $z$ as a function of $x$ and $y$, that is, $z=frac{x^2+y^2}{2}$ . With this I make the following parameterization: $O(x,y)=[x/2,y,z=frac{x^2+y^2}{2}]$ Then I calculated the norm of the normal vector of $O(x,y)$ that is $(-x,frac{-y}{2}, 1/2)$.
Is it okay what I did? How did I continue the exercise?






calculus integration multivariable-calculus surfaces surface-integrals







share|cite|improve this question











$endgroup$




The exercise is the text in the title. I'm studying surface integrals. To start I thougt to make a change to cartesian coordinates with $z$ as a function of $x$ and $y$, that is, $z=frac{x^2+y^2}{2}$ . With this I make the following parameterization: $O(x,y)=[x/2,y,z=frac{x^2+y^2}{2}]$ Then I calculated the norm of the normal vector of $O(x,y)$ that is $(-x,frac{-y}{2}, 1/2)$.
Is it okay what I did? How did I continue the exercise?






calculus integration multivariable-calculus surfaces surface-integrals




calculus integration multivariable-calculus surfaces surface-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 11:23









Robert Z

101k1069142




101k1069142










asked Dec 18 '18 at 10:57









AyescaAyesca

1999




1999












  • $begingroup$
    I don't understand your approach. Moreover are you sure that you have written all the formulas correctly? Have you a final resul for this exercise?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 11:18












  • $begingroup$
    @RobertZ What I did was parameterize trivially. My parameterization in principle was $L (x, y)=(x,y,(x^+y^)/2)$ but since the module of the vector product of the partial derivatives oficina L(x,y) did not allow me to use using the change in polar corrdinates x^2+(y^2)/2=r^2 in the integral of surface area I decided to use O(x,y) that does allow me to make this change to polar
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:42


















  • $begingroup$
    I don't understand your approach. Moreover are you sure that you have written all the formulas correctly? Have you a final resul for this exercise?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 11:18












  • $begingroup$
    @RobertZ What I did was parameterize trivially. My parameterization in principle was $L (x, y)=(x,y,(x^+y^)/2)$ but since the module of the vector product of the partial derivatives oficina L(x,y) did not allow me to use using the change in polar corrdinates x^2+(y^2)/2=r^2 in the integral of surface area I decided to use O(x,y) that does allow me to make this change to polar
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:42
















$begingroup$
I don't understand your approach. Moreover are you sure that you have written all the formulas correctly? Have you a final resul for this exercise?
$endgroup$
– Robert Z
Dec 18 '18 at 11:18






$begingroup$
I don't understand your approach. Moreover are you sure that you have written all the formulas correctly? Have you a final resul for this exercise?
$endgroup$
– Robert Z
Dec 18 '18 at 11:18














$begingroup$
@RobertZ What I did was parameterize trivially. My parameterization in principle was $L (x, y)=(x,y,(x^+y^)/2)$ but since the module of the vector product of the partial derivatives oficina L(x,y) did not allow me to use using the change in polar corrdinates x^2+(y^2)/2=r^2 in the integral of surface area I decided to use O(x,y) that does allow me to make this change to polar
$endgroup$
– Ayesca
Dec 18 '18 at 12:42




$begingroup$
@RobertZ What I did was parameterize trivially. My parameterization in principle was $L (x, y)=(x,y,(x^+y^)/2)$ but since the module of the vector product of the partial derivatives oficina L(x,y) did not allow me to use using the change in polar corrdinates x^2+(y^2)/2=r^2 in the integral of surface area I decided to use O(x,y) that does allow me to make this change to polar
$endgroup$
– Ayesca
Dec 18 '18 at 12:42










1 Answer
1






active

oldest

votes


















1












$begingroup$

According to the definition of surface area you have to evaluate
$$A=iint_Ssqrt{1+f_x^2+f_y^2}dxdy=iint_Ssqrt{1+x^2+y^2}dxdy$$
where $z=f(x,y)=frac{x^2+y^2}{2}$ (a paraboloid) and
$S={(x,y):x^2+frac{y^2}{4}leq 1}$ (the interior of an ellipse).
After letting $X=x$, and $Y=y/2$ you should be able to use the polar coordinates. Unfortunately in the final result elliptic integrals are involved and there is no closed form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What would the Jacobin module look like in this case? (when make the change to polar coordinates)
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:45






  • 2




    $begingroup$
    The integral does not have a closed form. Are you sure that the statement of the problem is correct? What is the source of this exercise. Have a result?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 12:52












  • $begingroup$
    I see. It is an exercise of a mathematical analysis II practice of the University of Buenos Aires
    $endgroup$
    – Ayesca
    Dec 18 '18 at 13:05






  • 1




    $begingroup$
    OK. If we replace $x^2+y^2/4=1$ with $x^2+y^2=1$ then the result is $-(2/3)pi+(4/3)sqrt{2}pi$.
    $endgroup$
    – Robert Z
    Dec 18 '18 at 13:11











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

According to the definition of surface area you have to evaluate
$$A=iint_Ssqrt{1+f_x^2+f_y^2}dxdy=iint_Ssqrt{1+x^2+y^2}dxdy$$
where $z=f(x,y)=frac{x^2+y^2}{2}$ (a paraboloid) and
$S={(x,y):x^2+frac{y^2}{4}leq 1}$ (the interior of an ellipse).
After letting $X=x$, and $Y=y/2$ you should be able to use the polar coordinates. Unfortunately in the final result elliptic integrals are involved and there is no closed form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What would the Jacobin module look like in this case? (when make the change to polar coordinates)
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:45






  • 2




    $begingroup$
    The integral does not have a closed form. Are you sure that the statement of the problem is correct? What is the source of this exercise. Have a result?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 12:52












  • $begingroup$
    I see. It is an exercise of a mathematical analysis II practice of the University of Buenos Aires
    $endgroup$
    – Ayesca
    Dec 18 '18 at 13:05






  • 1




    $begingroup$
    OK. If we replace $x^2+y^2/4=1$ with $x^2+y^2=1$ then the result is $-(2/3)pi+(4/3)sqrt{2}pi$.
    $endgroup$
    – Robert Z
    Dec 18 '18 at 13:11
















1












$begingroup$

According to the definition of surface area you have to evaluate
$$A=iint_Ssqrt{1+f_x^2+f_y^2}dxdy=iint_Ssqrt{1+x^2+y^2}dxdy$$
where $z=f(x,y)=frac{x^2+y^2}{2}$ (a paraboloid) and
$S={(x,y):x^2+frac{y^2}{4}leq 1}$ (the interior of an ellipse).
After letting $X=x$, and $Y=y/2$ you should be able to use the polar coordinates. Unfortunately in the final result elliptic integrals are involved and there is no closed form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What would the Jacobin module look like in this case? (when make the change to polar coordinates)
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:45






  • 2




    $begingroup$
    The integral does not have a closed form. Are you sure that the statement of the problem is correct? What is the source of this exercise. Have a result?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 12:52












  • $begingroup$
    I see. It is an exercise of a mathematical analysis II practice of the University of Buenos Aires
    $endgroup$
    – Ayesca
    Dec 18 '18 at 13:05






  • 1




    $begingroup$
    OK. If we replace $x^2+y^2/4=1$ with $x^2+y^2=1$ then the result is $-(2/3)pi+(4/3)sqrt{2}pi$.
    $endgroup$
    – Robert Z
    Dec 18 '18 at 13:11














1












1








1





$begingroup$

According to the definition of surface area you have to evaluate
$$A=iint_Ssqrt{1+f_x^2+f_y^2}dxdy=iint_Ssqrt{1+x^2+y^2}dxdy$$
where $z=f(x,y)=frac{x^2+y^2}{2}$ (a paraboloid) and
$S={(x,y):x^2+frac{y^2}{4}leq 1}$ (the interior of an ellipse).
After letting $X=x$, and $Y=y/2$ you should be able to use the polar coordinates. Unfortunately in the final result elliptic integrals are involved and there is no closed form.






share|cite|improve this answer











$endgroup$



According to the definition of surface area you have to evaluate
$$A=iint_Ssqrt{1+f_x^2+f_y^2}dxdy=iint_Ssqrt{1+x^2+y^2}dxdy$$
where $z=f(x,y)=frac{x^2+y^2}{2}$ (a paraboloid) and
$S={(x,y):x^2+frac{y^2}{4}leq 1}$ (the interior of an ellipse).
After letting $X=x$, and $Y=y/2$ you should be able to use the polar coordinates. Unfortunately in the final result elliptic integrals are involved and there is no closed form.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 13:17

























answered Dec 18 '18 at 11:05









Robert ZRobert Z

101k1069142




101k1069142












  • $begingroup$
    What would the Jacobin module look like in this case? (when make the change to polar coordinates)
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:45






  • 2




    $begingroup$
    The integral does not have a closed form. Are you sure that the statement of the problem is correct? What is the source of this exercise. Have a result?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 12:52












  • $begingroup$
    I see. It is an exercise of a mathematical analysis II practice of the University of Buenos Aires
    $endgroup$
    – Ayesca
    Dec 18 '18 at 13:05






  • 1




    $begingroup$
    OK. If we replace $x^2+y^2/4=1$ with $x^2+y^2=1$ then the result is $-(2/3)pi+(4/3)sqrt{2}pi$.
    $endgroup$
    – Robert Z
    Dec 18 '18 at 13:11


















  • $begingroup$
    What would the Jacobin module look like in this case? (when make the change to polar coordinates)
    $endgroup$
    – Ayesca
    Dec 18 '18 at 12:45






  • 2




    $begingroup$
    The integral does not have a closed form. Are you sure that the statement of the problem is correct? What is the source of this exercise. Have a result?
    $endgroup$
    – Robert Z
    Dec 18 '18 at 12:52












  • $begingroup$
    I see. It is an exercise of a mathematical analysis II practice of the University of Buenos Aires
    $endgroup$
    – Ayesca
    Dec 18 '18 at 13:05






  • 1




    $begingroup$
    OK. If we replace $x^2+y^2/4=1$ with $x^2+y^2=1$ then the result is $-(2/3)pi+(4/3)sqrt{2}pi$.
    $endgroup$
    – Robert Z
    Dec 18 '18 at 13:11
















$begingroup$
What would the Jacobin module look like in this case? (when make the change to polar coordinates)
$endgroup$
– Ayesca
Dec 18 '18 at 12:45




$begingroup$
What would the Jacobin module look like in this case? (when make the change to polar coordinates)
$endgroup$
– Ayesca
Dec 18 '18 at 12:45




2




2




$begingroup$
The integral does not have a closed form. Are you sure that the statement of the problem is correct? What is the source of this exercise. Have a result?
$endgroup$
– Robert Z
Dec 18 '18 at 12:52






$begingroup$
The integral does not have a closed form. Are you sure that the statement of the problem is correct? What is the source of this exercise. Have a result?
$endgroup$
– Robert Z
Dec 18 '18 at 12:52














$begingroup$
I see. It is an exercise of a mathematical analysis II practice of the University of Buenos Aires
$endgroup$
– Ayesca
Dec 18 '18 at 13:05




$begingroup$
I see. It is an exercise of a mathematical analysis II practice of the University of Buenos Aires
$endgroup$
– Ayesca
Dec 18 '18 at 13:05




1




1




$begingroup$
OK. If we replace $x^2+y^2/4=1$ with $x^2+y^2=1$ then the result is $-(2/3)pi+(4/3)sqrt{2}pi$.
$endgroup$
– Robert Z
Dec 18 '18 at 13:11




$begingroup$
OK. If we replace $x^2+y^2/4=1$ with $x^2+y^2=1$ then the result is $-(2/3)pi+(4/3)sqrt{2}pi$.
$endgroup$
– Robert Z
Dec 18 '18 at 13:11


















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