How to evaluate a parameterized surface integral?












0














Suppose you have to evaluate the surface integral $$intint_S (x^2+y^2+4)space dS$$ where $S$ is the surface parameterized by $textbf{r} = <2uv, u^2-v^2, u^2+v^2>$ with $u^2+v^2 le 16.$



I know the equation to solve the surface integral is $$intint_S f space dS = intint_S f(textbf{r}(u,v))space |textbf{r}_u times textbf{r}_v |space du space dv$$ and I will have no trouble taking the cross product or evaluating the integral. However, how would one go about finding the limits for $S$? I don't even remotely know what $textbf{r} = <2uv, u^2-v^2, u^2+v^2>$ might be. Is there some way maybe to transform $textbf{r}$ to an explicit $z = f(x,y)$?



Thanks.






multivariable-calculus surface-integrals parametrization







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  • The domain in the right handed double integral is not the same of the left handed one. In fact, the domain in the right handed integral is the parametrization's domain.
    – DiegoMath
    Nov 22 at 21:05
















0














Suppose you have to evaluate the surface integral $$intint_S (x^2+y^2+4)space dS$$ where $S$ is the surface parameterized by $textbf{r} = <2uv, u^2-v^2, u^2+v^2>$ with $u^2+v^2 le 16.$



I know the equation to solve the surface integral is $$intint_S f space dS = intint_S f(textbf{r}(u,v))space |textbf{r}_u times textbf{r}_v |space du space dv$$ and I will have no trouble taking the cross product or evaluating the integral. However, how would one go about finding the limits for $S$? I don't even remotely know what $textbf{r} = <2uv, u^2-v^2, u^2+v^2>$ might be. Is there some way maybe to transform $textbf{r}$ to an explicit $z = f(x,y)$?



Thanks.






multivariable-calculus surface-integrals parametrization







share|cite|improve this question






















  • The domain in the right handed double integral is not the same of the left handed one. In fact, the domain in the right handed integral is the parametrization's domain.
    – DiegoMath
    Nov 22 at 21:05














0












0








0


1





Suppose you have to evaluate the surface integral $$intint_S (x^2+y^2+4)space dS$$ where $S$ is the surface parameterized by $textbf{r} = <2uv, u^2-v^2, u^2+v^2>$ with $u^2+v^2 le 16.$



I know the equation to solve the surface integral is $$intint_S f space dS = intint_S f(textbf{r}(u,v))space |textbf{r}_u times textbf{r}_v |space du space dv$$ and I will have no trouble taking the cross product or evaluating the integral. However, how would one go about finding the limits for $S$? I don't even remotely know what $textbf{r} = <2uv, u^2-v^2, u^2+v^2>$ might be. Is there some way maybe to transform $textbf{r}$ to an explicit $z = f(x,y)$?



Thanks.






multivariable-calculus surface-integrals parametrization







share|cite|improve this question













Suppose you have to evaluate the surface integral $$intint_S (x^2+y^2+4)space dS$$ where $S$ is the surface parameterized by $textbf{r} = <2uv, u^2-v^2, u^2+v^2>$ with $u^2+v^2 le 16.$



I know the equation to solve the surface integral is $$intint_S f space dS = intint_S f(textbf{r}(u,v))space |textbf{r}_u times textbf{r}_v |space du space dv$$ and I will have no trouble taking the cross product or evaluating the integral. However, how would one go about finding the limits for $S$? I don't even remotely know what $textbf{r} = <2uv, u^2-v^2, u^2+v^2>$ might be. Is there some way maybe to transform $textbf{r}$ to an explicit $z = f(x,y)$?



Thanks.






multivariable-calculus surface-integrals parametrization




multivariable-calculus surface-integrals parametrization






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asked Nov 22 at 20:58









Art

3127




3127












  • The domain in the right handed double integral is not the same of the left handed one. In fact, the domain in the right handed integral is the parametrization's domain.
    – DiegoMath
    Nov 22 at 21:05


















  • The domain in the right handed double integral is not the same of the left handed one. In fact, the domain in the right handed integral is the parametrization's domain.
    – DiegoMath
    Nov 22 at 21:05
















The domain in the right handed double integral is not the same of the left handed one. In fact, the domain in the right handed integral is the parametrization's domain.
– DiegoMath
Nov 22 at 21:05




The domain in the right handed double integral is not the same of the left handed one. In fact, the domain in the right handed integral is the parametrization's domain.
– DiegoMath
Nov 22 at 21:05










1 Answer
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active

oldest

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1














You do not need know how the surface is. Notice that the domain of the parametrizations is a disc of radius 4.



Hint: Since the domain is a disc, try integrate the right hand double integral by using polar coordinates.



I hope I have helped you.






share|cite|improve this answer





















  • So $S$ is not the entire surface?
    – Art
    Nov 22 at 21:07










  • $S$ is given by the parametrization. What do you mean with entire surface?
    – DiegoMath
    Nov 22 at 21:12












  • The limits for S in $intint_S$ are not for the entire surface?
    – Art
    Nov 22 at 21:14








  • 1




    You write this: where $S$ is the surface parameterized by $r=langle 2uv,u^2−v^2,u^2+v^2rangle$ with $u^2+v^2leq16$. So, your entire surface is given by this parametrization
    – DiegoMath
    Nov 22 at 21:16








  • 1




    For example, if your surface $S$ is given by $r=langle u,v,u+vrangle$ where $0leq uleq 1$ and $0leq vleq 1$, then the surface is a piece of a plane and this is the entire surface
    – DiegoMath
    Nov 22 at 21:19











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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














You do not need know how the surface is. Notice that the domain of the parametrizations is a disc of radius 4.



Hint: Since the domain is a disc, try integrate the right hand double integral by using polar coordinates.



I hope I have helped you.






share|cite|improve this answer





















  • So $S$ is not the entire surface?
    – Art
    Nov 22 at 21:07










  • $S$ is given by the parametrization. What do you mean with entire surface?
    – DiegoMath
    Nov 22 at 21:12












  • The limits for S in $intint_S$ are not for the entire surface?
    – Art
    Nov 22 at 21:14








  • 1




    You write this: where $S$ is the surface parameterized by $r=langle 2uv,u^2−v^2,u^2+v^2rangle$ with $u^2+v^2leq16$. So, your entire surface is given by this parametrization
    – DiegoMath
    Nov 22 at 21:16








  • 1




    For example, if your surface $S$ is given by $r=langle u,v,u+vrangle$ where $0leq uleq 1$ and $0leq vleq 1$, then the surface is a piece of a plane and this is the entire surface
    – DiegoMath
    Nov 22 at 21:19
















1














You do not need know how the surface is. Notice that the domain of the parametrizations is a disc of radius 4.



Hint: Since the domain is a disc, try integrate the right hand double integral by using polar coordinates.



I hope I have helped you.






share|cite|improve this answer





















  • So $S$ is not the entire surface?
    – Art
    Nov 22 at 21:07










  • $S$ is given by the parametrization. What do you mean with entire surface?
    – DiegoMath
    Nov 22 at 21:12












  • The limits for S in $intint_S$ are not for the entire surface?
    – Art
    Nov 22 at 21:14








  • 1




    You write this: where $S$ is the surface parameterized by $r=langle 2uv,u^2−v^2,u^2+v^2rangle$ with $u^2+v^2leq16$. So, your entire surface is given by this parametrization
    – DiegoMath
    Nov 22 at 21:16








  • 1




    For example, if your surface $S$ is given by $r=langle u,v,u+vrangle$ where $0leq uleq 1$ and $0leq vleq 1$, then the surface is a piece of a plane and this is the entire surface
    – DiegoMath
    Nov 22 at 21:19














1












1








1






You do not need know how the surface is. Notice that the domain of the parametrizations is a disc of radius 4.



Hint: Since the domain is a disc, try integrate the right hand double integral by using polar coordinates.



I hope I have helped you.






share|cite|improve this answer












You do not need know how the surface is. Notice that the domain of the parametrizations is a disc of radius 4.



Hint: Since the domain is a disc, try integrate the right hand double integral by using polar coordinates.



I hope I have helped you.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 21:02









DiegoMath

2,0491021




2,0491021












  • So $S$ is not the entire surface?
    – Art
    Nov 22 at 21:07










  • $S$ is given by the parametrization. What do you mean with entire surface?
    – DiegoMath
    Nov 22 at 21:12












  • The limits for S in $intint_S$ are not for the entire surface?
    – Art
    Nov 22 at 21:14








  • 1




    You write this: where $S$ is the surface parameterized by $r=langle 2uv,u^2−v^2,u^2+v^2rangle$ with $u^2+v^2leq16$. So, your entire surface is given by this parametrization
    – DiegoMath
    Nov 22 at 21:16








  • 1




    For example, if your surface $S$ is given by $r=langle u,v,u+vrangle$ where $0leq uleq 1$ and $0leq vleq 1$, then the surface is a piece of a plane and this is the entire surface
    – DiegoMath
    Nov 22 at 21:19


















  • So $S$ is not the entire surface?
    – Art
    Nov 22 at 21:07










  • $S$ is given by the parametrization. What do you mean with entire surface?
    – DiegoMath
    Nov 22 at 21:12












  • The limits for S in $intint_S$ are not for the entire surface?
    – Art
    Nov 22 at 21:14








  • 1




    You write this: where $S$ is the surface parameterized by $r=langle 2uv,u^2−v^2,u^2+v^2rangle$ with $u^2+v^2leq16$. So, your entire surface is given by this parametrization
    – DiegoMath
    Nov 22 at 21:16








  • 1




    For example, if your surface $S$ is given by $r=langle u,v,u+vrangle$ where $0leq uleq 1$ and $0leq vleq 1$, then the surface is a piece of a plane and this is the entire surface
    – DiegoMath
    Nov 22 at 21:19
















So $S$ is not the entire surface?
– Art
Nov 22 at 21:07




So $S$ is not the entire surface?
– Art
Nov 22 at 21:07












$S$ is given by the parametrization. What do you mean with entire surface?
– DiegoMath
Nov 22 at 21:12






$S$ is given by the parametrization. What do you mean with entire surface?
– DiegoMath
Nov 22 at 21:12














The limits for S in $intint_S$ are not for the entire surface?
– Art
Nov 22 at 21:14






The limits for S in $intint_S$ are not for the entire surface?
– Art
Nov 22 at 21:14






1




1




You write this: where $S$ is the surface parameterized by $r=langle 2uv,u^2−v^2,u^2+v^2rangle$ with $u^2+v^2leq16$. So, your entire surface is given by this parametrization
– DiegoMath
Nov 22 at 21:16






You write this: where $S$ is the surface parameterized by $r=langle 2uv,u^2−v^2,u^2+v^2rangle$ with $u^2+v^2leq16$. So, your entire surface is given by this parametrization
– DiegoMath
Nov 22 at 21:16






1




1




For example, if your surface $S$ is given by $r=langle u,v,u+vrangle$ where $0leq uleq 1$ and $0leq vleq 1$, then the surface is a piece of a plane and this is the entire surface
– DiegoMath
Nov 22 at 21:19




For example, if your surface $S$ is given by $r=langle u,v,u+vrangle$ where $0leq uleq 1$ and $0leq vleq 1$, then the surface is a piece of a plane and this is the entire surface
– DiegoMath
Nov 22 at 21:19


















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