Using Green's Theorem to find the flux
$begingroup$
$F(x,y)=langle y^2+e^x,x^2+e^yrangle$. Using green's theorem in its circulation and flux forms, determine the flux and circulation of $F$ around the triangle $T$, where $T$ is the triangle with vertices $(0,0),(1,0),$ and $(0,1)$, oriented counterclockwise.
My Try:
$f=y^2+e^x$
$g=x^2+e^y$
$$frac{partial f}{partial y}=2y$$
$$frac{partial g}{partial x}=2x$$
$$int Fcdot dr=intint_R(2x-2y)dA$$
$$int^{1}_{0}int_{0}^{1-x}(2x-2y)dydx=0$$
My quesiton: Is my above attempt correct? Because I got the answer as $0$
calculus integration multivariable-calculus definite-integrals greens-theorem
edited Dec 1 '18 at 21:27
Masacroso
13k41746
asked Dec 1 '18 at 21:19
user982787user982787
1117
$endgroup$
add a comment |
$begingroup$
$F(x,y)=langle y^2+e^x,x^2+e^yrangle$. Using green's theorem in its circulation and flux forms, determine the flux and circulation of $F$ around the triangle $T$, where $T$ is the triangle with vertices $(0,0),(1,0),$ and $(0,1)$, oriented counterclockwise.
My Try:
$f=y^2+e^x$
$g=x^2+e^y$
$$frac{partial f}{partial y}=2y$$
$$frac{partial g}{partial x}=2x$$
$$int Fcdot dr=intint_R(2x-2y)dA$$
$$int^{1}_{0}int_{0}^{1-x}(2x-2y)dydx=0$$
My quesiton: Is my above attempt correct? Because I got the answer as $0$
calculus integration multivariable-calculus definite-integrals greens-theorem
edited Dec 1 '18 at 21:27
Masacroso
13k41746
asked Dec 1 '18 at 21:19
user982787user982787
1117
$endgroup$
add a comment |
$begingroup$
$F(x,y)=langle y^2+e^x,x^2+e^yrangle$. Using green's theorem in its circulation and flux forms, determine the flux and circulation of $F$ around the triangle $T$, where $T$ is the triangle with vertices $(0,0),(1,0),$ and $(0,1)$, oriented counterclockwise.
My Try:
$f=y^2+e^x$
$g=x^2+e^y$
$$frac{partial f}{partial y}=2y$$
$$frac{partial g}{partial x}=2x$$
$$int Fcdot dr=intint_R(2x-2y)dA$$
$$int^{1}_{0}int_{0}^{1-x}(2x-2y)dydx=0$$
My quesiton: Is my above attempt correct? Because I got the answer as $0$
calculus integration multivariable-calculus definite-integrals greens-theorem
edited Dec 1 '18 at 21:27
Masacroso
13k41746
asked Dec 1 '18 at 21:19
user982787user982787
1117
$endgroup$
$F(x,y)=langle y^2+e^x,x^2+e^yrangle$. Using green's theorem in its circulation and flux forms, determine the flux and circulation of $F$ around the triangle $T$, where $T$ is the triangle with vertices $(0,0),(1,0),$ and $(0,1)$, oriented counterclockwise.
My Try:
$f=y^2+e^x$
$g=x^2+e^y$
$$frac{partial f}{partial y}=2y$$
$$frac{partial g}{partial x}=2x$$
$$int Fcdot dr=intint_R(2x-2y)dA$$
$$int^{1}_{0}int_{0}^{1-x}(2x-2y)dydx=0$$
My quesiton: Is my above attempt correct? Because I got the answer as $0$
calculus integration multivariable-calculus definite-integrals greens-theorem
calculus integration multivariable-calculus definite-integrals greens-theorem
edited Dec 1 '18 at 21:27
Masacroso
13k41746
asked Dec 1 '18 at 21:19
user982787user982787
1117
edited Dec 1 '18 at 21:27
Masacroso
13k41746
asked Dec 1 '18 at 21:19
user982787user982787
1117
edited Dec 1 '18 at 21:27
Masacroso
13k41746
edited Dec 1 '18 at 21:27
Masacroso
13k41746
edited Dec 1 '18 at 21:27
Masacroso
13k41746
13k41746
asked Dec 1 '18 at 21:19
user982787user982787
1117
asked Dec 1 '18 at 21:19
user982787user982787
1117
asked Dec 1 '18 at 21:19
user982787user982787
1117
1117
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The result is correct indeed we have that the region is symmetric with respect to $y=x$ and the integral for $2x$ is equal to the integral for $2y$ that is
$$intint_R 2x,dA=intint_R2y,dA implies intint_R(2x-2y)dA=0$$
answered Dec 1 '18 at 21:24
gimusigimusi
1
$endgroup$
$begingroup$
How did you get $int_0^x$? Can you please elaborate the answer.
$endgroup$
– user982787
Dec 1 '18 at 21:26
$begingroup$
@user982787 Sorry I had assumed the third vertex in $(1,1)$. Your result seems correct.
$endgroup$
– gimusi
Dec 1 '18 at 21:30
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The result is correct indeed we have that the region is symmetric with respect to $y=x$ and the integral for $2x$ is equal to the integral for $2y$ that is
$$intint_R 2x,dA=intint_R2y,dA implies intint_R(2x-2y)dA=0$$
answered Dec 1 '18 at 21:24
gimusigimusi
1
$endgroup$
$begingroup$
How did you get $int_0^x$? Can you please elaborate the answer.
$endgroup$
– user982787
Dec 1 '18 at 21:26
$begingroup$
@user982787 Sorry I had assumed the third vertex in $(1,1)$. Your result seems correct.
$endgroup$
– gimusi
Dec 1 '18 at 21:30
add a comment |
$begingroup$
The result is correct indeed we have that the region is symmetric with respect to $y=x$ and the integral for $2x$ is equal to the integral for $2y$ that is
$$intint_R 2x,dA=intint_R2y,dA implies intint_R(2x-2y)dA=0$$
answered Dec 1 '18 at 21:24
gimusigimusi
1
$endgroup$
$begingroup$
How did you get $int_0^x$? Can you please elaborate the answer.
$endgroup$
– user982787
Dec 1 '18 at 21:26
$begingroup$
@user982787 Sorry I had assumed the third vertex in $(1,1)$. Your result seems correct.
$endgroup$
– gimusi
Dec 1 '18 at 21:30
add a comment |
$begingroup$
The result is correct indeed we have that the region is symmetric with respect to $y=x$ and the integral for $2x$ is equal to the integral for $2y$ that is
$$intint_R 2x,dA=intint_R2y,dA implies intint_R(2x-2y)dA=0$$
answered Dec 1 '18 at 21:24
gimusigimusi
1
$endgroup$
The result is correct indeed we have that the region is symmetric with respect to $y=x$ and the integral for $2x$ is equal to the integral for $2y$ that is
$$intint_R 2x,dA=intint_R2y,dA implies intint_R(2x-2y)dA=0$$
answered Dec 1 '18 at 21:24
gimusigimusi
1
edited Dec 1 '18 at 21:29
answered Dec 1 '18 at 21:24
gimusigimusi
1
answered Dec 1 '18 at 21:24
gimusigimusi
1
answered Dec 1 '18 at 21:24
gimusigimusi
1
1
$begingroup$
How did you get $int_0^x$? Can you please elaborate the answer.
$endgroup$
– user982787
Dec 1 '18 at 21:26
$begingroup$
@user982787 Sorry I had assumed the third vertex in $(1,1)$. Your result seems correct.
$endgroup$
– gimusi
Dec 1 '18 at 21:30
add a comment |
$begingroup$
How did you get $int_0^x$? Can you please elaborate the answer.
$endgroup$
– user982787
Dec 1 '18 at 21:26
$begingroup$
@user982787 Sorry I had assumed the third vertex in $(1,1)$. Your result seems correct.
$endgroup$
– gimusi
Dec 1 '18 at 21:30
$begingroup$
How did you get $int_0^x$? Can you please elaborate the answer.
$endgroup$
– user982787
Dec 1 '18 at 21:26
$begingroup$
How did you get $int_0^x$? Can you please elaborate the answer.
$endgroup$
– user982787
Dec 1 '18 at 21:26
$begingroup$
@user982787 Sorry I had assumed the third vertex in $(1,1)$. Your result seems correct.
$endgroup$
– gimusi
Dec 1 '18 at 21:30
$begingroup$
@user982787 Sorry I had assumed the third vertex in $(1,1)$. Your result seems correct.
$endgroup$
– gimusi
Dec 1 '18 at 21:30
add a comment |
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