Mixing
How to solve the surface integral...
$begingroup$
Consider the unit sphere $S={(x,y,z)in {R^3}:x^2+y^2+z^2=1}$ and the unit normal vector $bar{n}$ at each point $(x,y,z)$ on $S$. Then the value of the surface integral $$intint_{S}{(frac{2x}{pi}+sin(y^2))x+(e^z-frac{y}{pi})y+(frac{2z}{pi}+sin^2y)z}dsigma$$ is____?
Ok so this is what I tried to do:
Since this is a surface integral over a sphere, therefore I thought to convert it into a triple integral by using Gauss's Theorem, viz,
$$intint_{S}(f ~dy ~dz+g~dz~dx+h~dx~dy)=intintint_{E}(f_x+g_y+h_z)~dx~dy~dz$$
So $f=(e^z-frac{y}{pi})y+(frac{2z}{pi}+sin^2y)z$, $~g=0$ (since there is no term involving $~dz~dx$), and $h=(frac{2x}{pi}+sin(y^2))x$. At this point I am stuck. This way doesnot provide any sensible conclusion.
Can anyone provide some other way? How to solve this problem? Thanks.
calculus surface-integrals
asked Jan 19 at 3:16
Kushal BhuyanKushal Bhuyan
4,99121245
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add a comment |
$begingroup$
Consider the unit sphere $S={(x,y,z)in {R^3}:x^2+y^2+z^2=1}$ and the unit normal vector $bar{n}$ at each point $(x,y,z)$ on $S$. Then the value of the surface integral $$intint_{S}{(frac{2x}{pi}+sin(y^2))x+(e^z-frac{y}{pi})y+(frac{2z}{pi}+sin^2y)z}dsigma$$ is____?
Ok so this is what I tried to do:
Since this is a surface integral over a sphere, therefore I thought to convert it into a triple integral by using Gauss's Theorem, viz,
$$intint_{S}(f ~dy ~dz+g~dz~dx+h~dx~dy)=intintint_{E}(f_x+g_y+h_z)~dx~dy~dz$$
So $f=(e^z-frac{y}{pi})y+(frac{2z}{pi}+sin^2y)z$, $~g=0$ (since there is no term involving $~dz~dx$), and $h=(frac{2x}{pi}+sin(y^2))x$. At this point I am stuck. This way doesnot provide any sensible conclusion.
Can anyone provide some other way? How to solve this problem? Thanks.
calculus surface-integrals
asked Jan 19 at 3:16
Kushal BhuyanKushal Bhuyan
4,99121245
$endgroup$
$begingroup$
You mean $x^2+y^2+z^2=1$ right?
$endgroup$
– Jacky Chong
Jan 19 at 3:19
$begingroup$
Change to spherical coordinates.
$endgroup$
– lightxbulb
Jan 19 at 3:19
$begingroup$
@JackyChong yes. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:20
$begingroup$
@lightxbulb the integrand inside the double integral?
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:21
$begingroup$
All variables. And the limits. And the Jacobian.
$endgroup$
– lightxbulb
Jan 19 at 3:23
add a comment |
$begingroup$
Consider the unit sphere $S={(x,y,z)in {R^3}:x^2+y^2+z^2=1}$ and the unit normal vector $bar{n}$ at each point $(x,y,z)$ on $S$. Then the value of the surface integral $$intint_{S}{(frac{2x}{pi}+sin(y^2))x+(e^z-frac{y}{pi})y+(frac{2z}{pi}+sin^2y)z}dsigma$$ is____?
Ok so this is what I tried to do:
Since this is a surface integral over a sphere, therefore I thought to convert it into a triple integral by using Gauss's Theorem, viz,
$$intint_{S}(f ~dy ~dz+g~dz~dx+h~dx~dy)=intintint_{E}(f_x+g_y+h_z)~dx~dy~dz$$
So $f=(e^z-frac{y}{pi})y+(frac{2z}{pi}+sin^2y)z$, $~g=0$ (since there is no term involving $~dz~dx$), and $h=(frac{2x}{pi}+sin(y^2))x$. At this point I am stuck. This way doesnot provide any sensible conclusion.
Can anyone provide some other way? How to solve this problem? Thanks.
calculus surface-integrals
asked Jan 19 at 3:16
Kushal BhuyanKushal Bhuyan
4,99121245
$endgroup$
Consider the unit sphere $S={(x,y,z)in {R^3}:x^2+y^2+z^2=1}$ and the unit normal vector $bar{n}$ at each point $(x,y,z)$ on $S$. Then the value of the surface integral $$intint_{S}{(frac{2x}{pi}+sin(y^2))x+(e^z-frac{y}{pi})y+(frac{2z}{pi}+sin^2y)z}dsigma$$ is____?
Ok so this is what I tried to do:
Since this is a surface integral over a sphere, therefore I thought to convert it into a triple integral by using Gauss's Theorem, viz,
$$intint_{S}(f ~dy ~dz+g~dz~dx+h~dx~dy)=intintint_{E}(f_x+g_y+h_z)~dx~dy~dz$$
So $f=(e^z-frac{y}{pi})y+(frac{2z}{pi}+sin^2y)z$, $~g=0$ (since there is no term involving $~dz~dx$), and $h=(frac{2x}{pi}+sin(y^2))x$. At this point I am stuck. This way doesnot provide any sensible conclusion.
Can anyone provide some other way? How to solve this problem? Thanks.
calculus surface-integrals
calculus surface-integrals
asked Jan 19 at 3:16
Kushal BhuyanKushal Bhuyan
4,99121245
asked Jan 19 at 3:16
Kushal BhuyanKushal Bhuyan
4,99121245
edited Jan 19 at 3:19
Kushal Bhuyan
asked Jan 19 at 3:16
Kushal BhuyanKushal Bhuyan
4,99121245
asked Jan 19 at 3:16
Kushal BhuyanKushal Bhuyan
4,99121245
asked Jan 19 at 3:16
Kushal BhuyanKushal Bhuyan
4,99121245
4,99121245
$begingroup$
You mean $x^2+y^2+z^2=1$ right?
$endgroup$
– Jacky Chong
Jan 19 at 3:19
$begingroup$
Change to spherical coordinates.
$endgroup$
– lightxbulb
Jan 19 at 3:19
$begingroup$
@JackyChong yes. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:20
$begingroup$
@lightxbulb the integrand inside the double integral?
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:21
$begingroup$
All variables. And the limits. And the Jacobian.
$endgroup$
– lightxbulb
Jan 19 at 3:23
add a comment |
$begingroup$
You mean $x^2+y^2+z^2=1$ right?
$endgroup$
– Jacky Chong
Jan 19 at 3:19
$begingroup$
Change to spherical coordinates.
$endgroup$
– lightxbulb
Jan 19 at 3:19
$begingroup$
@JackyChong yes. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:20
$begingroup$
@lightxbulb the integrand inside the double integral?
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:21
$begingroup$
All variables. And the limits. And the Jacobian.
$endgroup$
– lightxbulb
Jan 19 at 3:23
$begingroup$
You mean $x^2+y^2+z^2=1$ right?
$endgroup$
– Jacky Chong
Jan 19 at 3:19
$begingroup$
You mean $x^2+y^2+z^2=1$ right?
$endgroup$
– Jacky Chong
Jan 19 at 3:19
$begingroup$
Change to spherical coordinates.
$endgroup$
– lightxbulb
Jan 19 at 3:19
$begingroup$
Change to spherical coordinates.
$endgroup$
– lightxbulb
Jan 19 at 3:19
$begingroup$
@JackyChong yes. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:20
$begingroup$
@JackyChong yes. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:20
$begingroup$
@lightxbulb the integrand inside the double integral?
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:21
$begingroup$
@lightxbulb the integrand inside the double integral?
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:21
$begingroup$
All variables. And the limits. And the Jacobian.
$endgroup$
– lightxbulb
Jan 19 at 3:23
$begingroup$
All variables. And the limits. And the Jacobian.
$endgroup$
– lightxbulb
Jan 19 at 3:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recall
begin{align}
iint_Sigma mathbf{F}cdotmathbf{n} dS = iiint_D operatorname{div}mathbf{F} dV
end{align}
Here $mathbf{F} = (frac{2x}{pi}+sin(y^2))mathbf{i}+(e^z-frac{y}{pi})mathbf{j}+(frac{2z}{pi}+sin^2(y))mathbf{k}$ where $operatorname{div}mathbf{F} = frac{3}{pi}$.
answered Jan 19 at 3:26
Jacky ChongJacky Chong
19k21129
$endgroup$
$begingroup$
Ok so the volume of the sphere is $frac{4}{3}pi$, which will give the value of the triple integral to be 4. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:30
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
$begingroup$
Recall
begin{align}
iint_Sigma mathbf{F}cdotmathbf{n} dS = iiint_D operatorname{div}mathbf{F} dV
end{align}
Here $mathbf{F} = (frac{2x}{pi}+sin(y^2))mathbf{i}+(e^z-frac{y}{pi})mathbf{j}+(frac{2z}{pi}+sin^2(y))mathbf{k}$ where $operatorname{div}mathbf{F} = frac{3}{pi}$.
answered Jan 19 at 3:26
Jacky ChongJacky Chong
19k21129
$endgroup$
$begingroup$
Ok so the volume of the sphere is $frac{4}{3}pi$, which will give the value of the triple integral to be 4. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:30
add a comment |
$begingroup$
Recall
begin{align}
iint_Sigma mathbf{F}cdotmathbf{n} dS = iiint_D operatorname{div}mathbf{F} dV
end{align}
Here $mathbf{F} = (frac{2x}{pi}+sin(y^2))mathbf{i}+(e^z-frac{y}{pi})mathbf{j}+(frac{2z}{pi}+sin^2(y))mathbf{k}$ where $operatorname{div}mathbf{F} = frac{3}{pi}$.
answered Jan 19 at 3:26
Jacky ChongJacky Chong
19k21129
$endgroup$
$begingroup$
Ok so the volume of the sphere is $frac{4}{3}pi$, which will give the value of the triple integral to be 4. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:30
add a comment |
$begingroup$
Recall
begin{align}
iint_Sigma mathbf{F}cdotmathbf{n} dS = iiint_D operatorname{div}mathbf{F} dV
end{align}
Here $mathbf{F} = (frac{2x}{pi}+sin(y^2))mathbf{i}+(e^z-frac{y}{pi})mathbf{j}+(frac{2z}{pi}+sin^2(y))mathbf{k}$ where $operatorname{div}mathbf{F} = frac{3}{pi}$.
answered Jan 19 at 3:26
Jacky ChongJacky Chong
19k21129
$endgroup$
Recall
begin{align}
iint_Sigma mathbf{F}cdotmathbf{n} dS = iiint_D operatorname{div}mathbf{F} dV
end{align}
Here $mathbf{F} = (frac{2x}{pi}+sin(y^2))mathbf{i}+(e^z-frac{y}{pi})mathbf{j}+(frac{2z}{pi}+sin^2(y))mathbf{k}$ where $operatorname{div}mathbf{F} = frac{3}{pi}$.
answered Jan 19 at 3:26
Jacky ChongJacky Chong
19k21129
answered Jan 19 at 3:26
Jacky ChongJacky Chong
19k21129
answered Jan 19 at 3:26
Jacky ChongJacky Chong
19k21129
answered Jan 19 at 3:26
Jacky ChongJacky Chong
19k21129
19k21129
$begingroup$
Ok so the volume of the sphere is $frac{4}{3}pi$, which will give the value of the triple integral to be 4. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:30
add a comment |
$begingroup$
Ok so the volume of the sphere is $frac{4}{3}pi$, which will give the value of the triple integral to be 4. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:30
$begingroup$
Ok so the volume of the sphere is $frac{4}{3}pi$, which will give the value of the triple integral to be 4. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:30
$begingroup$
Ok so the volume of the sphere is $frac{4}{3}pi$, which will give the value of the triple integral to be 4. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:30
add a comment |
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$begingroup$
You mean $x^2+y^2+z^2=1$ right?
$endgroup$
– Jacky Chong
Jan 19 at 3:19
$begingroup$
Change to spherical coordinates.
$endgroup$
– lightxbulb
Jan 19 at 3:19
$begingroup$
@JackyChong yes. Thanks.
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:20
$begingroup$
@lightxbulb the integrand inside the double integral?
$endgroup$
– Kushal Bhuyan
Jan 19 at 3:21
$begingroup$
All variables. And the limits. And the Jacobian.
$endgroup$
– lightxbulb
Jan 19 at 3:23