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Flux through a side of a cylinder

1

$begingroup$

My troubles come with calculating the flux perpendicular to the cylinder's axis (ie, radial direction; $S_3$) through the surface. What I'd do is:
$$iint_R v cdot n fracdxdz = int_0^3 int_0^2 (frac4x^2y - 2y^2) dxdz$$

But it doesn't yield $48pi$.

The book provides another method which indeed yields the expected solution:

Why am I wrong?

I don't really understand the book's method; so if you want to provide an explanation on that as well I'd be grateful for it.

Thanks

multivariable-calculus surface-integrals

share|cite|improve this question

asked Mar 30 at 17:22

JD_PMJD_PM

18711

$endgroup$

add a comment | 

1

$begingroup$

My troubles come with calculating the flux perpendicular to the cylinder's axis (ie, radial direction; $S_3$) through the surface. What I'd do is:
$$iint_R v cdot n fracdxdz = int_0^3 int_0^2 (frac4x^2y - 2y^2) dxdz$$

But it doesn't yield $48pi$.

The book provides another method which indeed yields the expected solution:

Why am I wrong?

I don't really understand the book's method; so if you want to provide an explanation on that as well I'd be grateful for it.

Thanks

multivariable-calculus surface-integrals

share|cite|improve this question

asked Mar 30 at 17:22

JD_PMJD_PM

18711

$endgroup$

add a comment | 

$begingroup$

My troubles come with calculating the flux perpendicular to the cylinder's axis (ie, radial direction; $S_3$) through the surface. What I'd do is:
$$iint_R v cdot n fracdxdz = int_0^3 int_0^2 (frac4x^2y - 2y^2) dxdz$$

But it doesn't yield $48pi$.

The book provides another method which indeed yields the expected solution:

Why am I wrong?

I don't really understand the book's method; so if you want to provide an explanation on that as well I'd be grateful for it.

Thanks

multivariable-calculus surface-integrals

share|cite|improve this question

asked Mar 30 at 17:22

JD_PMJD_PM

18711

$endgroup$

share|cite|improve this question

asked Mar 30 at 17:22

JD_PMJD_PM

18711

share|cite|improve this question

asked Mar 30 at 17:22

JD_PMJD_PM

18711

asked Mar 30 at 17:22

JD_PMJD_PM

18711

asked Mar 30 at 17:22

JD_PMJD_PM

18711

1 Answer
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1

$begingroup$

You posed well the integral, but some things have to be fixed: the range for $x$ is $-2leq xleq 2$; the integral has to be done for $y=sqrt4-x^2$, one half of the cylinder, and for $y=-sqrt4-x^2$, the other half and, further, we are dealing with the absolute value of $y$ in $|n cdot j|$, so we have to be careful with the signs in some expressions: $y^3/|y|=y^2$ if $ygeq0$ but $y^3/|y|=-y^2$ if $ylt0$

$$iint_R v cdot n fracdxdz = int_0^3 int_-2^2 left(frac4x^2y - 2y^2right) dxdz+int_0^3 int_-2^2 left(frac4x^2-y + 2y^2right) dxdz=$$

$$= int_0^3 int_-2^2 left(frac4x^2sqrt4-x^2 - 2(4-x^2)right) dxdz+int_0^3 int_-2^2 left(frac4x^2sqrt4-x^2 + 2(4-x^2)right) dxdz=$$

$$=2int_0^3dz int_-2^2 left(frac4x^2sqrt4-x^2right) dx=48pi$$

The solution you cited uses cylindrical coordinates, far more easier as they adapt to the symmtry the problem has.

share|cite|improve this answer

answered Mar 30 at 21:11

Rafa BudríaRafa Budría

5,9721825

$endgroup$

add a comment | 

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1

$begingroup$

You posed well the integral, but some things have to be fixed: the range for $x$ is $-2leq xleq 2$; the integral has to be done for $y=sqrt4-x^2$, one half of the cylinder, and for $y=-sqrt4-x^2$, the other half and, further, we are dealing with the absolute value of $y$ in $|n cdot j|$, so we have to be careful with the signs in some expressions: $y^3/|y|=y^2$ if $ygeq0$ but $y^3/|y|=-y^2$ if $ylt0$

$$iint_R v cdot n fracdxdz = int_0^3 int_-2^2 left(frac4x^2y - 2y^2right) dxdz+int_0^3 int_-2^2 left(frac4x^2-y + 2y^2right) dxdz=$$

$$= int_0^3 int_-2^2 left(frac4x^2sqrt4-x^2 - 2(4-x^2)right) dxdz+int_0^3 int_-2^2 left(frac4x^2sqrt4-x^2 + 2(4-x^2)right) dxdz=$$

$$=2int_0^3dz int_-2^2 left(frac4x^2sqrt4-x^2right) dx=48pi$$

The solution you cited uses cylindrical coordinates, far more easier as they adapt to the symmtry the problem has.

share|cite|improve this answer

answered Mar 30 at 21:11

Rafa BudríaRafa Budría

5,9721825

$endgroup$

add a comment | 

1

$begingroup$

You posed well the integral, but some things have to be fixed: the range for $x$ is $-2leq xleq 2$; the integral has to be done for $y=sqrt4-x^2$, one half of the cylinder, and for $y=-sqrt4-x^2$, the other half and, further, we are dealing with the absolute value of $y$ in $|n cdot j|$, so we have to be careful with the signs in some expressions: $y^3/|y|=y^2$ if $ygeq0$ but $y^3/|y|=-y^2$ if $ylt0$

$$iint_R v cdot n fracdxdz = int_0^3 int_-2^2 left(frac4x^2y - 2y^2right) dxdz+int_0^3 int_-2^2 left(frac4x^2-y + 2y^2right) dxdz=$$

$$= int_0^3 int_-2^2 left(frac4x^2sqrt4-x^2 - 2(4-x^2)right) dxdz+int_0^3 int_-2^2 left(frac4x^2sqrt4-x^2 + 2(4-x^2)right) dxdz=$$

$$=2int_0^3dz int_-2^2 left(frac4x^2sqrt4-x^2right) dx=48pi$$

The solution you cited uses cylindrical coordinates, far more easier as they adapt to the symmtry the problem has.

share|cite|improve this answer

answered Mar 30 at 21:11

Rafa BudríaRafa Budría

5,9721825

$endgroup$

add a comment | 

$begingroup$

You posed well the integral, but some things have to be fixed: the range for $x$ is $-2leq xleq 2$; the integral has to be done for $y=sqrt4-x^2$, one half of the cylinder, and for $y=-sqrt4-x^2$, the other half and, further, we are dealing with the absolute value of $y$ in $|n cdot j|$, so we have to be careful with the signs in some expressions: $y^3/|y|=y^2$ if $ygeq0$ but $y^3/|y|=-y^2$ if $ylt0$

$$iint_R v cdot n fracdxdz = int_0^3 int_-2^2 left(frac4x^2y - 2y^2right) dxdz+int_0^3 int_-2^2 left(frac4x^2-y + 2y^2right) dxdz=$$

$$= int_0^3 int_-2^2 left(frac4x^2sqrt4-x^2 - 2(4-x^2)right) dxdz+int_0^3 int_-2^2 left(frac4x^2sqrt4-x^2 + 2(4-x^2)right) dxdz=$$

$$=2int_0^3dz int_-2^2 left(frac4x^2sqrt4-x^2right) dx=48pi$$

The solution you cited uses cylindrical coordinates, far more easier as they adapt to the symmtry the problem has.

share|cite|improve this answer

answered Mar 30 at 21:11

Rafa BudríaRafa Budría

5,9721825

$endgroup$

share|cite|improve this answer

answered Mar 30 at 21:11

Rafa BudríaRafa Budría

5,9721825

answered Mar 30 at 21:11

Rafa BudríaRafa Budría

5,9721825

answered Mar 30 at 21:11

Rafa BudríaRafa Budría

5,9721825