Flux integral

**Killer** · February 4th 2011, 04:22 PM

Compute the flux of the vector field $F(x,y,z)=(x,y,z)$ through the portion of the parabolic cylinder $z=x^2$ bounded by the planes $z=a^2$ with $a>0,\,y=0,\,y=b>0$ oriented so that component $z$ of the normal is negative.

I need to solve this by not using first Divergence Theorem, and then by using it.

I think by using it the triple integral (Divergenge Theorem) could be $\displaystyle\int_{-a}^{a}{\int_{0}^{{{a}^{2}}}{\int_{0}^{b}{3\,dy}\,d z}\,dx}.$ Is it the correct set up?

How to solve it by not using Gauss?

**TheEmptySet** · February 4th 2011, 05:24 PM

Originally Posted by Killer

Compute the flux of the vector field $F(x,y,z)=(x,y,z)$ through the portion of the parabolic cylinder $z=x^2$ bounded by the planes $z=a^2$ with $a>0,\,y=0,\,y=b>0$ oriented so that component $z$ of the normal is negative.

I need to solve this by not using first Divergence Theorem, and then by using it.

I think by using it the triple integral (Divergenge Theorem) could be $\displaystyle\int_{-a}^{a}{\int_{0}^{{{a}^{2}}}{\int_{0}^{b}{3\,dy}\,d z}\,dx}.$ Is it the correct set up?

How to solve it by not using Gauss?

$\displaystyle \iint_S \vec{F}(x,y,z) \cdot d\vec{S}=\int_{0}^{b}\int_{-a}^{a}\left( x\vec{i}+y\vec{j}+x^2\vec{k}\right)\cdot \left( 2x\vec{i}-\vec{k}\right)dxdy =\left( \int_{-a}^{a}x^2dx\right)\left( \int_{0}^{b}dy\right)=\frac{2a^3b}{3}$

To use the divergence theorem you need to close the surface with the plane that is gives. The flux though the plane is easy to calculate(you don't need an integral for a constant vector field though a flat surface). So find the flux with the divergence theorem then subtract off the plane to get the flux of the parabolic part.

P.S you limits of integration are incorrect on the triple integral z goes from the plane to the parabolic cylinder.

**Killer** · February 4th 2011, 06:58 PM

Originally Posted by TheEmptySet

$\displaystyle \iint_S \vec{F}(x,y,z) \cdot d\vec{S}=\int_{0}^{b}\int_{-a}^{a}\left( x\vec{i}+y\vec{j}+x^2\vec{k}\right)\cdot \left( 2x\vec{i}-\vec{k}\right)dxdy$

Could you explain how you went from the left side to the right side?

It's just that I don't have either remember much formulae since I don't have my calculus notebook.

As for using the Divergence Theorem, can you please show me how to do it?

**TheEmptySet** · February 4th 2011, 07:21 PM

Originally Posted by Killer

Could you explain how you went from the left side to the right side?

It's just that I don't have either remember much formulae since I don't have my calculus notebook.

As for using the Divergence Theorem, can you please show me how to do it?

Flux is defined to be

$\iint_S \vec{F}\cdot d\vec{S}=\iint_{S}\vec{F}\cdot \vec{n}dS$

So a normal to the surface is given by the gradiant

$z=x^2 \iff x^2 -z=0 \implies G(x,y,z)=x^2-z$

$\nabla G=2x\vec{i}-\vec{k}$ we choose this normal because the z component is negative.

$||\nabla G||=\sqrt{4x^2+1}$ so the unit normal vector is

$\displaystyle \vec{n}=\frac{\nabla G}{||\nabla G||}$

The surface area element is given by

$\displaystyle dS=\sqrt{1+\left( \frac{\partial z}{\partial x}\right)^2+\left( \frac{\partial z}{\partial y}\right)^2}=\sqrt{4x^2+1}dA$

Putting all of this together gives

$\displaystyle \iint_S \vec{F}\cdot d\vec{S}=\iint_{S}\vec{F}\cdot \vec{n}dS \iint_D F(x,y,x^2) \cdot\left (-\frac{\partial z}{\partial x}\vec{i}-\frac{\partial z}{\partial y}\vec{j}+\vec{k} \right)dA$

Please write out where you are stuck with the divergence theorem.

**Killer** · March 12th 2011, 03:13 PM

Originally Posted by TheEmptySet

Flux is defined to be

$\iint_S \vec{F}\cdot d\vec{S}=\iint_{S}\vec{F}\cdot \vec{n}dS$

So a normal to the surface is given by the gradiant

$z=x^2 \iff x^2 -z=0 \implies G(x,y,z)=x^2-z$

$\nabla G=2x\vec{i}-\vec{k}$ we choose this normal because the z component is negative.

$||\nabla G||=\sqrt{4x^2+1}$ so the unit normal vector is

$\displaystyle \vec{n}=\frac{\nabla G}{||\nabla G||}$

The surface area element is given by

$\displaystyle dS=\sqrt{1+\left( \frac{\partial z}{\partial x}\right)^2+\left( \frac{\partial z}{\partial y}\right)^2}=\sqrt{4x^2+1}dA$

Putting all of this together gives

$\displaystyle \iint_S \vec{F}\cdot d\vec{S}=\iint_{S}\vec{F}\cdot \vec{n}dS \iint_D F(x,y,x^2) \cdot\left (-\frac{\partial z}{\partial x}\vec{i}-\frac{\partial z}{\partial y}\vec{j}+\vec{k} \right)dA$

Okay, perfect, I'm almost there! The only thing I didn't get is when you have

$\displaystyle\iint_{D}{F\left( x,y,{{x}^{2}} \right)}\cdot \left( -\frac{\partial z}{\partial x}\overrightarrow{i}-\frac{\partial z}{\partial y}\overrightarrow{j}+\overrightarrow{k} \right)dA,$ why is it $F(x,y,x^2)$ and not $F(x,y,z)$ ?

Now, does $\displaystyle-\frac{\partial z}{\partial x}\overrightarrow{i}-\frac{\partial z}{\partial y}\overrightarrow{j}+\overrightarrow{k}=2x\overrig htarrow{i}-\overrightarrow{k}$ ? If so, how? I don't get the signs. Also, where does $\displaystyle-\frac{\partial z}{\partial x}\overrightarrow{i}-\frac{\partial z}{\partial y}\overrightarrow{j}+\overrightarrow{k}$ come from?

Originally Posted by TheEmptySet

Please write out where you are stuck with the divergence theorem.

Okay, I think this is the correct set up $\displaystyle\int_{-a}^{a}{\int_{{{x}^{2}}}^{{{a}^{2}}}{\int_{0}^{b}{3 \,dy}\,dz}\,dx}.$

**TheEmptySet** · March 13th 2011, 10:27 AM

Originally Posted by Killer

Okay, perfect, I'm almost there! The only thing I didn't get is when you have

$\displaystyle\iint_{D}{F\left( x,y,{{x}^{2}} \right)}\cdot \left( -\frac{\partial z}{\partial x}\overrightarrow{i}-\frac{\partial z}{\partial y}\overrightarrow{j}+\overrightarrow{k} \right)dA,$ why is it $F(x,y,x^2)$ and not $F(x,y,z)$ ?

In the surface $z=x^2$ so they are the same!

Now, does $\displaystyle-\frac{\partial z}{\partial x}\overrightarrow{i}-\frac{\partial z}{\partial y}\overrightarrow{j}+\overrightarrow{k}=2x\overrig htarrow{i}-\overrightarrow{k}$ ? If so, how? I don't get the signs.

As above the surface is We are finding a normal vector to the surface

$z=x^2 \iff x^2-z=0$
Notice that the z coefficient is negative because you needed the downward orientation. Now if you take its gradient you get

$\displaystyle \frac{\partial }{\partial x}\left( x^2-z\right)\vec{i} +\frac{\partial }{\partial y}\left( x^2-z\right)\vec{j}+\frac{\partial }{\partial z}\left( x^2-z\right)\vec{k}$

Also, where does $\displaystyle-\frac{\partial z}{\partial x}\overrightarrow{i}-\frac{\partial z}{\partial y}\overrightarrow{j}+\overrightarrow{k}$ come from?

This can also be found by using the trivial parametrization.

$\vec{r}(x,y)=x\vec{i}+y\vec{j}+z\vec{k}=x\vec{i}+y \vec{j}+x^2\vec{k}$

Now just find $|\vec{r}_x \times \vec{r}_y|dxdy$
The magnitude of the cross product

**Killer** · March 13th 2011, 11:07 AM

Okay, I got it, but divergence theorem gives a different result when calculating the double integral, how do you achieve the same value?

**Killer** · March 14th 2011, 07:34 AM

Okay, I was solving four flux integrals, for $z=x^2,\,z=a^2,\,y=0,\,y=b.$

For $z=x^2$ is done, for $z=a^2$ I should take my $G(x,y,z)$ as $a^2-z$ so $\nabla G=-\vec k,$ so I'd have to calculate $\displaystyle\iint_S F(x,y,a^2)\cdot(0,0,-1) dx\,dy,$ but my professor said is wrong since I'd have to take $(0,0,1)$ to make it, why is this?

Hope you can help me, I have my exam tomorrow!

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