Surface Integral Check: \(\mathbf{F}\cdot\hat{\mathbf{n}} \int_0^1\int_0^1dxdy\)

[Dustinsfl]

I haven't done a surface integral in a while so I am asking to get this checked.

\(\mathbf{F} = \langle x, y, z\rangle\) and the surface is \(z = xy + 1\) where \(0\leq x\leq 1\) and \(0\leq y\leq 1\).

\(\hat{\mathbf{n}} = \nabla f/ \lvert\nabla f\rvert = \frac{1}{\sqrt{3}}\langle 1, 1, 1\rangle\)

\(dS = \frac{\lvert\nabla f\rvert dxdy}{\frac{\partial f}{\partial z}} = \sqrt{3}dxdy\)

\(\mathbf{F}\cdot\hat{\mathbf{n}} = \frac{1}{\sqrt{3}}(x+y+z) = \frac{1}{\sqrt{3}}(x+y+xy + 1)\)

\[
\int_0^1\int_0^1(x + y + xy + 1)dxdy = \frac{9}{4}
\]

So is this the correct integral I should I obtain or is there a mistake some where?

 

[Dustinsfl]

Based on this post: http://mathhelpboards.com/geometry-11/volume-triangle-type-shape-square-bottom-6305.html my solution must be incorrect since
\[
\int_S(\mathbf{F}\cdot\hat{\mathbf{n}})dS = \int(\nabla\cdot\mathbf{F})dV
\]
and \(\nabla\cdot\mathbf{F} = 3\)
so
\[
\int 3dV = 3V = 3\frac{5}{4} = \frac{15}{4}
\]
(Is this integral correct?)
So what went wrong with the surface integral calculation?

 

[Dustinsfl]

We can calculate \(\unit{n}\) by
\(\unit{n} = \frac{\mathbf{U}_x\times\mathbf{U}_y}
{\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert}\).
\begin{align*}
\mathbf{U}_x &= \unit{i} + y\unit{k}\\
\mathbf{U}_y &= \unit{j} + x\unit{k}\\
\mathbf{U}_x\times\mathbf{U}_y &= -y\unit{i} - x\unit{j} + \unit{k}\\
\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert &= \sqrt{1 + x^2 + y^2}
\end{align*}
So \(\unit{n} = \frac{-y\unit{i} - x\unit{j} + \unit{k}}{\sqrt{1 + x^2
+ y^2}}\).
\begin{align*}
\int_S\mathbf{F}\cdot\unit{n}dA &= \int_0^1\int_0^1
(x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot\unit{n}
\sqrt{1 + x^2 + y^2}dxdy\\
&= \int_0^1\int_0^1 (x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot
(-y\unit{i} - x\unit{j} + \unit{k})dxdy\\
&= \int_0^1\int_0^1 (-2xy + xy + 1)dxdy\\
&= \int_0^1\int_0^1 (1 - xy)dxdy\\
&= \int_0^1\left[1 - \frac{1}{2}y\right]dy\\
&= \frac{3}{4}
\end{align*}

So now I want to show that using the divergence theorem leads to the same answer. However, I have been un-successful.

Note \unit = \hat{\mathbf{#1}}.

 

[I like Serena]

I haven't done a surface integral in a while so I am asking to get this checked.

\(\mathbf{F} = \langle x, y, z\rangle\) and the surface is \(z = xy + 1\) where \(0\leq x\leq 1\) and \(0\leq y\leq 1\).

\(\hat{\mathbf{n}} = \nabla f/ \lvert\nabla f\rvert = \frac{1}{\sqrt{3}}\langle 1, 1, 1\rangle\)

Are F and f supposed to be the same function?

Anyway, suppose we define g(x,y,z)=z-xy-1, then your surface is given by g(x,y,z)=0.
Then:
$$\mathbf{\hat n} = \frac{\nabla g}{|\nabla g|} = \frac {(-y,-x,1)}{\sqrt{x^2+y^2+1}}$$
which is different from what you have.

Based on this post: http://mathhelpboards.com/geometry-11/volume-triangle-type-shape-square-bottom-6305.html my solution must be incorrect since
\[
\int_S(\mathbf{F}\cdot\hat{\mathbf{n}})dS = \int(\nabla\cdot\mathbf{F})dV
\]
and \(\nabla\cdot\mathbf{F} = 3\)
so
\[
\int 3dV = 3V = 3\frac{5}{4} = \frac{15}{4}
\]
(Is this integral correct?)
So what went wrong with the surface integral calculation?

Don't forget that you're supposed to integrate over the entire closed surface. In particular that includes the sides where x=1 respectively y=1.

 

[Dustinsfl]

Are F and f supposed to be the same function?

Anyway, suppose we define g(x,y,z)=z-xy-1, then your surface is given by g(x,y,z)=0.
Then:
$$\mathbf{\hat n} = \frac{\nabla g}{|\nabla g|} = \frac {(-y,-x,1)}{\sqrt{x^2+y^2+1}}$$
which is different from what you have.

You neglected to look at post 3.

 

[I like Serena]

You neglected to look at post 3.

Heh. Yes, I can see there is a $\sqrt{x^2+y^2+1}$ in there, but no mention that you discovered the mistake in the first post.
And to be honest, I stopped reading it due to all the distracting red \units in there.

Are you satisfied then with what you found out?

It appears you're still not integrating the sides where x=1 respectively y=1.

 

[Dustinsfl]

Heh. Yes, I can see there is a $\sqrt{x^2+y^2+1}$ in there, but no mention that you discovered the mistake in the first post.
And to be honest, I stopped reading it due to all the distracting red \units in there.

Are you satisfied then with what you found out?

It appears you're still not integrating the sides where x=1 respectively y=1.

\unit is my notation for \hat{\mathbf{#1}}. I refuse to constantly type that.

So are you saying that 3/4 in post 3 is incorrect? What needs to be changed about that integral?

 

[MarkFL]

\unit is my notation for \hat{\mathbf{#1}}. I refuse to constantly type that...

(Wondering)

It seems to me it would be just as easy, if not easier, to copy/paste that command after you type it once, as to type "\unit" each time.

Letting things be more difficult to read by others because you perceive it to be easier for yourself is not a path I would take personally. (Speechless) (Shake)

 

[Dustinsfl]

(Wondering)

It seems to me it would be just as easy, if not easier, to copy/paste that command after you type it once, as to type "\unit" each time.

Letting things be more difficult to read by others because you perceive it to be easier for yourself is not a path I would take personally. (Speechless) (Shake)

If you use LaTeX enough, it shouldn't even phase you.

 

[I like Serena]

\unit is my notation for \hat{\mathbf{#1}}. I refuse to constantly type that.

So are you saying that 3/4 in post 3 is incorrect? What needs to be changed about that integral?

Oh man, be a little bit creative!
In particular you only need to type it once per post.
Just put in "\newcommand{unit}[1]{\mathbf{\hat #1}} \unit n" once, like this: $\newcommand{unit}[1]{\mathbf{\hat #1}} \unit n$, and the rest of the post, even quoted, transforms.
If you want help, it helps if you make it attractive for people to help you.

We can calculate \(\unit{n}\) by
\(\unit{n} = \frac{\mathbf{U}_x\times\mathbf{U}_y}
{\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert}\).
\begin{align*}
\mathbf{U}_x &= \unit{i} + y\unit{k}\\
\mathbf{U}_y &= \unit{j} + x\unit{k}\\
\mathbf{U}_x\times\mathbf{U}_y &= -y\unit{i} - x\unit{j} + \unit{k}\\
\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert &= \sqrt{1 + x^2 + y^2}
\end{align*}
So \(\unit{n} = \frac{-y\unit{i} - x\unit{j} + \unit{k}}{\sqrt{1 + x^2
+ y^2}}\).
\begin{align*}
\int_S\mathbf{F}\cdot\unit{n}dA &= \int_0^1\int_0^1
(x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot\unit{n}
\sqrt{1 + x^2 + y^2}dxdy\\
&= \int_0^1\int_0^1 (x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot
(-y\unit{i} - x\unit{j} + \unit{k})dxdy\\
&= \int_0^1\int_0^1 (-2xy + xy + 1)dxdy\\
&= \int_0^1\int_0^1 (1 - xy)dxdy\\
&= \int_0^1\left[1 - \frac{1}{2}y\right]dy\\
&= \frac{3}{4}
\end{align*}

So now I want to show that using the divergence theorem leads to the same answer. However, I have been un-successful.

Note \unit = \hat{\mathbf{#1}}.

What appears to be missing is that you have a surface at x=1.
Its normal vector is (1,0,0). Together with the function F(1,y,z) = (1,y,z) this yields a dot product of 1, meaning you get a contribution that is as large as the surface at x=1.
The same holds for y=1.
You can ignore the surfaces at x=0, y=0, respectively z=0, since they have a dot product that is 0.

 

[Dustinsfl]

Oh man, be a little bit creative!
In particular you only need to type it once per post.
Just put in "\newcommand{unit}[1]{\mathbf{\hat #1}} \unit n" once, like this: $\newcommand{unit}[1]{\mathbf{\hat #1}} \unit n$, and the rest of the post, even quoted, transforms.

I wasn't aware mathjax would allow me to enter in newcommands and trust me LaTeX can get creative.

 

[I like Serena]

I wasn't aware mathjax would allow me to enter in newcommands and trust me LaTeX can get creative.

I have been aware of this feature since before mathjax.
It appears that vBulletin somehow keeps track of all $\LaTeX$ within 1 post.

 

[Ackbach]

$\newcommand{\unit}[1]{\mathbf{\hat{#1}}}$

Just re-typing dwsmith's post with the newcommand.

We can calculate \(\unit{n}\) by
\(\unit{n} = \frac{\mathbf{U}_x\times\mathbf{U}_y}
{\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert}\).
\begin{align*}
\mathbf{U}_x &= \unit{i} + y\unit{k}\\
\mathbf{U}_y &= \unit{j} + x\unit{k}\\
\mathbf{U}_x\times\mathbf{U}_y &= -y\unit{i} - x\unit{j} + \unit{k}\\
\lvert\mathbf{U}_x\times\mathbf{U}_y\rvert &= \sqrt{1 + x^2 + y^2}
\end{align*}
So \(\unit{n} = \frac{-y\unit{i} - x\unit{j} + \unit{k}}{\sqrt{1 + x^2
+ y^2}}\).
\begin{align*}
\int_S\mathbf{F}\cdot\unit{n}dA &= \int_0^1\int_0^1
(x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot\unit{n}
\sqrt{1 + x^2 + y^2}dxdy\\
&= \int_0^1\int_0^1 (x\unit{i} + y\unit{j} + (xy + 1)\unit{k})\cdot
(-y\unit{i} - x\unit{j} + \unit{k})dxdy\\
&= \int_0^1\int_0^1 (-2xy + xy + 1)dxdy\\
&= \int_0^1\int_0^1 (1 - xy)dxdy\\
&= \int_0^1\left[1 - \frac{1}{2}y\right]dy\\
&= \frac{3}{4}
\end{align*}

So now I want to show that using the divergence theorem leads to the same answer. However, I have been un-successful.