Multivariable Calculus, Help with pre-Bachelos Homework I need help with a couple of problems for homework, btw I need to finish this in less than 12 hours, I'll appreciate your help, and srry if something is wrong or is confusing I'm not native speaker.

The problems are the following:


Let $F:\ U= \mathbb{R}^3$\ {axis Z} $\subset \mathbb{R}^3$ $\rightarrow$ $\mathbb{R}^3$ defined as:


$F(x,y,z) = (\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0)$

Determine if $F$ it a solenoid field

2.Porve Green's Theorem using Stokes's Theorem

Plus: How do I calculate the mass of thin layers wich form correspond to a certain surface $\Sigma$ and a certain density function $\rho$
Let $F(x,y,z)=F(x,y)=f(x,y)\mathbf{i}+g(x,y)\mathbf{j}+0\mathbf{k}$

A solenoid field means the field is "circular", specifically its divergence is $0$

(1) Show that

$$\nabla \cdot F=0$$

Green's theorem states that given a vector field $F \in\mathbb{R}^2$

$$\oint_CF\cdot dr=\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

Stokes' theorem states that given a vector field $G\in\mathbb{R}^3$

$$\oint_CG\cdot dr=\iint_S\left(\nabla \times G\right)\cdot dS=\iint_D\left(\left(\nabla \times G\right)\cdot n\right) dA$$

(2) Show that

$$\nabla \times F = \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\mathbf{k}$$

And reason why, in this case,

$$n=\mathbf{k}$$

(Plus)

Given a density function, $\rho(x,y,z)$, the mass of a thin surface is given by a surface integral

$$m=\iint_S\rho(x,y,z)dS=\iint_D\rho(x,y,z)|n|dA$$

Where the surface has center of mass coordinates $(\bar{x},\bar{y},\bar{z})$

$$\bar{x}=\frac{1}{m}\iint_Sx\rho(x,y,z)dS$$

$$\bar{y}=\frac{1}{m}\iint_Sy\rho(x,y,z)dS$$

$$\bar{z}=\frac{1}{m}\iint_Sz\rho(x,y,z)dS$$

Notes: $n$ is the normal vector, $|n|$ is the normal vector's magnitude, and $S=\Sigma$

Good luck on the exam! :)



edit:

$F$ is a $3$-dim vector field but behaves like a field in $\mathbb{R}^2$ since its $\mathbf{k}$ component is $0$ which is why we can consider the $2$-dim vector field case of Green's theorem. 

To show it, take the curl of $F$ to begin Stokes' theorem

$$\nabla \times F=\begin{vmatrix} \mathbf{i} &amp; \mathbf{j} &amp; \mathbf{k}\\ \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \\ f(x,y) &amp; g(x,y) &amp; 0\end{vmatrix}=\mathbf{i}\left(\frac{\partial}{\partial y}0-\frac{\partial g}{\partial z}\right)-\mathbf{j}\left(\frac{\partial}{\partial x}0-\frac{\partial f}{\partial z}\right)+\mathbf{k}\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)$$

$$\oint_CF\cdot dr=\iint_S\nabla\times F\cdot dS=\iint_D\left(\nabla\times F\cdot n\right)dA$$

$\displaystyle\frac{\partial g}{\partial z}=\frac{\partial f}{\partial z}=0$ since both are independent of $z$ and the partial of $0$ is $0$

$$\iint_D\left(\nabla\times F\cdot n\right)dA=\iint_D\left&lt;0,0,\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\right&gt;\cdot\left&lt;a,b,c\right&gt;dA=c\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

But $n$ is a unit normal vector so if $C\in\mathbb{R}^2$ then $c = 1$ and we get Green's theorem

$$\oint_CF\cdot dr=\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

I hope that answered your question



In fact, we know the answer regardless of what $C$ is

$$\frac{\partial g}{\partial x}=\frac{(x^2+y^2)-2x^2}{\left(x^2+y^2\right)^2}=\frac{y^2-x^2}{\left(y^2+x^2\right)^2}$$

$$\frac{\partial f}{\partial y}=-\frac{(x^2+y^2)-2y^2}{\left(x^2+y^2\right)^2}=\frac{y^2-x^2}{\left(y^2+x^2\right)^2}$$

$$\frac{\partial g}{\partial x}=\frac{\partial f}{\partial y}$$

Since its z-component is $0$ and $f$ and $g$ are independent of $z$, it's unnecessary to check the remaining equations.

$\therefore F$ is a conservative field.

$$c\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA=c\iint_D(0)dA=0$$

$$\oint_C F\cdot dr=\oint_C \nabla f\cdot dr=f\left(\gamma(b)\right)-f\left(\gamma(a)\right)=0$$

Since $C$ is closed so $\gamma(b)=\gamma(a)$ for any closed path $C\in\mathbb{R}^3$
Let $F(x,y,z)=F(x,y)=f(x,y)\mathbf{i}+g(x,y)\mathbf{j}+0\mathbf{k}$

A solenoid field means the field is "circular", specifically its divergence is $0$

(1) Show that

$$\nabla \cdot F=0$$

Green's theorem states that given a vector field $F \in\mathbb{R}^2$

$$\oint_CF\cdot dr=\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

Stokes' theorem states that given a vector field $G\in\mathbb{R}^3$

$$\oint_CG\cdot dr=\iint_S\left(\nabla \times G\right)\cdot dS=\iint_D\left(\left(\nabla \times G\right)\cdot n\right) dA$$

(2) Show that

$$\nabla \times F = \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\mathbf{k}$$

And reason why, in this case,

$$n=\mathbf{k}$$

(Plus)

Given a density function, $\rho(x,y,z)$, the mass of a thin surface is given by a surface integral

$$m=\iint_S\rho(x,y,z)dS=\iint_D\rho(x,y,z)|n|dA$$

Where the surface has center of mass coordinates $(\bar{x},\bar{y},\bar{z})$

$$\bar{x}=\frac{1}{m}\iint_Sx\rho(x,y,z)dS$$

$$\bar{y}=\frac{1}{m}\iint_Sy\rho(x,y,z)dS$$

$$\bar{z}=\frac{1}{m}\iint_Sz\rho(x,y,z)dS$$

Notes: $n$ is the normal vector, $|n|$ is the normal vector's magnitude, and $S=\Sigma$

Good luck on the exam! :)



edit:

$F$ is a $3$-dim vector field but behaves like a field in $\mathbb{R}^2$ since its $\mathbf{k}$ component is $0$ which is why we can consider the $2$-dim vector field case of Green's theorem. 

To show it, take the curl of $F$ to begin Stokes' theorem

$$\nabla \times F=\begin{vmatrix} \mathbf{i} &amp; \mathbf{j} &amp; \mathbf{k}\\ \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} \\ f(x,y) &amp; g(x,y) &amp; 0\end{vmatrix}=\mathbf{i}\left(\frac{\partial}{\partial y}0-\frac{\partial g}{\partial z}\right)-\mathbf{j}\left(\frac{\partial}{\partial x}0-\frac{\partial f}{\partial z}\right)+\mathbf{k}\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)$$

$$\oint_CF\cdot dr=\iint_S\nabla\times F\cdot dS=\iint_D\left(\nabla\times F\cdot n\right)dA$$

$\displaystyle\frac{\partial g}{\partial z}=\frac{\partial f}{\partial z}=0$ since both are independent of $z$ and the partial of $0$ is $0$

$$\iint_D\left(\nabla\times F\cdot n\right)dA=\iint_D\left&lt;0,0,\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\right&gt;\cdot\left&lt;a,b,c\right&gt;dA=c\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

But $n$ is a unit normal vector so if $C\in\mathbb{R}^2$ then $c = 1$ and we get Green's theorem

$$\oint_CF\cdot dr=\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA$$

I hope that answered your question



In fact, we know the answer regardless of what $C$ is

$$\frac{\partial g}{\partial x}=\frac{(x^2+y^2)-2x^2}{\left(x^2+y^2\right)^2}=\frac{y^2-x^2}{\left(y^2+x^2\right)^2}$$

$$\frac{\partial f}{\partial y}=-\frac{(x^2+y^2)-2y^2}{\left(x^2+y^2\right)^2}=\frac{y^2-x^2}{\left(y^2+x^2\right)^2}$$

$$\frac{\partial g}{\partial x}=\frac{\partial f}{\partial y}$$

Since its z-component is $0$ and $f$ and $g$ are independent of $z$, it's unnecessary to check the remaining equations.

$\therefore F$ is a conservative field.

$$c\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)dA=c\iint_D(0)dA=0$$

$$\oint_C F\cdot dr=\oint_C \nabla f\cdot dr=f\left(\gamma(b)\right)-f\left(\gamma(a)\right)=0$$

Since $C$ is closed so $\gamma(b)=\gamma(a)$ for any closed path $C\in\mathbb{R}^3$