Multivariable Calculus – Change of Bound Help!

I am having trouble following these steps in a reading on multivariable calculus.

Due to a change of variables: 

$ \displaystyle\int_t^T \int_t^s \theta_v dv \, ds = \int_t^T \int_s^T \theta_s dv \, ds$

Could anyone explain how you get the right equation? What is the ‘change of variable’ that is applied? Thanks!! Also $ t \leq T$ is given.

Squeeze theorem of multivariable functions

My Question is the following: Let F: U $ \subset \mathbb{R}^n$ $ \rightarrow \mathbb{R}^m$ where U is an open set containing a punctured neighbourhood of a. Let G : $ U \subset \mathbb{R}^n \rightarrow \mathbb{R}^{\geq 0}$ . Assume $ 0 \leq |f(x)| \leq h(x)$ ( Note, here |.| denotes the magnitude of the vector. Show that if limit of h(x) as x approaches a is 0 then the limit of f(x) as x approaches a is 0. Note that I get the general idea; however, what i’m stuck on is showing that
|f(x)|$ \leq$ h(x)$ \leq$ $ |h(x)|$ . I’m not sure why that follows and what it means because h(x) is a vector, and the absolute value is the length of the vector..

Critical points of multi-variable function.

I have the expression:

$ (x^3-y^2 )(x-1)$

and have to find the critical points and their nature. The points found are:

$ (1,1) = $ saddle point

$ (1,-1) = $ saddle point

$ (\frac 3 4 ,0) = $ global minimum

$ (0,0) = $ inconclusive, however, when I take $ 0.1$ and $ -0.1$ as $ 2$ points slightly higher or lower to find the nature of the point, I get a value positive or negative value for both the $ fxx$ and the Hessian Determinant i.e. $ D < 0$ at $ 0.1$ and $ D > 0$ at $ -0.1$ .

I understand, for a single variable function, that two different signs for values higher or lower than the points, would mean the point $ (0,0)$ is an inflection point. But since it is a multi-variable function, how can I determine the nature of $ (0,0)$ since the Hessian Determinant must be $ < 0 $ for the points to be saddle points?

Finding the Wedgeproduct of Two Multivariable Vectors and Making Sense of It

I was attempting to solve: $ $ -2dx_{1} \land dx_{4}( \begin{bmatrix} 2 \ 3 \ -5 \ 2 \end{bmatrix}, \begin{bmatrix} 2 \ 3 \ 4 \ -5 \end{bmatrix}) $ $ I solved this by pulling out the scalar and then finding the determinant of the matrix given by (let $ u$ be the first column vector and $ v$ be the following one): $ $ det\begin{bmatrix} -2*dx_1(u) & -2*dx_1(v) \ dx_4(u) & dx_4(v) \end{bmatrix}=-2*det\begin{bmatrix} dx_1(u) & dx_1(v) \ dx_4(u) & dx_4(v) \end{bmatrix}=-2*det\begin{bmatrix} 2 & 2 \ 2 & -5\end{bmatrix}=-2(-14)=28$ $ We then have to explain what this number represents and that is where my confusion comes in. It is my understanding that this represents the area of the parallelogram formed from $ u$ and $ v$ being projected onto the $ \partial_{x_1}\partial_{x_4}$ plane. Should this area be positive or negative tho? Without the scalar $ -2$ the oriented area would flip since going from $ u$ to $ v$ would be clockwise but with the scalar it would be counterclockwise so you wouldn’t switch the sign? I guess my main confusion is coming from whether or not the determinant has the orientation built in and whether or not the scalar on one of the one-forms affects the orientation?

Why are complex functions allowed to have the same derivative in all directions while multivariable functions aren’t?

Comsider a multivariable function from $ R^2$ to $ R^2$ which is $ f(x,y)=<g(x,y), h(x,y)>$ . We’re initially at $ (x,y)$ and we change the argument by a small vector of magnitude $ dr$ in two different directions. In this case, the rate of change of $ f(x,y)$ will depend on the direction even if the ‘small change’ is of the same magnitude.

This is because the rate of change is given by gradient•u. So we’re bound to have different rates of change in different directions. It’s just not possible to change by the same rate in different directions.

But if we have a complex differentiable function $ c(z)$ . If we’re initially at $ z$ and change by $ dz$ in two different directions, then the rate of change is the same.

How do complex functions make this work? My math level is high school. Don’t use advanced math in the answer.

Flux. Multivariable Calculus

Find the flux of F(x, y, z) = ⟨-x, -y, z^3⟩ through the surface S when S is the part of the

cone z =√x^2 + y^2

that lies between the planes z = 1 and z = 3, oriented upwards.

If I use the usual method, Im getting 1712pi/15 If I use divergence theorem, Im getting 1916pi/15

Is the divergence theorem not valid in here? thanks

Global extremas of multivariable function on circle : are my results correct?

I wanted to know if my method and my results are correct. There isn’t a solution to that question, so I don’t know if it’s correct.

Given $ f(x,y)=3x^2-y^3$ and I need to find its global extremas on the circle $ x^2+y^2 \leq 25$ .

So inside the domain, I have $ \nabla f =(6x, -3y^2)= (0,0)$ which yields the trivial case $ x=y=0$
Now on the border, we have $ x^2+y^2=25$ . I use lagrange multiplier : $ \nabla f = \lambda \nabla g$ where $ g(x,y) =x^2+y^2-25$
This gives $ (6x, -3y^2)= \lambda (2x, 2y)$ .
If $ x \neq 0$ and $ y \neq 0$ , then $ \lambda = 3$ , $ y=-2$ and $ x= \pm \sqrt{21}$ . If $ x=0, y = \pm 5$ . If $ y=0, x = \pm 5$ . So if we plug our values into our function, we get $ 0, 71, -125, 125, 75$ So there is a global maximum of 125 at $ (0,-5)$ and a global minimum of $ -125$ at $ (0, 5)$ .
Are my solutions correct ?

Also, I guess I could have used polar coordinates on the border given that it’s a circle. But if we do that ($ r=\sqrt{25}=5$ ) and try to take the derivative of our function in polar coordinates, we get $ 5^2\frac{d}{d \theta}(3cos^2(\theta)-5sin^3(\theta))$ which isn’t nice to derivate. Do you agree that it’s much simpler to use Lagrange multiplier in this case ?

Thanks for your help !