Recent content by slider142

These are all correct methodologies. Good job!

Definitely. I do not know many good free books that go in depth the way that introductory books should, so unfortunately you will have to either pay for these texts or find them at a local library. The best introductory textbook path I know to calculus and vector calculus is to first go...

There is no problem doing that. However, the volume element is then different: you can calculate it using classical geometry or using the Jacobian of your new coordinate system. If you integrate the old volume element using your new coordinate system, what you are integrating will no longer...

The first bolded statement means that if you use a real number as an input value, then the output value will also be a real number. It is not a prescription that bars complex numbers from being used as input values. It simply has nothing to say about what happens if you use complex numbers that...

You've got it. You can use the remainder to find the sign of f(k), since the remainder is f(k), but not necessarily the sign of the quotient, which depends on the value of x.

Yes, assuming you mean ##dx\, dy\, dz\, dt##. One reason for using ##d^4x## instead is to save space, and to not emphasize any particular order for the iterated integrals. (There is also a little matter of the fact that the integral over a region is only equivalent to iterated integration over a...

1) Yes. 2) No. The quotient is a polynomial, plus a fraction whose denominator is the factor ##(x-k)##. In particular, if your synthetic division is: $$\begin{array}{r|rrrrr}\frac{7}{2} & 3 & -27 & 177 & 1347 & 420\\ & & \frac{21}{2} & -\frac{231}{4} & \frac{3339}{8} & \frac{98805}{16}\\\hline...

It is a straightforward generalization of the single-variable derivative to a multivariable function. Recall that if ##f:\mathbb{R}\rightarrow\mathbb{R}##, then $$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$ If we try to generalize this to two-variable functions, we have a problem with the...

No problem, but the value of sin(0) is indeed 0. :-) Edit: Oh, never mind. I see you meant to write "sin(0) was 1" now. No worries! It happens to all of us!

You've got it. However, one of your dot products was not evaluated properly. You should have: ##\alpha (s)=\left< s,0 \right> \quad \dot { \alpha } =\left< 1,0 \right> \quad 0\le s\le x\\ \beta (t)=\left< x,t \right> \quad \dot { \beta } =\left< 0,1 \right> \quad 0\le t\le y\\ f=\int _{ 0 }^{...

Not quite! We can also take any two points in 3-dimensional space without the origin and join them with a path that avoids the hole at the origin. Simply connected means more than just being able to join every pair of points with paths: it also means the absence of holes. Holes can be detected...

Not quite! The boundary occurs when both surfaces have the same z coordinate, as we can see from the diagram. Equating the z values of the variables in both equations, we see that the x and y coordinates of points on the boundary must satisfy the equation: \sqrt{x^2 + y^2} = \sqrt{8 - x^2 - y^2}...

What is the boundary at which the cone and spherical cap intersect? As such, between which two x coordinates does the volume lie?

Unfortunately, it does not! A closed vector field only implies that it is an exact vector field if its domain is simply connected. A closed vector field defined over a domain with a hole, such as a vector field which is undefined at the origin, is not exact. In addition, there exist exact vector...

Hi mr-feeno, Invertible means the same thing as in "invertible function". We are seeking a "reverse" function that can give us back the original vector from the image of a vector in the range of the matrix transformation. That is, consider the matrix A as a function f(v) = Av, where Av is...