Recent content by LCKurtz

Since ##f(0^+) = -\infty##, your function doesn't satisfy the Dirichlet conditions, so the usual theorem about convergence doesn't apply.

Of course, if you use KNMP paint, there is no paradox. Definition: The xy plane is painted if for any ##r > 0##, the circular area at the origin of radius ##r## is covered. Theorem: A single drop of KNMP paint is sufficient to paint the ##xy## plane. Proof: Suppose a drop of KNMP paint has...

For (b), think about a vector parallel to each of the parallel lines i.e., a common direction vector ##\vec D## for the intersection lines. What is the relation of the three normals to the planes to ##\vec D##? What does that tell you? For (c), What direction would ##\vec N_1 \times \vec N_2##...

##r## going from ##0## to ##\cos\theta## is correct. I get ##\pi - 2## for the answer.

##(\cos t, \sin t )## traverses the unit circle in the xy plane counterclockwise. If you want to go the other way, reverse the parameter, ##t \rightarrow -t##. What does that do to the expression?

Imagine the area being actually cut and pulled slightly apart at the cut. Start to go around it. There's no way to walk around it without going the opposite direction on the inner circle. Just follow the arrows.

It's a bit tricky to see geometrically. But imagine that cone was much shallower, almost flat. In fact, suppose in the extreme it was flat, and you are looking down on two concentric circles, with the normal pointing away from you (downward), By the right hand rule, the outer circle would be...

Let me add that when I read the OP's title and post, I assumed he wanted to verify Stoke's theorem, i.e., work both sides and show they are equal. If you can use Stoke's theorem, there is indeed an easy method to evaluate the flux integral, appropriate for a 3rd semester calculus class, which I...

I think you may be helping me make my point. No third semester calculus course I ever taught covered 3d rotations. I wouldn't expect a typical 3rd semester calculus student to know how to find the equations for your ##x',~y',~z'##.

Well, yes, but, playing Devil's advocate for the OP, assuming he knows how to make vectors ##\vec v_1,~\vec v_2##, the real problem is setting up the appropriate line and surface integrals with correct limits.

I think you have picked a random very tricky problem. First of all, the problem is stated poorly. I presume they are talking about the portion of the sphere above the plane. Note that the circular intersection of the plane and sphere is not a great circle and does not stay in the first octant...

Did you check all the sides of the volume when you did the surface integrals? They each have their own normal and some of them may not be zero.

I think your answer is correct. Also, looking at the answer choices, I would bet the first answer is supposed to be ##180##, given the pattern in the answer choices. Probably just a typo.

You didn't specify whether this is a 2D or 3D problem. You also didn't specify that ##\vec u, ~ \vec v,~ m## are given constants and the unknowns are ##\vec x## and ##\vec y##. Is that correct? Certainly, if it is a 3D problem there can be many solutions. For example if ##\vec v = \langle...

The shape of the island is an upside-down paraboloid. The integral looks like everything is OK except the ##r## limits should be reversed. You want to integrate from the smallest to largest values of ##r## to get a positive answer.