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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.

I'm not quite sure what you actually want, but I don't think your equation for ##L_r## satisfies your definition for line ##r##.

What is the ##^*31## operation?

Here's another way to see the new limits. Think about ##u## and ##v## in polar coordinates. Then ##u = r^2\cos^2\theta - r^2\sin^2\theta = r^2\cos(2\theta)## and ##v = 2r^2\cos\theta \sin\theta = r^2\sin(2\theta)##. Now think about a ray in the first quadrant at angle ##\theta##. Hold ##\theta##...

I gather you have successfully changed the integral to ##u,~v## variables and are just asking about the limits. Am I correct about that? For doing the limits on this problem, what I have noticed is that ##u_x = v_y## and ##v_x = -u_y## suggesting that ##z = x^2 -y^2 + i(2xy)## is an analytic...

Your answer is not correct as you can verify that for your ##\phi##, ##\phi_x = P,~\phi_y = Q## fails. The problem is you have an extra ##\frac x y## in your answer. You really need to read the link I mentioned in post #7. In particular the last example.

The slanted plane leans up against the coordinate planes in the first octant. It is the only octant where the plane bounds a finite volume with the coordinate planes. So the 1/8 makes no sense. There is no larger volume that this is a portion of.

So you are going to integrate in the ##z## direction first, the ##x## direction second, and the ##y## direction last. Ok, that means ##dV = dzdxdy## in that order. Why the ##\frac 1 8##? That shouldn't be there. And it should be ##dV=dzdxdy## (a typo?) Should be$$ V= \int_{y_1=0}^{y_2 = 2}...

Look here for help on computing potential functions: https://www.physicsforums.com/insights/how-to-find-potential-functions-a-10-minute-introduction/ Use it on your ##\int \frac{xy+1}{y}~dx + \frac {2y-x}{y^2}~dy##

Inflection points? Really?

How do you get your answer for b? What happens at an inflection point? For c, do you mean 2 local maxes and mins, or 2 of each local maxes and mins? Which is correct?

Remember, ##\int_C \vec F \cdot d\vec r = \int_t \vec F(\vec r(t))\cdot \vec r'(t)~dt##. So the obvious hint is plug the formulas in and see what happens. Then come back to show us where you are stuck. There is no substitute for getting your hands dirty.

Your teacher's statement that the dashed line is not a boundary of either circle is beyond ridiculous. You have a region that contains part of each circle and the dashed line is a boundary of each subarea. You have done the problem absolutely correctly. Nicely done. Your teacher deserves a demerit.

How else to get students through calculus who can't do algebra?

Look at the statement of the problem in post #1.

I don't want to slog through all the details to look for typos etc., I'm sure you can find them yourself. One more point I would give you. When you separated variables and had the equation for ##Y(y)## in the form ##Y''(y) = k^2Y## and ##X(x)Y(0) = 0## that gives you ##Y(0) = 0##. At that point...

Probably just a typo, but that ##x## should be a ##y##.

I didn't check the details, but it looks pretty good to here and the below equation is what I would expect. Not too lost I think. You are almost done. Let's call ##A_n\sinh(n\pi) = a_n##. So your equation is $$ \sum^{\infty}_{n=1} a_n sin(\frac{n\pi} S x)= T_0$$Hopefully, if you were given that...

This reply is about using ##\{\sinh(kx), \cosh(kx)\}## pair instead of the ##\{e^{kx}, e^{-kx}\}## pair. When are solving ##y'' - k^2 y = 0##, with characteristic equation ##r^2 - k^2 = 0## and roots ##r = \pm k## that gives you an independent solution pair ##\{ e^{kx}, e^{-kx}\}## and you would...

But it isn't in best factored form yet.

@Athenian : One thing that will simplify you work at this point is to use sinh and cosh functions instead of exponentials. So your ##Y_n(y)## would be ##C_n\cosh(k_ny) + D_n\sinh(k_ny)##. Then when you apply the boundary condition ##Y(0) = 0## you get ##C_n = 0##. So your solution will be a...

I supose all those denominators that look sort of like an ##\alpha## are actually ##2##'s.Your figure shows the inner distance between the squares as ##1## unit and the outer distance between them is ##\frac 1 2 - (-\frac 1 2) = 1## so the squares have sides of length ##0##, so they aren't there.

Your graph is incorrect. The y intercept should be positive and the x intercept negative. And my answer (posting when tired) in post #1 (and yours) is incorrect . The correct answer should be ##\frac {21}{\sqrt{29}}##.

Have you drawn a graph? 13.6 looks quite a bit too big. I get about 6.8 with a different method. Edit: See below.

It depends on the boundary conditions. For a string with fixed ends, you might want a periodic extension. The idea is if you have an infinite string where a point never moves, that is indistinguishable from the string being tied down at that point. Yes. It is that interference that gives what...

You want to consider the odd extension of the function. Not the odd periodic extension. So your initial problem looks like an infinite string with initial displacement the odd reflection of the triangle and zero outside of ##[-2L,2L]## and released from rest. Solve that infinite string problem...

Please post your image right side up.

Yes. I inadvertently left out the ##r## going from ##0## to ##1##. And for the OP's benefit, I think he should get ##6\pi## for the answer to both.

What did you get for ##\nabla \times \vec A##? Think about parameterizing the surface with ##x = 3\cos\theta,~y=2\sin\theta## and similarly for the boundary curve.

It's hard to tell from your graphic, but I'm not sure what you are describing is even a cone, slanted or not. Is the base curve a circle? Ellipse? Look at the following figure: You should have a base and the cross sections should be similar as in the figure. The volume of any such solid is the...

I would stop right there. It's factored.

No. The limit wasn't ##1##. It was ##|2x-1|\cdot 1##.

The ratio test gave you ##|2x - 1| < 1##. Factor out a ##2## and divide both sides by it giving ##|x - \frac 1 2| < \frac 1 2##. That tells you the center of the series is ##\frac 1 2## and the radius of convergence is ##\frac 1 2##.

Obviously a typo, he means 1/2.

It is pretty easy to show that both ##\frac{k+1}{k}## and ##\frac{\log(k+2)}{\log(k+1)}## go to ##1##.

All these answers to a 4 year old post by someone who hasn't been active for 2 years.