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The center of mass, in rectangular coordinates, is NOT located at (−0.147, 0.015, 0.467) as these coordinates should be divided by M = \pi/4.

You should also be careful about when answers are written as equalities and when they are approximate.

The precise answers are

M_{yz} = - \frac{3 \pi}{64}

M_{xz} = \frac{3 \pi}{640}

M_{xy} = \frac{1903 \pi}{12800}

So the centre of mass is located at (-3/16, 3/160, 1903/3200).

There is an error in the theorem for the p-series test. In particular, the currently stated theorem is false in the special case when a=0.

The current theorem states that

A general $p$--series $\ds\sum_{n=1}^\infty \frac{1}{(an+b)^p}$ will converge if, and only if, $p>1$.

The error could be corrected by changing this to

A general $p$--series $\ds\sum_{n=1}^\infty \frac{1}{(an+b)^p}$ with $a\neq 0$ will converge if, and only if, $p>1$.

Alternatively, the error could be corrected by changing the definition of a general p-series to require that a is nonzero. (Note that the definition of a general p-series is just a few lines before the theorem for the p-series test.)

Hello, the file README.md in APEXCalculusV5 still refers to version 4.0 and should probably be updated.

Sorry that I don't have a more substantial contribution to make at this point. Thank you for your work!

$5\vec r(t) = \langle 5\cos t+1,5\sin t+1.5\rangle$
should be
$5\vec r(t) = \langle 5\cos t+t,5\sin t+\frac32 t\rangle$

Somehow the sine is multiplied by -1 in the asy files.

In the text 14_Line_Integral_Intro.tex you have:

134: {Find the area under the surface $f(x,y) =\cos(x)+\sin(y)+2$ over the curve $C$, which is the segment of the line $y=2x+1$ on $-1\leq x\leq 1$, as shown in Figure \ref{fig:linescalarfield2}.

138: We find the values of $f$ over $C$ as $f(x,y) = f(t,2t+1) = \cos(t)+\sin(2t+1) + 2$.

while in the figures figlinescalarfield2_3D.asy and figlinescalarfield2b_3D.asy you have

4: // The surface is z=cos(x)+sin(y)+2; the path is the line
38: //Draw the surface z=cos(x)+sin(y)=2
40: return (t.x,t.y,-sin(t.y)+cos(t.x)+2);
48: triple g(real t) {return (t,2*t+1,-sin(2t+1)+cos(t)+2);}

and

4: // The surface is z=cos(x)+sin(y)=2; the path is the line
37://Draw the surface z=cos(x)+sin(y)=2
39: return (t.x,2*t.x+1,t.y*(-sin(t.x*2+1)+cos(t.x)+2));
47:triple g(real t) {return (t,2*t+1,-sin(2t+1)+cos(t)+2);}

Typo: 1+4t^2 on line 214 became 1+t^2 on line 218 of 14_Line_Integral_Intro.tex

214: {We parametrize our curve $C$ as $\vrt = \langle t,t^2\rangle$ for $-1\leq t\leq 1$; we find $\norm{\vrp(t)} = \sqrt{1+4t^2}$, so $ds = \sqrt{1+4t^2}\ dt$.

218: \int_C f(s)\ ds &= \int_{-1}^1 \Big(1-\cos(t)\sin\big(t^2\big)\Big)\sqrt{1+t^2}\ dt.\\

"Logarithmic definitions of Inverse Hyperbolic Functions" and "Integrals Involving Inverse Hyperbolic Functions" both show how to define $\csch^{-1}x$ in terms of $\ln$. But they disagree by a negative sign when $x<0$. "Definitions" seems correct, not "Integrals". Playing around with Desmos, that negative sign can go in several different places: https://www.desmos.com/geometry-beta/8dnclvtqdt

Unfortunately, I've not thought of a nice way of combining the two fractions from "Definitions" without introducing the sgn function.

This isn't technically a mistake, but \ds\frac2y, looks very similar to \ds\frac2{y'}. I'm going to try \ds\frac2y; and change to semicolons for the adjacent problems.

Unless it's really aiming to be tricky, I think you're wanting f(x)=\sin x to approximate \sin\pi, not to approximate \cos\pi.