On an interval with $x\ne 0$, the equation is a separable differential equation. For $y\ne 0$,Taking antiderivatives gives $-1/y=-4/x-x^3/3+C$. Renaming the constant, the explicit general nonzero solution isIt is defined on any interval avoiding $x=0$ and zeros of its denominator. The separation divided by $y^2$, so the additional solution $\boxed{y(x)\equiv 0}$ must be included.

The handle and the two end pieces have disjoint interiors; their common boundary circles contribute no volume. Hence the total mass is $M=M_h+2M_e$. With $\delta =\alpha -\gamma$, the results above giveEquivalently, since $\gamma +\delta =\alpha$,

Use spherical coordinates $(r,\theta ,\varphi )$ centred on the right-hand sphere, with polar axis pointing outwards along $s$. The Jacobian determinant is $r^2\sin \theta$, and the retained region is $r\le R$ with $r\cos \theta \ge -d$.

For $0\le r\le d$, every polar angle $0\le \theta \le \pi$ is allowed. For $d<r\le R$, the cut requires $0\le \theta \le \arccos (-d/r)$. Splitting the volume integral at $r=d$ therefore givesSubstitution of $R=\beta a$ and $d=(\alpha -\gamma )a$ reproduces the mass in part (c). The radial split avoids integrating over a polar angle range independent of $r$, which would incorrectly include part of the removed spherical cap.

Centre cylindrical coordinates on the right-hand sphere, with the axial coordinate $s=z-\alpha a$ positive outwards. Using $R=\beta a$ and $d=(\alpha -\gamma )a$, the retained solid satisfiesIn particular, it contains more than a hemisphere; it is the small cap on the inward side that has been removed. The volume integral for its mass, including the Jacobian determinant $r$, isThus each end piece has massEquivalently this is the volume of a spherical cap with height $R+d$, multiplied by the uniform mass density.

Use cylindrical coordinates $(r,\varphi ,z)$ about the dumbbell axis. Their Jacobian determinant contributes the volume factor $r$, so the handle's mass is the volume integralInserting the radius from part (a) yields

Take the dumbbell axis to be the $z$ axis, with its midpoint at $z=0$. The right-hand sphere has centre $z=\alpha a$ and radius $R=\beta a$, while its junction with the cylinder is at $z=\gamma a$. Put $d=(\alpha -\gamma )a$, so $0<d<R$. The Pythagorean theorem in a meridional cross-section gives the handle radius $H$ from $H^2+d^2=R^2$. Thus

For $\theta \notin 2\pi \mathbb{Z}$, use the finite geometric series with ratio $e^{i\theta }$:The second equality follows by factoring $1-e^{it}=-2i{e}^{it/2}\sin (t/2)$ in numerator and denominator. Taking real parts proves the finite trigonometric sum identityWhen $\theta =2p\pi$, every cosine is $1$, so the sum is $\boxed{N}$. This also gives the continuous limiting value at the excluded arguments.

Since $x=\cos y$ with real $y$, necessarily $-1\le x\le 1$. Use Euler's formula and set $w=e^{iy}\ne 0$. Thenso the quadratic equation gives $w=x\pm i\sqrt{1-x^2}$. Let $q=x+i\sqrt{1-x^2}$. It lies on the upper half of the unit circle, so $q=e^{i\theta }$ with $0\le \theta \le \pi$. Choose the complex logarithm value $\ln q=i\theta$, taking $\ln (-1)=i\pi$ at $x=-1$. The two roots are $e^{\pm i\theta }$, so $e^{iy}=e^{\pm i\theta }$ gives $y=\pm \theta +2n\pi$.

Because $i\ln q=-\theta$, this set of real solutions is exactlyThe formula includes $x=\pm 1$: the two sign families then coincide after adjusting $n$. No real $y$ exists if $|x|>1$.

A counterclockwise rotation through $\alpha$ about the origin in the complex plane sends $z_j$ to $e^{i\alpha }{z}_j$. By part (a), and because the complex conjugate of $e^{i\alpha }$ is $e^{-i\alpha }$,Thus $Q$ is invariant under the common rotation.

The given complex numbers have positive real parts, so $\arg z_1=\arctan A$ and $\arg z_2=\arctan B$. Their product isBy the assumed complex argument ranges, this product also has positive real part, hence $1-AB>0$. Its complex argument is consequently $\arctan [(A+B)/(1-AB)]$, with no extra multiple of $\pi$. Since multiplication adds complex arguments,This proves the arctangent addition formula under the stated branch restriction. For arbitrary real $A,B$, the principal inverse tangent formula can require adding or subtracting $\pi$ when $AB>1$.

The sine is nonnegative on $[0,\pi ]$, so the geometric area equals the definite integral, without cancellation:

For integration by parts, take $u=x$ and $dv=\cos xdx$. Then $du=dx$ and $v=\sin x$, giving the antiderivative