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The horizontal cross-section is a triangle, so its area is half its width times its depth:Thus the volume integral, with $t=z/H$, isFor the substitution rule, put $u=1-t/2$. The endpoint values are $1$ and $1/2$, and $1-t=2u-1$, givingTherefore $\boxed{V=\frac{(4+2\sqrt{2})LDH}{15}}$.

The volume form in cylindrical coordinates is $dV=rdrd\varphi dz$. Integrating around the axis and over the height gives cylindrical-shell volume $2\pi hrdr$. Hence the volume integral isFor unit density, this is the kinetic energy of the rotational motion.

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

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 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.

Here $y\ge 0$, so the absolute value equals $y$. The volume integral, evaluated in Cartesian coordinates, isThe integrand is constant with respect to $z$, giving

Use spherical coordinates $x=s\sin \theta \cos \varphi$, $y=s\sin \theta \sin \varphi$, $z=s\cos \theta$. The density becomes ${\rho }_{0}s\sin \theta$, and the Jacobian determinant supplies $dV=s^2\sin \theta dsd\theta d\varphi$. Therefore the mass is the volume integralUsing ${\sin }^2\theta =(1-\cos 2\theta )/2$, the three integrals are $2\pi$, $\pi /2$ and $1/4$. Hence

Direct differentiation givesOn the unit disk $D$, polar coordinates give ${\int }_D{y}^{2}dA=\pi /4$, ${\int }_{D}xdA=0$ and ${\int }_{D}dA=\pi$. Consequently the volume integral isFor an independent surface integral calculation, the relevant cross product isThe bottom face contributes ${\int }_{D}xydA=0$ and the top contributes ${\int }_D(-h{y}^2-xy)dA=-\pi h/4$. On the cylindrical side, put $x=\cos \theta$, $y=\sin \theta$, $\hat{\mathbf{n}}=(x,y,0)$ and $dS=d\theta dz$. The flux density is $x^3+2xy{z}^2+2y^{2}z-x{y}^2$. All terms except $2y^{2}z$ integrate to zero because they are odd functions of $x$ or $y$, givingThus minus the total outward flux is $-(\pi h^2-\pi h/4)$, in agreement with the volume integral.

Let $a=\sinh 1$ and $c=\cosh 1$. The region is the first-octant portion of the ellipsoid $(x^2+y^2)/a^2+z^2/c^2\le 1$. In cylindrical coordinates, write $\rho =\sqrt{x^2+y^2}$ and $0\le \varphi \le \pi /2$. Away from the axis, the divergence isThe field is singular on the axis, so applying the classical divergence theorem there without qualification would be invalid. Instead remove $\rho <\epsilon$ and apply it to the remaining region. On the added cylindrical boundary the unit normal is $-{\mathbf{e}}_{\rho }$, andIts area element is $\epsilon d\varphi dz$, so its flux is $O(\epsilon |\log \epsilon |)\to 0$. The axial component of the field has zero dot product with this normal. The volume integral of the divergence is locally integrable, because its $1/\rho$ factor cancels the cylindrical measure. Thus removing a singular axis in a flux integral gives the desired improper surface integral with no extra axial term.

The divergence theorem now reduces the calculation toThe integral over $s$ is the area of a quarter unit disk. All boundary fluxes are interpreted as convergent improper integrals at the axis.

For a continuously differentiable vector field $\mathbf{u}$ on a bounded region $V\subset {\mathbb{R}}^3$ with piecewise smooth closed boundary $S=\partial V$, the divergence theorem statesHere $\nabla \cdot \mathbf{u}$ is its divergence, $dV$ is the volume element, $dS$ is the surface area element, $\hat{\mathbf{n}}$ is the outward unit normal, and $d\mathbf{S}=\hat{\mathbf{n}}dS$. Thus a volume integral of the divergence equals the outward flux through the boundary.

Use Einstein summation convention and write ${\partial }_j={\nabla }_j$. Since ${\partial }_j{r}_i={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta, the product rule givesThis proves $\boxed{P_i={\partial }_j(r_i{w}_j)-w_i}$. Now take $w_j={\partial }_j\psi$. Integrating this identity and applying the divergence theorem to the first term givesPart (b) changes the last volume integral of the gradient into a surface integral, yieldingHere $d{S}_i={\hat{n}}_{i}dS$; the Laplacian is ${\nabla }^2\psi ={\partial }_j{\partial }_j\psi$.

Close the hyperboloid segment with the two disks at $z=\pm 1$, and let $V$ be its interior. The flux through either disk is zero because $A_z=0$. Thus the desired surface integral is the total outward flux through the closed boundary. Sincethe divergence theorem and cylindrical coordinates giveThe extra factor $r$ is the volume integral measure in cylindrical coordinates; omitting it would give the wrong flux.

For an arbitrary material volume $V(t)$, Newton's second law equates the rate of change of linear momentum to the resultant contact force, since there is no body force:Use the Reynolds transport theorem on the left. Expanding the product and using the continuity equation leavesOn the right, the divergence theorem applied to each row of the Cauchy stress tensor givesConsequently the volume integral of $\rho D{u}_i/Dt-{\partial }_j{\sigma }_{ij}$ vanishes on every material volume. Localizing as in part (i) gives the Cauchy momentum equation:The left side is mass density times material acceleration; the right side is the divergence of the Cauchy stress tensor.

A vector field singular on an axis need not satisfy the classical divergence theorem on the original closed region. Remove a thin tube $\rho <\epsilon$, apply the theorem to the remaining smooth region, then examine the added cylindrical boundary. If its flux tends to zero and the remaining volume integral converges, the improper boundary flux equals the improper integral of the divergence. For $\mathbf{F}=(\log {\rho }^2,\log {\rho }^2,z/\rho )$ with $\rho =\sqrt{x^2+y^2}$, the cylindrical unit normal has zero $z$ component, so the added flux is $O(\epsilon |\log \epsilon |)$ on a bounded axial interval. The divergence $(2\cos \varphi +2\sin \varphi +1)/\rho$ is locally integrable under the cylindrical coordinates measure $\rho d\rho d\varphi dz$. Hence no finite line-source correction survives, despite the field itself being unbounded.

A volume integral integrates a scalar or componentwise vector field over a three-dimensional region with its volume form. In spherical polar coordinates the Euclidean volume form is $r^2\sin \theta drd\theta d\varphi$. The divergence theorem relates a volume integral of a divergence to a boundary flux integral.

Let a bounded planar region $D$ have positive area $A$ and centroid $\overline{\mathbf{x}}=A^{-1}{\int }_D\mathbf{x}dA$. If a solid has nonnegative thickness $h(\mathbf{x})=c+\mathbf{v}\cdot \mathbf{x}$ above $D$, its volume integral givesThus the average of an affine function over a region is its value at the centroid. The proof is the linearity of integration. For a disk the centroid is its centre, making the formula particularly useful for a cylindrical bore cut through a solid whose width varies linearly with height.

For a solid spherical cap of height $0\le h\le 2R$ cut from a sphere of radius $R$,Centre the sphere at the origin with the cap on the side $z\ge R-h$. Its cross-section at height $z$ is a circle of squared radius $R^2-z^2$. A volume integral in cylindrical coordinates givesThe formula includes caps larger than a hemisphere, when $h>R$, and the whole solid sphere, when $h=2R$. For the retained region $z\ge -d$ with $0<d<R$, set $h=R+d$ to obtain $V=\pi (2R^3/3+R^{2}d-d^3/3)$. Multiplying by uniform mass density gives its mass.