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Apply the divergence theorem to $\varphi \mathbf{c}$, using ii. Because the vector $\mathbf{c}$ is constant, it can be taken outside both integrals:This holds for every $\mathbf{c}$, in particular each of the three Cartesian basis vectors. Equality of the three components proves the gradient volume-to-boundary identityIt suffices that $\varphi$ be a continuously differentiable function on a neighbourhood of the closure of $V$.

For a bounded region $V\subset {\mathbb{R}}^3$ with piecewise smooth boundary, suppose $\mathbf{F}$ is a continuously differentiable vector field on an open neighbourhood of its closure. The divergence theorem, with outward unit normal vector $\mathbf{n}$, isThe surface integral measures total outward flux. Edges and isolated vertices have zero surface area and do not affect the integral.

Use cylindrical coordinates. The real square root restricts the projection to $0\le r\le 1$, and there $r-1\le 0\le \sqrt{1-r^2}$. Since $\nabla x={\mathbf{e}}_x$, the volume integral isFor an independent boundary calculation, split $\partial V$ into an upper hemisphere $S_{+}$ and a lower right circular conical surface $S_{-}$. They meet in the circle $r=1,z=0$ and there is no additional planar face.

On $S_{+}$, use the parametrizationThe outward unit normal vector is $\mathbf{R}$, and the surface area element is $\sin \alpha d\alpha d\theta$. The $y$ and $z$ components of ${\iint }_{S_{+}}x\mathbf{n}dS$ vanish by the angular integrals of $\sin \theta \cos \theta$ and $\cos \theta$. The $x$ component isOn $S_{-}$, the parametrization $(r\cos \theta ,r\sin \theta ,r-1)$ has outward vector area elementIts negative vertical component points out below the solid. Multiplying by $x=r\cos \theta$ and using the same angular symmetry in integration givesTherefore the independently evaluated surface integral isin agreement with the volume integral. The isolated apex of the cone does not contribute to the surface integral.

Start with the solid between the elliptic paraboloid $z=x^2+y^2$ and the upper plane $z=10-2y$. Its projection is the disksince their vertical separation is $11-x^2-(y+1{)}^2$. The second lower bound, $z\ge 2-2x$, removes the part between the elliptic paraboloid and this lower plane whereverCall this second disk $E$. Its centre is distance $\sqrt{2}$ from the centre of $D$, andbecause $5+2\sqrt{6}<11$. Thus $E$ lies wholly inside $D$. In particular the lower plane never exceeds the upper plane on $E$, so the subtraction describes exactly the requested solid.

The nested-disk subtraction for paraboloid volumes therefore gives a difference of two double integrals:Translate polar coordinates to the centre of each disk. A general integral of this form isUsing $A=11$ and $A=3$ gives

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

Enforce the fixed-end Dirichlet boundary condition by taking the odd extension of the initial velocity to the whole line:This agrees with the specified initial condition for $s\ge 0$. To derive the relevant form of D'Alembert's formula, the zero initial displacement in b gives $G=-F$, up to a cancelling constant. The initial velocity then gives $-2c{F}^{\prime }=v_o$. HenceAn antiderivative, valid across $s=0$, isEvaluation therefore gives the solution throughout $x>0,t>0$:For $ct\le x$, replace $|x-ct|$ by $x-ct$; for $ct\ge x$, replace it by $ct-x$. Thus the same expression includes the reflected wave after it reaches the fixed end.

At $x=0$ the two terms agree, so $y(0,t)=0$. At $t=0$ they again agree, so $y(x,0)=0$. Differentiating the integral with respect to $t$ giveswhich recovers $y_t(x,0)=\kappa x{e}^{-x/L}$. Since $v_o$ is a continuously differentiable function, the formula is twice continuously differentiable even across $x=ct$, and satisfies the wave equation there as well.

For $0<y<\pi$, define $J(y)={\int }_0^{\pi }{u}_y(x,y)dx$. Differentiate under the integral and use Laplace's equation:Thus $J$ is constant. The top Neumann boundary condition gives $J(\pi )=0$, soFor the included endpoint $y^{\ast }=0$, the integral is interpreted as an improper integral at the bottom corners if necessary. The Fourier series constructed in c gives, on $0<x<\pi$,The second series is absolutely and uniformly convergent; the first term has only integrable logarithmic singularities. Its integral is zero since ${\int }_0^{\pi }\log (2\sin x)dx=0$, while every cosine term also has zero integral. This proves the endpoint as well:For completeness, the logarithmic integral follows without any endpoint differentiation. Put $I={\int }_0^{\pi /2}\log (\sin x)dx$ and substitute $x\mapsto \pi /2-x$ to obtain $I={\int }_0^{\pi /2}\log (\cos x)dx$. Adding and using $\sin x\cos x=\sin (2x)/2$ gives $2I=I-(\pi /2)\log 2$, hence $I=-(\pi /2)\log 2$ and the required full-interval result.

For separation of variables, set $u=X(x)Y(y)$, with $X^{\prime \prime }/X=-Y^{\prime \prime }/Y=-\lambda$. The side Neumann boundary conditions give $X^{\prime }(0)=X^{\prime }(\pi )=0$. As in the standard Sturm-Liouville problem, integration by parts shows $\lambda \ge 0$. The zero eigenvalue gives constant $X$, and $Y^{\prime \prime }=0$ together with $Y^{\prime }(\pi )=0$ makes $Y$ constant too.

For positive eigenvalues, the side conditions give $\lambda =n^2$, $X_n=\cos (nx)$, $n\ge 1$. The top condition chooses the hyperbolic cosine factor $\cosh (n(\pi -y))$. Normalize it to one at $y=0$, obtainingThe orthogonality of integer Fourier modes and the permitted interchange of summation and integration giveFor the remaining Fourier coefficients, integrate by parts twice. The sine boundary term vanishes, and with $f=x(\pi -x)$,The last integral vanishes because $f^{\prime \prime }=-2$ and $n$ is a positive integer. Thus $a_n=-2(1+(-1{)}^n)/n^2$: only even Fourier modes survive, as expected from reflection symmetry about $x=\pi /2$. Writing $n=2m$ givesThe factor involving hyperbolic cosine lies between zero and one on the closed rectangle. Comparison with $\sum m^{-2}$ proves uniform convergence, so the solution is continuous on the entire boundary and takes the prescribed bottom values. On every strip $y\ge \epsilon >0$, the Fourier coefficients have additional exponential decay, allowing differentiation and verifying Laplace's equation and all three Neumann boundary conditions.

To justify the endpoint calculation in b, differentiate at $0<x<\pi$, $y=0$ in the interior-edge sense:Subtracting $1$ from the hyperbolic tangent leaves an absolutely convergent series. For the remaining part, the power series $-\log (1-z)={\sum }_{m\ge 1}{z}^m/m$ gives, by taking real parts at $z=r{e}^{2ix}$ and then $r\uparrow 1$,The same formula follows as the limit of the differentiated interior solution, since its leading multiplier is $e^{-2my}$. This establishes the boundary normal derivative used in b. It also explains why the failure of the corner compatibility condition for mixed boundary data affects first derivatives at the bottom corners without spoiling continuity of $u$ itself.

Use source spherical coordinates with radius $s$, polar angle $\alpha$ measured from the positive $z$-axis, and azimuth $\theta$. The lower half of the spherical shell has $a\le s\le 2a$, $\pi /2\le \alpha \le \pi$, and $0\le \theta \le 2\pi$. Its volume element is $s^2\sin \alpha dsd\alpha d\theta$. At the observation point $(0,0,h)$, $h>0$, the separation is $\sqrt{s^2+h^2-2sh\cos \alpha }$, soThe angular integral is elementary: put $q=\cos \alpha$ to obtainWe used $\sqrt{(s+h{)}^2}=s+h$ since both $s,h$ are positive. HenceThus the axial potential of a uniform hemispherical shell, at every point of the positive $z$-axis, isThere is no need to split into $h<a$, $a<h<2a$, and $h>2a$: the observation point is outside the lower hemisphere in all three cases. As checks, $\varphi (h)\to -3\pi G\rho a^2$ as $h\downarrow 0$, and at large $h$,The last factor is exactly the total mass of the lower half of the spherical shell.

Write the observation point as $(0,0,h)$ with $h>0$, reserving source coordinates for the integral. For the straight segment, $d{s}^{\prime }=d{x}^{\prime }$ and the separation is $\sqrt{x^{\prime 2}+h^2}$. Therefore its Newtonian gravitational potential isEquivalently $\varphi (h)=-2G\gamma \mathrm{arsinh}(a/h)$. At large $h$ it approaches $-G(2a\gamma )/h$, the Newtonian potential of a point mass with the segment's total mass.

Put $E_n={\sup }_{t\in [a,b]}|f(t)-{\varphi }_n(t)|\to 0$, so the hypothesis is uniform convergence. Every step function is bounded and Riemann integrable: away from its finitely many partition points it is constant, and those exceptional points can be covered by intervals of arbitrarily small total length. A partition of an interval containing their boundaries makes the difference of its upper Darboux sum and lower Darboux sum arbitrarily small.

Choose one $n$ with $E_n<\infty$. Boundedness of ${\varphi }_n$ and $|f|\le |{\varphi }_n|+E_n$ show that $f$ is a bounded function. Assume $a<b$, since for $a=b$ all the integrals are zero. Given $\eta >0$, choose $n$ so that $2E_n(b-a)<\eta /2$, and then a partition of an interval $P$ such that $U({\varphi }_n,P)-L({\varphi }_n,P)<\eta /2$. On every partition subinterval, $|f-{\varphi }_n|\le E_n$ givesThusThe Riemann integrability criterion proves that $f$ is Riemann integrable. Finally, linearity of the integral and monotonicity of the Riemann integral giveThereforeThis proves the relevant case of uniform convergence preserves integrals on a compact interval.

Use the polar radius $r=at\in [0,1]$ as the parameter of the Archimedean spiral. Its angle in polar coordinates is $r/a$. The radial and angular contributions to its arc length are orthogonal, soThere is no need to evaluate this integral. The simple bounds $r\le \sqrt{a^2+r^2}\le r+a$ implyThe squeeze theorem therefore gives the scaled length of an Archimedean spiral:Geometrically, the radial contribution becomes negligible after multiplying the arc length by $a$; the remaining contribution is the area under the function graph $r\mapsto r$ on $[0,1]$.

By symmetry, the centre of mass is $G=(0,0,\ell /2)$. Choose horizontal axes through $G$ and the vertical symmetry axis as the third axis, writing $Z=z-\ell /2$. With uniform mass density $\rho$, the inertia tensor has entriesReflection in each coordinate plane makes all the mixed integrals $\int xydm$, $\int xZdm$, and $\int yZdm$ vanish. Rotational symmetry about the vertical axis gives $I_{11}=I_{22}=A$. Thus these are principal axes andThe axial moment of inertia of a hollow cylinder isIts mass is $M=\pi \rho \ell (b^2-a^2)$. Dividing and factoring the difference of squares gives

The triangle is $0\le y\le x\le 1$. In polar coordinates this becomesbecause the outer side is $x=r\cos \theta =1$. The vector field and the Jacobian determinant giveEvaluate the double integral component by component:Consequently,The origin causes no ambiguity in this double integral: the continuous extension from part (a) is bounded there, and changing a value at a single point does not change the integral.

Pair each point $\mathbf{x}$ of the square's boundary with $-\mathbf{x}$. The vector field obeys $\mathbf{f}(-\mathbf{x})=\mathbf{f}(\mathbf{x})$, whereas the outward unit normal vector changes sign: $\mathbf{n}(-\mathbf{x})=-\mathbf{n}(\mathbf{x})$. Corresponding arc length elements are equal. Thus the two contributions to the flux cancel, proving without evaluating any boundary integral thatThe same cancellation appears in the divergence, which is an odd function of $x$ over a region symmetric under $x\mapsto -x$. To justify using the planar divergence theorem despite the lack of differentiability at the origin, remove a disk of radius $\epsilon <1$. The vector field is continuously differentiable on the remaining region, its divergence has zero double integral by this symmetry, and the magnitude of the inner flux is at most $2\pi {\epsilon }^2$, since $\Vert \mathbf{f}\Vert \le \epsilon$ there. Letting $\epsilon \to 0$ recovers the result. This is flux around a removable point singularity.

For any fixed subregion $\Omega$, with no internal heat generation, conservation of energy says that the rate of stored thermal energy equals the incoming heat flux:The minus sign is needed because $\mathbf{q}\cdot \mathbf{n}>0$ denotes outgoing heat flux. Substitute Fourier's law $\mathbf{q}=-k\nabla T$ and apply the divergence theorem:Here differentiation under the integral uses a fixed region and sufficient regularity of $T$. Since the identity holds for every $\Omega$, continuity of the integrands gives the local heat equationFor spatially varying thermal conductivity, the product rule gives $\nabla \cdot (k\nabla T)=k{\nabla }^{2}T+\nabla k\cdot \nabla T$; the second term cannot generally be discarded.

First let $b>0$. Substitute $x=-y$ in the negative half of the integral. Since $g$ is an odd function,The two halves cancel, soFor $b=0$ both limits coincide. For $b<0$, reverse the limits and apply the result with $-b>0$. Thus the identity holds for every real $b$.

The real Fourier series of $f$ is writtenwhere the Fourier coefficients areBy part (a)(i), any other interval of length $2\pi$ gives the same Fourier coefficients, because each integrand is a periodic function with period $2\pi$. The notation $\sim$ associates $f$ with its Fourier series; existence of the integrals alone does not assert pointwise convergence.

On $(-\pi ,\pi )$ the function is odd, so its mean and all its cosine Fourier coefficients vanish:The assigned endpoint value does not affect these integrals. The product of the two odd functions $x|x|$ and $\sin nx$ is even, hence the sine Fourier coefficients areTwo applications of integration by parts giveThereforeThe periodic extension has a jump discontinuity at every odd multiple of $\pi$: its left and right limits are ${\pi }^2$ and $-{\pi }^2$. Its assigned value at each such point is $-{\pi }^2$. Thus it is odd away from these points, but it is not a globally odd function pointwise. The Fourier series will take the midpoint value zero at these jumps.

For a nonzero integer $k$, direct integration giveswhereas ${\int }_{-\pi }^{\pi }1dx=2\pi$. The product-to-sum formulastherefore establish Fourier orthogonality: for positive integers $m,n$,Here ${\delta }_{mn}$ is the Kronecker delta. In particular the constant function is orthogonal to every nonconstant sine and cosine in this system. Integrating the given Fourier series, then multiplying it by $\cos (mx)$ or $\sin (mx)$ and integrating, isolates its Fourier coefficients. Using the permitted interchange of the sum and integral givesIf $f$ is an even function, then $f(x)\sin (nx)$ is an odd function, while $f(x)\cos (nx)$ is an even function. ThusThe Fourier series consequently reduces to a Fourier cosine series.