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Use cylindrical coordinates $x=r\cos \theta$, $y=r\sin \theta$. The upper plane gives $z=4-r(\cos \theta +\sin \theta )>0$ throughout the quarter disk, and the Jacobian determinant contributes $r$. Thereforeso

Multiply the two defining Gamma integrals. For $p,q>0$ the nonnegative product integrand allows Tonelli's theorem:Use $r=x+y$, $u=x/(x+y)$, so $x=ru$, $y=r(1-u)$ and the positive quadrant becomes $r>0$, $0<u<1$. The Jacobian determinant has absolute valueAfter including this factor, the transformed integral separates:This proves the beta-gamma identity,All changes of variables are made inside the open positive domains; their boundaries do not contribute to the integrals.

The liquid occupies $0\le z\le h$ and $0\le r\le \sqrt{az}$. A horizontal slice is a disk of area $\pi az$. Integrating these areas givesEquivalently, the triple integral is ${\int }_0^{2\pi }{\int }_0^h{\int }_0^{\sqrt{az}}rdrdzd\theta$, including the cylindrical Jacobian determinant factor $r$.

Substitute $x=r\cos \varphi$, $y=r\sin \varphi$ and include the Jacobian determinant factor $r$. The integrand times the area element becomesBoth terms containing one factor $\sin \varphi$ are odd in $\varphi$, while the radial bounds and angular interval are even. They therefore integrate to zero. The remaining double integral isThe domain stays away from $r=0$, so the denominator of the original integrand causes no singularity in this region.

Write $f(x,y)=F(u,v)$ in the proposed local coordinates. The multivariable chain rule givesConsequentlyAway from the origin the Jacobian determinant is $u_x{v}_y-u_y{v}_x=4(x^2+y^2)>0$, so the inverse function theorem justifies these local coordinates. Therefore $F_v=0$ and the local general solution isSubstitution verifies the result: $f_x=2x{\Phi }^{\prime }$ and $f_y=-2y{\Phi }^{\prime }$, whose weighted sum vanishes.

There is a global qualification. The coordinate map is two-to-one, and a level set of $x^2-y^2$ has separate characteristic curve branches. A single profile on the entire plane does not follow without a branch-identification assumption. For example, take $f=\exp [-1/(x^2-y^2{)}^2]$ on the wedge $x>0$, $x^2>y^2$, and zero elsewhere. This is smooth, satisfies the equation, and has different values at $(1,0)$ and $(-1,0)$ despite equal $u$. The characteristic components of a hyperbolic first-order equation explain this obstruction. At the origin the coordinate map is singular and the original equation reads $0=0$; the boxed answer is the intended local form on a regular branch, or a global form under an additional identification of branches.

In spherical coordinates, $\theta$ is the polar angle from the positive $z$ axis and $\varphi$ the azimuthal angle. The Jacobian determinant of the coordinate transformation gives the volume element

Scale the ellipsoid to a unit ball usingwith $0\le r\le 1$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$. This is spherical coordinates followed by a diagonal linear transformation, so the Jacobian determinant is $abc{r}^2\sin \theta$. The required volume integral is

For polar coordinates, $x=r\cos \theta$ and $y=r\sin \theta$. The Jacobian determinant is $r$, so the area element isThis accounts for the increasing length of an angular strip as its distance from the origin increases.

The volume element is the absolute Jacobian determinant times the coordinate differentials. In the Cartesian coordinate system, cylindrical coordinates $(r,\theta ,z)$ and spherical coordinates $(\rho ,\vartheta ,\phi )$, respectively,Here $\rho$ is distance from the origin, $\vartheta$ is the polar angle from the positive $z$ axis, and $\phi$ is the azimuthal angle. The spherical coordinates convention fixes which angle enters the sine factor.

Use cylindrical coordinates $x=r\cos \theta$, $y=r\sin \theta$. The Jacobian determinant givesThe inequalities imply $-d<z<0$ and $0\le r<a\sqrt{1+z/d}$. Thus the volume isEquivalently, integrating vertically first gives $V={\int }_0^{2\pi }{\int }_0^a{\int }_{d(r^2/a^2-1)}^{0}rdzdrd\theta$. The region is bounded by an elliptic paraboloid and its top disc; its volume is half that of the cylinder with base radius $a$ and height $d$. Including or excluding the boundary surfaces does not change the volume.

The polar-coordinate Jacobian matrix and its Jacobian determinant areFor $r>0$, the corresponding area element is $dxdy=rdrd\theta$. The determinant vanishes at $r=0$, where the polar angle does not give a unique coordinate representation.

The PDF diagram includes a solid base of thickness $d$, occupying $-d\le z\le 0$ and $0\le r\le R$. Above it, $0\le z\le h$, the metal lies between the two specified curved surfaces. In cylindrical coordinates, the Jacobian determinant contributes a factor $r$, so the mass isThe base contributes $\pi \rho R^{2}d$. The wall contributesConsequentlyThe base thickness $d$ comes from the diagram; omitting it would omit part of the cup's volume.

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.

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

In Cartesian coordinates a rank-$r$ tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeysfor $Q\in SO(3)$, proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant $-1$ its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.

Contract the determinant identity with ${ϵ}_{ijk}$. Since ${ϵ}_{ijk}{ϵ}_{ijk}=3!=6$, one obtainsEvery index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently, $A^{\prime }=QA{Q}^T$ gives $\det A^{\prime }=(\det Q{)}^2\det A=\det A$.

For the matrix inverse, defineThis is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with $A_{ib}$ and use the determinant identity again:Here the contraction of two Levi-Civita symbols gives ${ϵ}_{ajk}{ϵ}_{ijk}=2{\delta }_{ai}$. Therefore $A{C}^T=(\det A)I$ and, when $\det A\ne 0$,These are the Levi-Civita formulas for a three-dimensional determinant and inverse.

Now differentiate the cubic determinant formula with respect to the individual entry $A_{ab}$. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. ThusSubstituting the inverse expression proves the entrywise derivative of the determinant:The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.

For the time-dependent Jacobian matrix, the stated smooth expansion impliesApplying the chain rule and the entrywise determinant derivative gives Jacobi's formula near $t=0$:At zero, $C=I$, so ${{\partial }_t\det A|}_0={\sum }_a{\partial }_a{u}_a=\nabla \cdot u$. A Taylor expansion therefore yieldsThe Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.

The coordinate tangent vectors are the partial derivatives of the position vector:Their pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, are $1,r,1+r\sin \alpha$. Since $R<1$, the last is positive throughout the body. Thus these circular torus coordinates are orthogonal wherever $r>0$; at $r=0$ the cross-section angle degenerates, as in ordinary polar coordinates.

The ordered Jacobian determinant isIts sign is positive for $r>0$, so the volume element is $dV=r(1+r\sin \alpha )drd\alpha d\beta$. The coordinate degeneracy on the central circle has zero volume and does not obstruct volume integration.

Only the quartic term contributes to the Laplacian. The mixed cubic has zero second derivative in each coordinate, while the exponential–trigonometric pairs are harmonic:Using $z=r\cos \alpha$ and the Jacobian determinant from part (a), the divergence theorem givesThe term containing $\sin \alpha {\cos }^2\alpha$ integrates to zero. Equivalently, the second vertical moment of a solid ring torus with major radius one is ${\int }_V{z}^{2}dV={\pi }^2{R}^4/2$.

Differentiate the position vector, keeping the other coordinates fixed in each derivative:The first two have zero dot product, since their two planar contributions cancel, and both are orthogonal to ${\mathbf{h}}_z$. Hence these are orthogonal coordinates away from coordinate degeneracies. Their scale factors, which are the lengths of these coordinate basis vectors, areFor example $|{\mathbf{h}}_u{|}^2=a^2[{\sinh }^{2}u{\cos }^{2}v+{\cosh }^{2}u{\sin }^{2}v]=a^2({\sinh }^{2}u+{\sin }^{2}v)$. The unit basis vectors are ${\mathbf{h}}_i/h_i$ when $h_i\ne 0$; the given coordinate basis vectors themselves are not normalized. The Jacobian determinant is $a^2({\sinh }^{2}u+{\sin }^{2}v)>0$ except on the focal lines. These are elliptic cylindrical coordinates.