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For $p,q>0$, express the product of the two Gamma integrals as the nonnegative double integralUse $r=x+y$ and $u=x/(x+y)$, with inverse $x=ru$, $y=r(1-u)$ and Jacobian determinant of absolute value $r$. The new domain of integration is $r>0$, $0<u<1$, and the integrand with this factor becomes $r^{p+q-1}{e}^{-r}{u}^{p-1}(1-u{)}^{q-1}$. Tonelli's theorem and separation of the two integrals therefore give $\Gamma (p+q)B(p,q)$. The positive real parameter hypothesis ensures that all the original integrals are finite.

For complex parameters with positive real parts, the same proof first gives the undivided identity $\Gamma (p+q)B(p,q)=\Gamma (p)\Gamma (q)$. Absolute integrability follows by taking real parts in the exponents, so Fubini's theorem applies. The nonvanishing of the Gamma function then permits division by $\Gamma (p+q)$.

The domain of integration is the set $D$ over which an integral is taken. For a multiple integral, inequalities or parameter ranges specify which points belong to $D$; a change of variables transforms both these bounds and the integration element through its Jacobian determinant. For example, a polar region $c\le r\le 1+\cos \varphi$ exists only at angles where $1+\cos \varphi \ge c$.

The set over which an integral is taken. For a double integral, boundary inequalities describe the allowed coordinate pairs, and slices of this set determine the limits in an iterated integral. The domain need not be a rectangle or connected. A change of variables transforms both the integrand and this set.

For a planar region $R$ of finite positive area $A$, its first moments of area about the coordinate axes are the displayed double integrals, provided they exist. Its centroid is $(\overline{x},\overline{y})=(M_y/A,M_x/A)$. This follows by applying the definition $\overline{\mathbf{r}}=A^{-1}{\iint }_R\mathbf{r}dA$ to each component. The subscript indicates the axis about which the moment is taken: the first moment about the $x$-axis weights by $y$, not $x$.

Green’s theorem converts circulation around a positively oriented planar boundary into the area integral of scalar curl.

Let $D$ be a bounded region in the Euclidean plane whose boundary is a finite union of disjoint, piecewise smooth simple closed curves. Suppose the vector field $F=(P,Q)$ is continuously differentiable on an open set containing the whole closed region. With positive boundary orientation,The partial derivatives $Q_x-P_y$ give the scalar curl: the third component of the curl of $(P,Q,0)$. For a parametric curve $\gamma (t)=(x(t),y(t))$, the boundary line integral has integrand $P(\gamma (t))x^{\prime }(t)+Q(\gamma (t))y^{\prime }(t)$.^[ref]

Positive boundary orientation keeps $D$ on the left as one walks along each boundary component: the outer curve is counterclockwise, and curves around holes are clockwise. Every component belongs in the line integral. The vector field must be regular throughout $D$, but need not extend into a hole.^[ref]

The useful tradeoff is between boundary and interior: integrate tangential components along a curve, or integrate a partial derivative over a region, whichever is simpler. The boundary formula for planar area uses this freedom in the opposite direction, recovering area from a line integral. As a planar case of Stokes theorem, Green theorem also explains why a local curl can determine a complete boundary measurement while leaving room for the global obstruction in Green theorem with holes.

A short proof starts with a rectangle $D=[a,b]\times [c,d]$. Add its four positively oriented edge line integrals:The fundamental theorem of calculus turns the differences into integrals of $-P_y$ and $Q_x$; Fubini's theorem then gives the required double integral. On subdividing a region, adjacent edge line integrals cancel because their directions oppose. For curved boundaries, the proof uses pieces bounded by graphs and double integrals; cancellation works the same way. This is the geometric reason only the region's complete boundary survives.^[ref]

Take the vector field $F=(-2y,3x)$. Its scalar curl is the constant $3-(-2)=5$, so Green theorem predicts that its positively oriented boundary line integral is five times the enclosed area. This one field makes both a round and a rectangular calculation transparent.

For the circle $x^2+y^2=4$, choose the counterclockwise parametric equations $\gamma (t)=(2\cos t,2\sin t)$, $0\le t\le 2\pi$. Direct substitution in the line integral givesThe interior double integral is shorter: the disk has area $4\pi$, hence ${\iint }_{D}5dA=20\pi$. Reversing the boundary orientation gives $-20\pi$.

For the rectangle $D=[0,3]\times [0,2]$, traverse its bottom, right, top, and left edges in that order. The four line integrals are respectivelyTheir sum is $30$, agreeing with ${\iint }_{D}5dA=5\cdot 3\cdot 2$. In both examples the path geometry changes the separate contributions, while constant curl makes only area matter to the final answer.

Suppose a solid is above the elliptic paraboloid $z=x^2+y^2$ and below a plane whose vertical separation from it is $A-|\mathbf{x}-\mathbf{p}{|}^2$, where $A>0$. A second lower plane exceeds the elliptic paraboloid by $B-|\mathbf{x}-\mathbf{q}{|}^2$, with $B>0$. If $|\mathbf{p}-\mathbf{q}|+\sqrt{B}\le \sqrt{A}$, the disk where the lower plane matters lies within the upper projection. Raising the bottom to this lower plane therefore removes exactly its excess volume, without changing the outer projection.

Translate polar coordinates separately to the two centres. The double integral over a disk of squared radius $A$ isSubtracting the analogous integral with $B$ proves the displayed volume formula. The containment condition is essential: if the two disks overlap only partly, subtract only over their intersection.

The proposed change of variables gives $u=x-3y$, $v=2x+y$, with inverse $x=(u+3v)/7$, $y=(v-2u)/7$. The Jacobian determinant isThe two lines of slope $-2$ become $v=1,4$, while the two lines of slope $1/3$ become $u=0,7$. Thus the parallelogram becomes $0\le u\le 7$, $1\le v\le 4$. Since $v$ stays positive, the transformed integrand has no singularity there. The double integral is

For a change of variables in a double integral, first solve for $x=x(u,v)$ and $y=y(u,v)$ and transform every boundary inequality to identify the new integration region $S$. Compute the Jacobian determinant, replace the integrand by its expression in $u,v$, and useA sufficient standard hypothesis is a continuously differentiable one-to-one transformation with nonzero Jacobian determinant in the interior, with harmless measure-zero boundaries. If one coordinate chart does not cover the region injectively, split into suitable pieces rather than silently counting points multiple times. The absolute value is essential for an ordinary area integral.

In the positive quadrant, the inverse transformation is single-valued:The two rays become $1\le u\le 2$ and the two hyperbolas become $1\le v\le 3$, so the curved original region becomes a rectangle. These are the middle two panels of the figure in part 2i. The forward Jacobian determinant isSince $x^{3}y=v^2/u$, the required double integral isThe signs of $x,y$ and $u>0$ justify the chosen square-root branch and the simplified absolute Jacobian determinant.

Interpret the cube root as the real cube root. The curved boundary is $y=-x^3$, so the bounded region is $0\le x\le 3$, $-x^3\le y\le 0$. In particular $y$ is negative inside it. The double integral is the convergent improper integralFor negative real $y$, an antiderivative is $(3/2)y^{2/3}=(3/2)|y{|}^{2/3}$. The inner integral is therefore $-3x^2/2$. HenceThe negative sign is essential. Absolute convergence follows by the same calculation with the sign removed, so the singular edge at $y=0$ causes no ambiguity.

SetBecause the region lies in the positive quadrant, its four inequalities becomeThus this change of variables sends $D$ to a rectangle in the $uv$-plane. Its Jacobian determinant issoSince $x^2=uv$, the double integral is

The original double integral covers the triangle $0\le y\le x\le 1$. By Fubini's theorem, after reversing the order, $y$ runs from $0$ to $1$ and $x$ from $y$ to $1$. Since a continuous function is integrable on this bounded region,For the square pyramid, a horizontal section at height $z$ is a square of side $2(1-z)$: each vertex moves linearly from the base towards $(1,1,1)$. Hence its cross-sectional area is $4(1-z{)}^2$, and its volume is $4{\int }_0^1(1-z{)}^{2}dz$. To use the identity just proved, choose $f(y)=1-y$:

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

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.

The region of integration is $0\le y\le 1$, $y\le x\le 2-y$. It is the closed triangle with vertices $(0,0)$, $(2,0)$, and $(1,1)$, bounded by $y=0$, $x=y$, and $x=2-y$. Each horizontal slice runs from the rising left edge to the falling right edge.

Integrate with respect to $x$ first:The triangle has area one, but the nonconstant integrand makes the double integral a weighted area.

Let $I={\int }_0^{\infty }{e}^{-x^2}dx$. Positivity allows the squared integral to be written as a first-quadrant double integral. In polar coordinates the Jacobian determinant is $r$, soUsing integration by parts, with the boundary term $x{e}^{-x^2}$ vanishing at both endpoints,The substitution $u=x^2$ also gives the two antiderivativesThese values also follow from the Gaussian moment recurrence and can be checked by differentiation.

The upper boundary follows $y=x$ until its intersection with $y=1/x$ at $(1,1)$, then follows the hyperbola to $(2,1/2)$. The lower boundary is $y=0$. Therefore the two vertical-slice descriptions are $0\le x\le 1$, $0\le y\le x$, and $1\le x\le 2$, $0\le y\le 1/x$.

Integrating with respect to $y$ first reduces the double integral toUsing the antiderivatives from part (a),

The region lies in the first quadrant, inside the unit circle and above the positive branch of $xy=1/4$. In polar coordinates, $xy=(r^2/2)\sin 2\theta$, so its radial bounds are $(2\sin 2\theta {)}^{-1/2}\le r\le 1$. Such an interval exists when $\sin 2\theta \ge 1/2$, henceAt the intersections with the circle, $r=1$ and $\sin 2\theta =1/2$, giving exactly the two stated polar angles. Their Cartesian coordinates are $(\cos (\pi /12),\sin (\pi /12))$ and the swapped pair.

The Jacobian determinant contributes $dxdy=rdrd\theta$, while $xy(x^2+y^2)=r^4\sin 2\theta /2$. Thus the double integral isIts antiderivative is $-\cos 2\theta /24+\cot 2\theta /192$. Evaluating the bounds gives

In first-quadrant polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the Jacobian determinant contributes $r$. The integrand and area element together become $r^2{e}^{-r^2}\sin \theta {\cos }^2\theta drd\theta$. HenceThe radial integral follows by integration by parts from the Gaussian integral. The isolated undefined ratio at the origin does not affect this convergent double integral.