The cross derivatives are both $2y$, so the equation is exact. The function $x\cos x+x{y}^2$ differentiates to the given one-form. HenceDifferentiating this implicit equation checks both coefficients of the exact differential directly.

Here $M_y=2y$ whereas $N_x=0$, so the equation is not exact. The identity $M_y-N_x=N$ gives the integrating factor for a differential one-form $\mu (x)=e^x$. Multiplying makes the left-hand side the exact differential of $e^x(y^2-x+1)$. ThusFor real graph solutions, the right-hand side must be nonnegative; a smooth choice of square-root branch is made on the interval concerned. Finding the factor did not require dividing by $y$ at zero.

With $M=2x{y}^2+4$ and $N=2x^{2}y-6$, the cross derivatives agree: $M_y=N_x=4xy$. This is an exact first-order ordinary differential equation. Integrating $M$ with respect to $x$ gives a potential $x^2{y}^2+4x+g(y)$; matching $N$ gives $g^{\prime }=-6$. The implicit solution is

For $n<x<n+1$, monotonicity gives $1/x<1/n<1/(x-1)$. Integrating over that unit interval proves the strict inequalities, equivalently $\ln (1+1/n)<1/n<\ln (n/(n-1))$ for $n>1$.

Let $H_N$ be the harmonic number and define $u_N=H_N-\ln N$, $v_N=H_N-\ln (N+1)$. The same integral comparison gives $u_{N+1}-u_N=1/(N+1)-\ln (1+1/N)<0$ and $v_{N+1}-v_N=1/(N+1)-\ln (1+1/(N+1))>0$. Also $v_N<u_N$ and $u_N-v_N=\ln (1+1/N)\to 0$. The monotone convergence theorem for sequences therefore gives a common finite limit, Euler's constant $\gamma$.

The integral bounds for Euler's constant start at $v_1=1-\ln 2$ and $u_1=1$. Strict monotonicity, already at the next index, makes the limit bounds strict:Positivity is explicit because $\ln 2={\int }_1^{2}dx/x<1$. This establishes the required limit without assuming its decimal value.

For $x>0$, the kernel is smooth between the two positive endpoints. The Leibniz integral rule gives both endpoint terms and the parameter derivative:The remaining integral equals $[\cos (x^3)-\cos (x^4)]/x$, soThe substitution $u=xt$ gives the same result directly from limits $x^3,x^4$. Negative $x$ would make the original interval cross the nonintegrable $1/t$ singularity, so it is not part of this ordinary-integral domain.

Integrate the horizontal-disk expression over $\pi /2<\theta <\pi$. With $u=\cos \theta$ and $\sin \theta d\theta =-du$,Evaluating the integral of this polynomial gives

For horizontal slices of a surface of revolution in spherical coordinates, use disk area times vertical thickness. Here the cylindrical radius is $\rho =a(1-\cos \theta )\sin \theta$ and the height is $z=a(\cos \theta -{\cos }^2\theta )$. For $\theta >\pi /2$, $dz/d\theta =-a(1-2\cos \theta )\sin \theta <0$. Hence a positive angular increment givesThese are horizontal disks in the lower part of the solid. A spherical angular-wedge element $r^2\sin \theta drd\theta d\varphi$ describes a different infinitesimal slice.

In the plane $x=0$, the right half of the cross-section has coordinates $y=a(1-\cos \theta )\sin \theta$, $z=a(1-\cos \theta )\cos \theta$, $0\le \theta \le \pi$. Reflect across the $z$ axis for the other half. The cardioid has a cusp at $(y,z)=(0,0)$ for $\theta =0$, crosses $z=0$ at $(\pm a,0)$ for $\theta =\pi /2$, and reaches $(0,-2a)$ for $\theta =\pi$. The small upper lobes reach $z=a/4$; the widest section is below the equator.

Subtract the integrands before integrating: ${\mathbf{G}}_2-{\mathbf{G}}_1=(0,y^3,0)$. With the upward vector area element,This extends continuously to the lower edge, soThe individual flux integrals also converge: ${\mathbf{G}}_1\cdot d\mathbf{S}=x(3+2y)dxdy$ gives $I_1=5$, and hence $I_2=7$. Reversing the surface orientation changes both signs and leaves the requested magnitude unchanged.

Choose $f=x^2(1+y)+y^{2}z-1$. Since $f_z=y^2>0$ on the stated open domain, its upward vector area element isThe graph becomes unbounded near parts of $y=0$; the flux integrals below are understood as limits from $y>0$.

Where $f_z\ne 0$, the implicit function theorem makes the surface a local graph $z=h(x,y)$ with $h_x=-f_x/f_z$ and $h_y=-f_y/f_z$. Its upward vector area element is $(-h_x,-h_y,1)dxdy$. Therefore the vector area element of an implicit surface isDividing by $f_z$ fixes the upward orientation even if the initially chosen gradient points downward. The restriction excludes a locally vertical graph, where this particular projected-area representation cannot be used.