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The curl of a gradient is zero pointwise, so its surface integral over this sphere isFor this particular polynomial, $\nabla \Phi =2(x,y,z{)}^T$, whose curl vanishes directly as well.

Reflection in the $y$-axis preserves the disk but changes the sign of the integrand $xy$. The double integral is therefore its own negative and must beEquivalently, for each fixed $y$ the integrand is an odd function of $x$ on a symmetric interval.

Normalize the direction to the unit vector $\hat{\mathbf{u}}=(1,1,0{)}^T/\sqrt{2}$. The directional derivative is the dot product of this unit vector with the gradient:

The gradient is $(x-1,y-1,z-1{)}^T$. At the origin this gives

Here $f(1)=-1$ and $f^{\prime }(x)=1+1/x$, so $f^{\prime }(1)=2$. The Newton root-finding iteration gives

Setting the tangent line approximation equal to zero gives the Newton root-finding iterationThe step requires a nonzero derivative at the starting point.

The cross product has magnitude $|\mathbf{a}||\mathbf{b}|\sin \theta$, where $\theta$ is their angle. Therefore

The area is zero if the vectors are parallel or either is zero. The corresponding triangle has half this area.

The vector triple product identity isConsequently a choice of the requested coefficients isThese are the standard identity's coefficients, even if the particular three vectors are linearly dependent and a representation need not be unique.

For $u=XY$, the separated equations areOn a region containing the prescribed points, take $0<x<\pi$ and $0<y<\pi /2$. Integration givesThe two specified values, verified in the original PDF, giveDividing these relations eliminates $C$:The right-hand side is the square of the positive base on the left, which differs from one. Thus $\lambda =2$, and either original condition gives $C=1$. ThereforeIts partial derivatives satisfy $u_x/\sin x=2u$ and $\tan y{u}_y=2u$, verifying the equation wherever its coefficients are defined. Substitution at the two assigned points gives $1/(4e^{\sqrt{3}})$ and $3/(4e)$ exactly. The formula extends to any region where $\sin x\ne 0$ and $\cos y\ne 0$.

With $u=XY$, divide the separated equation on $x\ne 0$, $y\ne 0$ to obtainThus $X^{\prime }/X=\lambda x{e}^{-x}$ and $Y^{\prime }/Y=\lambda /y$. Taking antiderivatives,Choose $\lambda =1$ and the signed branch $Y=y$. A nonconstant solution, smooth for every real $x,y$, isIndeed $u_x=x{e}^{-x}u$ and $u_y=\exp [-(x+1)e^{-x}]$, so $e^x{u}_x-xy{u}_y=0$. This verification includes the axes excluded temporarily during separation.

Seek a separated solution $u=X(x)Y(y)$. On a region where the factors do not vanish, the partial differential equation separates asThe resulting ordinary differential equations integrate to $X=A{e}^{\lambda x}$ and $Y=B|3-y{|}^{\lambda /2}$ on either side of $y=3$. Choose $\lambda =2$ and the signed linear branch $Y=3-y$, which extends smoothly across that line. This yields the nonconstant solutionDirect partial differentiation gives $u_x=2e^{2x}(3-y)$ and $u_y=-e^{2x}$, whose prescribed combination vanishes everywhere. This direct check also covers the line on which the separated factor is zero.

For spherical coordinates, the distance from the origin is $\sqrt{26}$ and the polar angle satisfies $\cos \theta =z/r=-1/\sqrt{26}$. The azimuth is the same as in part (a). ThusThe polar angle is in $(\pi /2,\pi )$ because the point lies below the $xy$-plane.

In the cylindrical coordinate system, $r=\sqrt{x^2+y^2}=5$ and $z=-1$. The point is in the third quadrant, so a representative azimuth in $[0,2\pi )$ is $\pi +\arctan (4/3)$. ThereforeThe angle is defined modulo $2\pi$.

Assume $\alpha <\beta$ and abbreviateThe integral of the nonnegative square expands to $A+2\lambda B+{\lambda }^{2}C\ge 0$ for every real $\lambda$. If $C>0$, completing the square, or choosing $\lambda =-B/C$, gives $A-B^2/C\ge 0$. Hence the Cauchy-Schwarz inequality isIf $C=0$, continuity and nonnegativity imply $g=0$ throughout the interval, so the same inequality is immediate.

When $C>0$, equality holds exactly when $\int (f-(B/C)g{)}^{2}dx=0$. The integrand is continuous and nonnegative, so it must vanish everywhere, giving $f=(B/C)g$. Including the case of a zero function, equality is equivalent to linear dependence of $f$ and $g$ as elements of the real vector space of continuous functions on the interval.

Evaluate the recurrence boundary terms at zero and $\pi /2$. For $n>1$, both $\cos x{\sin }^{n-1}x$ and $\sin x{\cos }^{n-1}x$ vanish at both endpoints. Therefore the Wallis integrals obeyThe starting value for both is ${\int }_0^{\pi /2}1dx=\pi /2$. Thus

Write $I_n=\int {\sin }^{n}xdx$. Apply integration by parts with ${\sin }^{n-1}x$ and $\sin xdx$:Using ${\cos }^{2}x=1-{\sin }^{2}x$ on the right gives $I_n=-\cos x{\sin }^{n-1}x+(n-1)(I_{n-2}-I_n)$. Rearranging provesFor $J_n=\int {\cos }^{n}xdx$, the analogous integration by parts uses ${\cos }^{n-1}x$ and $\cos xdx$, giving $J_n=\sin x{\cos }^{n-1}x+(n-1)(J_{n-2}-J_n)$. HenceAn arbitrary constant of integration absorbs the choices in the lower-order antiderivatives.

Starting from $I_0=J_0=x$, successive use of $n=2,4,6$ givesThese expressions follow directly from the requested recurrence, rather than requiring a separate trigonometric expansion.