The dot product is $1\cdot 0+2(-1)+(-1)\cdot 1=-3$, while the Euclidean norms are $\sqrt{6}$ and $\sqrt{2}$. Therefore the angle $\theta \in [0,\pi ]$ satisfies

The modulus is the Euclidean norm, obtained by taking the square root of the dot product of the vector with itself:

In three dimensions each contracted Kronecker delta has value three, while $b_k{b}_k=|\mathbf{b}{|}^2$. Thus

Contracting the Kronecker delta identifies $i=k$. The remaining two independent sums are dot products, so the Einstein summation convention givesThe index expression here is verified against the PDF because the OCR corrupts it completely.

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 $2\pi$.