At height $z$, the circular sector has radius $r(z)=R{e}^{-z/H}$ and angle $\Theta (z)=2\pi -\pi z/H$. Its area is $\Theta (z)r(z{)}^2/2$, so slicing by horizontal cross-sections gives the volume integralUse the substitution rule $t=z/H$. ThenAn integration by parts gives the antiderivative $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore

The denominator requires $z\ne 0$. Write $z^2=u+iv$, where $u=x^2-y^2$ and $v=2xy$. Multiplication by the nonzero denominator givesso $u=0$. Conversely, if $u=0$ and $z\ne 0$, then $v\ne 0$ and $v/(iv)=-i$. Thus the required Argand diagram consists exactly ofThese are two perpendicular straight lines, with the origin removed as shown in the figure in part (i).

Since $z^2=(x^2-y^2)+2ixy$, equating its real part and imaginary part gives $x^2-y^2=2xy$, orThe quadratic equation in $y$ factors as $[y-(-1+\sqrt{2})x][y-(-1-\sqrt{2})x]=0$. Thus the set is the union of two straight lines through the origin:Their slopes multiply to $-1$, so they are perpendicular. Their angles to the real axis are $\pi /8$ and $-3\pi /8$, modulo $\pi$. The origin is included.

Multiply numerator and denominator by the complex conjugate $4-3i$:Thus the real part is $\boxed{6/25}$ and the imaginary part is $\boxed{-17/25}$.

Direct multiplication, using $i^2=-1$, givesThus the real part is $\boxed{18}$ and the imaginary part is $\boxed{1}$.

The standard Fourier series iswith Fourier coefficientsThese follow by taking inner products with the individual basis vectors and using their orthogonality and squared norms.

With inner product $⟨u,v⟩={\int }_{-\pi }^{\pi }u(x)v(x)dx$, orthogonality meansThe constant function has squared norm $2\pi$, and each sine or cosine has squared norm $\pi$. Thus the displayed family is orthogonal but not normalized; these different lengths determine the coefficient factors.

A standard complete orthogonal family is the real Fourier basisCompleteness gives expansion in the L2 space for a square-integrable function. Mere integrability alone does not guarantee pointwise convergence of a Fourier series; the explicit function in part (b) is continuous and piecewise smooth, so this issue does not obstruct its expansion.

For $a\ne 1,5/2$, the matrix is invertible, so there is exactly one solution. To exhibit it, subtract the first linear equation from the third to obtain $(a-1)x=2$, then subtract twice the second from the first to obtain $-x+(5-2a)z=-3$. Back-substitution givesAt $a=1$, the first and third linear equations have identical left-hand sides but right-hand sides one and three, so there is no solution.

At $a=5/2$, subtracting the first from the third forces $x=4/3$, while the first minus twice the second forces $x=3$. These contradict each other, so there is again no solution. Thus there is one solution for $a\ne 1,5/2$, no solution for either exceptional value, and no parameter value with infinitely many solutions.

Subtract the first row from the third. Expansion along the resulting row $(a-1,0,0)$ gives the determinantHence

The matrix form is

For $k=1$, the vector field is $(x^{2}y,x{y}^2,z{)}^T$. On the cone,The first term has zero integral by symmetry. Using polar coordinates for the remaining double integral givesOn the paraboloid $z={\rho }^2-1$,Again the $xy$ term cancels by symmetry, givingTherefore direct surface integral yieldsNo separate disk at $z=0$ is included: that rim joins the two surfaces and is not an additional face of the closed boundary.

Both patches project onto the unit disk $D$ in the $xy$-plane. Their outward oriented surface elements areThe cone's apex has zero area and does not affect the surface integral. For $k=0$, the vector field is $(y,x,1{)}^T$, soThe terms involving $xy$ vanish by reflection symmetry of the disk. Thus the total outward flux isThis calculation evaluates the two surface integrals directly.

Here $C=0$ as well, but the branch through $5\pi /6$ is the other inverse-sine branch:Its derivative is $-1/\sqrt{1-x^2}$, and both the sign of $\tan y$ and the original equation agree with this choice.

Again $C=0$. Since the initial sine is nonzero, the local branch must satisfy $\sin y=x$. Selecting the inverse-sine branch through $y=\pi /6$ givesThis passes through the specified point and has $y^{\prime }=1/\sqrt{1-x^2}$, which verifies the original equation on this interval.

The initial data force $C=0$, so the implicit relation factors as $\sin y(x-\sin y)=0$. Near $(0,0)$ this permits either the constant branch or the sine branch:Both satisfy the original equation, including its degenerate value $0=0$ at the initial point. They have different slopes there, zero and one, so the initial value problem is not unique. There are no further differentiable local branches through that point: away from $x=0$ the two factors cannot switch, and joining the two different branches at zero would give unequal one-sided derivatives.

The nonconstant branch is defined as a solution for $-1<x<1$; the constant branch exists for every real $x$.

Let $N_A$ be the independent avocado count. There are three disjoint ways to have fewer than two total trees: no trees, only one avocado, or only one lemon. Their probabilities giveThe sum is generally a Poisson binomial distribution, rather than a single binomial distribution, because the two success probabilities may differ.

Take the complement of an event: fewer than two successes means zero or one. These mutually exclusive events have probabilities $(1-y{)}^5$ and $5y(1-y{)}^4$. Hence

Let $N_L$ count successful lemon seeds. By independence it has a binomial distribution with five trials and success probability $y$. Therefore

Put $q=2t-t^2$. The second coordinate is exactly $q^2$, so this is the same parabola as in part (i), with a different increasing parametrization. Since $q$ runs from zero to one and $dq=(2-2t)dt$, the line integral isEquivalently, substitute the original parameter to get ${\int }_0^{1}5(2t-t^2{)}^3(2-2t)dt$. This demonstrates that a line integral depends on the oriented path, not on its speed of traversal.