Use cylindrical coordinates $x=r\cos \theta$, $y=r\sin \theta$. The upper plane gives $z=4-r(\cos \theta +\sin \theta )>0$ throughout the quarter disk, and the Jacobian determinant contributes $r$. Thereforeso

The domain of integration is the upper half of the ellipse $x^2+4y^2\le 1$. Reversing the order makes $-1\le x\le 1$ and $0\le y\le \frac{1}{2}\sqrt{1-x^2}$. ThusThe inner antiderivative is $y^2/2$, so the square of the upper limit removes the square root.

The integrand is continuous on the compact rectangle, so Fubini's theorem allows integrating in $y$ first. For $x\ne 0$, ${\int }_0^{1}x{e}^{xy}dy=e^x-1$; the same formula holds by continuity at $x=0$. Hence

Under a uniform shuffle, every unordered five-card hand is equally likely, giving $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$ possible hands. Choose the triple's rank in 13 ways and its suits in $\left(\genfrac{}{}{0ex}{}{4}{3}\right)$ ways; then choose a different pair rank in 12 ways and its suits in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ ways. Each full house determines these choices uniquely, so there is no further ordering factor. ThereforeThe count uses binomial coefficients because the order of cards within a hand is irrelevant.

Take $\sigma >0$ and substitute $u=(x-\mu )/\sigma$. The density becomes the standard Gaussian distribution density $\phi (u)=e^{-u^2/2}/\sqrt{2\pi }$. Its normalization follows from the supplied Gaussian integral after scaling by $\sqrt{2}$. Since $u\phi (u)$ is an integrable odd function, its integral is zero, and therefore $\mathbb{E}X=\mu$.

For the second moment, ${\phi }^{\prime }=-u\phi$. Integration by parts givesThe endpoint term vanishes by Gaussian decay. Consequently

For a continuous random variable with probability density function $p$, its expected value and variance areThe mean requires an integrable first absolute moment, and a finite variance requires a finite second moment. When these hold, $\mathrm{Var}X=\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$.

From the preceding gradient, ${\partial }_x^2(r^{-1})=-r^{-3}+3x^2{r}^{-5}$. Summing the three Cartesian second partial derivatives givesThus the reciprocal radius is harmonic away from the origin. Its negative gradient isThe curl vanishes because $E$ is a gradient field and mixed second derivatives commute. The exclusion of the origin is essential: the potential and field are singular there.

For constant $m$, the curl of a cross product identity reduces toCommuting mixed second partial derivatives gives $(m\cdot \nabla )\nabla \varphi =\nabla (m\cdot \nabla \varphi )$. Hence the requested sum equals $m\Delta \varphi$, which is zero because $\varphi$ is harmonic. The argument assumes the indicated second derivatives are continuous on the domain.

Again the function is even, so all sine coefficients vanish. Its constant term is $a_0/2=\sinh 1$. With $k=n\pi$, an antiderivative isThus $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. In the requested factored notation,This is the period-two specialization of the Fourier series of periodic hyperbolic sine and cosine. The function values match at the endpoints, although its derivative has a jump in the periodic extension.

Apply Parseval's identity in the same coefficient convention:The left-hand side is ${\int }_{-1}^1{x}^{4}dx=2/5$. Thus $2/5=2/9+(16/{\pi }^4){\sum }_{n\ge 1}{n}^{-4}$, giving

Use the real Fourier series convention $g=a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$ for period two. Since $g$ is an even function, $b_n=0$, and $a_0={\int }_{-1}^1{x}^{2}dx=2/3$. Twice applying integration by parts, with $k=n\pi$, givesHenceThe periodic extension is continuous at its endpoints, so the series takes the specified endpoint values as well as the interior values.

The coefficient matrix is a Vandermonde matrix, with determinantTherefore a nonzero solution exists exactly whenFor example if $a=b$, the first two columns are equal and $(1,-1,0)$ is a solution. If exactly two parameters agree, the distinct remaining column is independent of the repeated one, so the kernel has dimension one. If all three agree, only $x+y+z=0$ remains, giving a two-dimensional kernel.