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For positive semiaxes $a,b$, the cross product of the two parametric tangent vectors gives the outward oriented surface element of an ellipsoid of revolution:Substituting the parametrization into the vector field and taking its dot product with this element givesThe middle term has zero integral over $0\le \varphi \le 2\pi$. In the other terms, ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$, ${\int }_0^{\pi }{\sin }^3\theta {\cos }^2\theta d\theta =4/15$, and ${\int }_0^{\pi }\sin \theta {\cos }^4\theta d\theta =2/5$. The directly evaluated flux is therefore

Use the real Fourier series convention $f(x)\sim a_0/2+{\sum }_{n\ge 1}[a_n\cos (n\pi x)+b_n\sin (n\pi x)]$ for period two. The Fourier coefficients areThe last two integrals follow by integration by parts. Thus the Fourier series of a half-wave rectified ramp isOn $(-1,1)$, $f(x)=|x|+x$. Its even part is $f_e(x)=|x|$ and its odd part is $f_o(x)=x$, so the cosine and sine terms separate:At a jump, the Fourier series converges to the average of the one-sided limits. Thus the first series for $f$ takes the value $1$ at odd integers, while the Fourier series of a periodic sawtooth for $f_o$ takes the value $0$ there. The even function has a continuous periodic extension.

At $x=0$, the formula for $f_e$ gives $0=1/2-(4/{\pi }^2){\sum }_{r\ge 0}(2r+1{)}^{-2}$. At $x=1/2$, the formula for $f_o$ has only odd-index terms and gives $1/2=(2/\pi ){\sum }_{r\ge 0}(-1{)}^r/(2r+1)$. Hence

The integrand in the preceding expression is nonnegative. Tonelli's theorem permits squaring as a quadrant integral; then polar coordinates giveThe defining integral is positive, fixing the sign of the square root. ThereforeThis is the positive-quadrant proof of the Gaussian integral.

Choose the upward orientation. The boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise from above. By Stokes theorem, the surface integral is ${\oint }_{\partial S}G\cdot dr$. On this boundary $dz=0$ and $z^2=1$, soThe first term integrates to zero around a closed curve. By Green's theorem, the second equals the disk integral ${\int }_{x^2+y^2\le 4}(-3y)dxdy=0$, by oddness in $y$. ThusHere reversing orientation leaves the zero value unchanged.

The function $x^2\sin x$ is an odd function, since $x^2$ is even and the sine is odd. Its integral on the symmetric interval therefore cancels:

Split the integrand into its three coordinate contributions. Each integrates independently over the rectangular box:For example, the last factor is ${\int }_{-3}^3{z}^{2}dz=18$. No spherical or cylindrical Jacobian is needed for this Cartesian integral.

The region is the triangle $0\le y\le x\le 1$. Reversing the integration order makes the inner integral elementary:The integrand is continuous on this compact triangle, so Fubini theorem justifies exchanging the order.

The gradient part contributes zero around a closed path by the fundamental theorem for line integrals:Along the parametrization, $x^2+z^2=1$, so the first two components of $\mathbf{w}$ vanish wherever the printed expression is defined. The remaining component gives $\mathbf{w}\cdot d\mathbf{x}=xdz={\cos }^2\theta d\theta$. ThusThe printed field needs a qualification: $\mathrm{arctanh}(x^2)$ is undefined at the isolated parameters where $x^2=1$, so $0\mathrm{arctanh}(1)$ cannot literally be evaluated. The result above is the improper line integral, or equivalently the integral of the continuous extension of the restricted field along the curve. That restricted field tends to $(0,0,\cos \theta )$ at each omitted parameter. No extension of the full field to a neighbourhood is being assumed.

Writing $r^2=x^2+y^2+z^2$, the gradient is $\mathbf{v}=2e^{r^2}(x,y,z)$. Its divergence, equivalently the Laplacian of $\varphi$, isOn the unit sphere this scalar is the constant $10e$. The lower hemisphere has surface area $2\pi$, soThis is a scalar surface integral of the divergence, not the electric- or vector-field flux integral. There is no normal component to insert.

Use the integration by substitution $x=\sin \theta$, with $-\pi /6\le \theta \le \pi /6$. On this interval $\cos \theta >0$, so it cancels the positive denominator square root. The transformed integral isIts integrand is even. Therefore the answer in the PDF's modulus-amplitude convention for the elliptic integral of the second kind isThe modulus is $2/3$, not the squared parameter $4/9$.

The integrand separates into a radial and an angular factor, soThe differential printed in the PDF is already $drd\theta$. No further polar Jacobian is to be inserted into this given iterated integral.

The projection onto the $x$-$y$ plane is the disk $x^2+y^2\le 4$. In polar coordinates, the height is $4-r^2$ and the polar area element is $rdrd\theta$. The volume is thereforeThe radial factor here is required because this integral is being constructed from the geometric volume, unlike the already specified integral in (a).

Multiply the two displayed equations by $e^{-x}$ and subtract the one with indices reversed. The product terms combine into the Lagrange identity for a Sturm-Liouville operator:Integrating from zero to infinity givesAll factors $L_j,L_j^{\prime }$ are polynomials. The endpoint term at zero vanishes because of the factor $x$, and the one at infinity vanishes because the exponential function dominates every polynomial. Therefore, if $m\ne n$,The integral is absolutely convergent. This is orthogonality in the weighted inner product with weight $e^{-x}$; the boundary cancellation is the self-adjoint mechanism behind the Laguerre differential operator on polynomials.

Apply the fundamental theorem of calculus separately to the upper and lower limits. No evaluation of the integral is needed:The first result has no $y$ dependence and the second no $x$ dependence, so the mixed partial derivative is

Set $s=\sin u$, so $ds=\cos udu$, and write $J_n=I_{n,1}(\pi /2,0)$. The integration by substitution gives $J_n={\int }_0^1(1-s^3{)}^{n}ds$. For $n>0$, integration by parts yieldsBoth endpoint terms vanish. Use $s^3=1-(1-s^3)$ inside the remaining integral:ConsequentlyThis is the $p=3$ instance of the power-complement integral recurrence, with initial value $J_0=1$.

Apply the power-complement integral recurrence three times, starting from $I_{0,1}(\pi /2,0)=1$:Each factor corresponds to reducing $n$ by one; the final unreduced integral is just ${\int }_0^{\pi /2}\cos udu=1$.

Write the kernel as $K(s,t)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$, with $t>0$. Its partial derivatives areThus the heat kernel itself satisfies $K_t=K_{ss}$. The proposed convolution consequently givesThis verifies the heat equation. The parameter is positive time; the real integral formula is not defined at negative time.

The passage of derivatives through the integral needs justification. A sufficient condition is the Gaussian-weighted absolute integrability in heat-kernel convergence for Gaussian-integrable initial data; it holds, in particular, for data of polynomial growth. On compact positive-time intervals strictly below an integrable later time, the differentiated kernels have an integrable Gaussian bound, so differentiation under the integral sign applies.

Even if the PDF's assumed convergence is read as ordinary conditional convergence, one can justify the same calculation. Choose a later time $T>t$ and let $H(\xi )$ be a primitive of $\phi (\xi )e^{-{\xi }^2/(4T)}$. Convergence at spatial point zero and time $T$ makes $H$ bounded on both tails. Put $q(x,\xi ,t)=K(x-\xi ,t)e^{{\xi }^2/(4T)}$. This function, its spatial and time derivatives, and their $\xi$-derivatives decay as Gaussians when $t<T$. Integration by parts rewrites the convolution as $-\int H{q}_{\xi }d\xi$, with vanishing endpoint terms. Boundedness of $H$ now permits differentiation under this absolutely convergent integral; since $q_t=q_{xx}$, the same equation follows. Thus one need not silently equate conditional and absolute convergence.

Insert the linear initial profile into the transformed integral from (b):The constant part integrates to $ax$. The other part is an absolutely integrable odd function, so its integral vanishes. HenceAs a direct check, $u_t=u_{xx}=0$ and $u(x,0)=ax$. Thus a linear temperature profile is stationary on the whole line. More generally, every affine function is preserved by this heat-kernel solution.

The chain rule identifies an antiderivative directly:The upper limit in the PDF is $\sqrt{\pi /2}$. By the fundamental theorem of calculus, the required integral is

The probability density function is nonnegative for $\alpha >0$. Its integral over the support isso it is normalized for every $\alpha >0$. This is the unit-threshold Pareto distribution.