Multiple, Line and Surface Integrals: Very Difficult Problems with Solutions

Solution:
The region lies in the first quadrant ([tex]x^2=3y[/tex] forces [tex]y\ge 0[/tex] and [tex]y^2=x[/tex] forces [tex]x\ge 0[/tex]) and is described by the inequalities
[tex]3\le\frac{x^2}{y}\le 5,\qquad 1\le\frac{y^2}{x}\le 2[/tex]
Introduce the new variables [tex]u=\frac{x^2}{y}[/tex], [tex]v=\frac{y^2}{x}[/tex]. Then [tex]G[/tex] corresponds to the rectangle [tex]3\le u\le 5,\ 1\le v\le 2[/tex], and the map is one-to-one in the open first quadrant, since [tex]u^2v=x^3[/tex] and [tex]uv^2=y^3[/tex]. Its Jacobian is
[tex]\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}\frac{2x}{y} & -\frac{x^2}{y^2}\\ -\frac{y^2}{x^2} & \frac{2y}{x}\end{vmatrix}=4-1=3,\qquad dx\,dy=\frac{1}{3}\,du\,dv[/tex]
Moreover, the integrand in the new variables is
[tex]xy=\frac{x^2}{y}\times \frac{y^2}{x}=uv[/tex]
Therefore
[tex]\iint_G xy\,dx\,dy=\frac{1}{3}\int_{3}^{5}u\,du\int_{1}^{2}v\,dv=\frac{1}{3}\times 8\times \frac{3}{2}=4[/tex]

Solution:
The inequality [tex]x^2+y^2\le 2y[/tex], i.e. [tex]x^2+(y-1)^2\le 1[/tex], describes the disk with center [tex](0, 1)[/tex] and radius [tex]1[/tex]; it lies in the half-plane [tex]y\ge 0[/tex] and its boundary passes through the origin. In polar coordinates [tex]x=r\cos\varphi,\ y=r\sin \varphi[/tex] (Jacobian [tex]r[/tex]) this disk is
[tex]0\le r\le 2\sin \varphi,\qquad 0\le\varphi\le\pi[/tex]
The condition [tex]y\le\sqrt{3}\,x[/tex] together with [tex]y\ge 0[/tex] forces [tex]x\ge 0[/tex] and [tex]\tan \varphi\le\sqrt{3}[/tex], i.e. [tex]0\le\varphi\le\frac{\pi}{3}[/tex]. On [tex]D[/tex] we have [tex]r^2\le 4\sin ^2\frac{\pi}{3}=3\lt 6[/tex], so the integrand is continuous. The inner integral is
[tex]\int_{0}^{2\sin \varphi}\frac{r\,dr}{\left(6-r^2\right)^2}=\left[\frac{1}{2\left(6-r^2\right)}\right]_{0}^{2\sin \varphi}=\frac{1}{2\left(6-4\sin ^2\varphi\right)}-\frac{1}{12}[/tex]
For the remaining integral substitute [tex]t=\tan \varphi[/tex], so that [tex]\sin ^2\varphi=\frac{t^2}{1+t^2}[/tex] and [tex]d\varphi=\frac{dt}{1+t^2}[/tex]:
[tex]\int_{0}^{\pi/3}\frac{d\varphi}{6-4\sin ^2\varphi}=\int_{0}^{\sqrt{3}}\frac{dt}{6+2t^2}=\frac{1}{2\sqrt{3}}\left[\arctan\frac{t}{\sqrt{3}}\right]_{0}^{\sqrt{3}}=\frac{\pi}{8\sqrt{3}}[/tex]
Hence
[tex]I=\frac{1}{2}\times \frac{\pi}{8\sqrt{3}}-\frac{1}{12}\times \frac{\pi}{3}=\frac{\pi\sqrt{3}}{48}-\frac{\pi}{36}=\frac{\pi\left(3\sqrt{3}-4\right)}{144}[/tex]

Solution:
The integrand is continuous on the rectangle [tex]\left[0, a\right]×\left[0, b\right][/tex] (as a maximum of continuous functions), so we may split the rectangle into parts and integrate over each part in any order. The diagonal [tex]y=\frac{b}{a}x[/tex] from [tex](0, 0)[/tex] to [tex](a, b)[/tex] divides it into two triangles. Below the diagonal ([tex]0\le y\le\frac{b}{a}x[/tex]) we have [tex]ay\le bx[/tex], hence [tex]a^2y^2\le b^2x^2[/tex] and the integrand is [tex]e^{b^2x^2}[/tex]; above the diagonal it is [tex]e^{a^2y^2}[/tex].
The function [tex]e^{x^2}[/tex] has no elementary antiderivative, so in each triangle we first integrate with respect to the variable that does not occur in the exponent. Lower triangle:
[tex]I_1=\int_0^a dx\int_0^{bx/a}e^{b^2x^2}\,dy=\int_0^a\frac{b}{a}\,x\,e^{b^2x^2}\,dx=\frac{b}{a}\times \frac{e^{b^2x^2}}{2b^2}\Big|_0^a=\frac{e^{a^2b^2}-1}{2ab}[/tex]
Upper triangle ([tex]0\le y\le b,\ 0\le x\le\frac{a}{b}y[/tex]):
[tex]I_2=\int_0^b dy\int_0^{ay/b}e^{a^2y^2}\,dx=\int_0^b\frac{a}{b}\,y\,e^{a^2y^2}\,dy=\frac{a}{b}\times \frac{e^{a^2y^2}}{2a^2}\Big|_0^b=\frac{e^{a^2b^2}-1}{2ab}[/tex]
Hence the integral equals
[tex]I_1+I_2=\frac{e^{a^2b^2}-1}{ab}[/tex]

Solution:
Let [tex]R=20[/tex], so [tex]|S|=4\pi R^2[/tex]. Expanding the squares,
[tex]T=x^2+2y^2+z^2+2xy-2yz[/tex]
The sphere and the area element [tex]dS[/tex] are invariant under [tex]x\mapsto -x[/tex], while [tex]xy[/tex] changes sign, so [tex]\iint_S xy\,dS=0[/tex]; in the same way (with [tex]z\mapsto -z[/tex]) [tex]\iint_S yz\,dS=0[/tex]. The sphere is also invariant under permutations of the coordinates, and [tex]x^2+y^2+z^2=R^2[/tex] on [tex]S[/tex], hence
[tex]\iint_S x^2\,dS=\iint_S y^2\,dS=\iint_S z^2\,dS=\frac{1}{3}\iint_S\left(x^2+y^2+z^2\right)dS=\frac{1}{3}R^2\times 4\pi R^2=\frac{4\pi R^4}{3}[/tex]
Therefore [tex]\iint_S T\,dS=(1+2+1)\times \frac{4\pi R^4}{3}=\frac{16\pi R^4}{3}[/tex] and the mean temperature is
[tex]\frac{1}{4\pi R^2}\times \frac{16\pi R^4}{3}=\frac{4R^2}{3}=\frac{4\times 400}{3}=\frac{1600}{3}[/tex]

Solution:
The surface [tex]S[/tex] is closed and piecewise smooth; it bounds the solid [tex]V:\ x\ge 0,\ y\ge 0,\ x^2+y^2\le 1,\ 0\le z\le x^2+y^2[/tex], and the functions [tex]P=xz[/tex], [tex]Q=x^2y[/tex], [tex]R=y^2z[/tex] have continuous partial derivatives. By the Gauss-Ostrogradsky theorem (for the outer side)
[tex]\iint_S P\,dy\,dz+Q\,dz\,dx+R\,dx\,dy=\iiint_V\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dx\,dy\,dz=\iiint_V\left(z+x^2+y^2\right)dx\,dy\,dz[/tex]
In cylindrical coordinates [tex]x=r\cos\varphi,\ y=r\sin \varphi,\ z=z[/tex] (Jacobian [tex]r[/tex]) the solid is [tex]0\le\varphi\le\frac{\pi}{2},\ 0\le r\le 1,\ 0\le z\le r^2[/tex], so
[tex]I=\int_{0}^{\pi/2}d\varphi\int_{0}^{1}r\,dr\int_{0}^{r^2}\left(z+r^2\right)dz=\frac{\pi}{2}\int_{0}^{1}r\left(\frac{r^4}{2}+r^4\right)dr=\frac{\pi}{2}\times \frac{3}{2}\times \frac{1}{6}=\frac{\pi}{8}[/tex]
Direct check: the flux through the planes [tex]x=0[/tex], [tex]y=0[/tex], [tex]z=0[/tex] is zero (there [tex]P=0[/tex], [tex]Q=0[/tex], [tex]R=0[/tex], respectively), the cylinder gives [tex]\frac{3\pi}{16}[/tex] and the paraboloid gives [tex]-\frac{\pi}{16}[/tex], in total [tex]\frac{\pi}{8}[/tex].

Solution:
The substitution [tex]t=x^2[/tex], [tex]dt=2x\,dx[/tex], [tex]x\in(0, +\infty)[/tex], reduces the problem to a single integral:
[tex]\Gamma\left(\frac{1}{2}\right)=\int_0^{+\infty}t^{-1/2}e^{-t}\,dt=\int_0^{+\infty}\frac{1}{x}\,e^{-x^2}\,2x\,dx=2I,\qquad I=\int_0^{+\infty}e^{-x^2}\,dx[/tex]
The integral [tex]I[/tex] converges, because [tex]0\lt e^{-x^2}\le e^{-x}[/tex] for [tex]x\ge 1[/tex]. So it suffices to prove that [tex]I=\frac{\sqrt{\pi}}{2}[/tex].
For [tex]A\gt 0[/tex] let [tex]Q_A=\left[0, A\right]×\left[0, A\right][/tex] be a square and [tex]K_R=\left\{x\ge 0,\ y\ge 0,\ x^2+y^2\le R^2\right\}[/tex] a quarter disk. Then [tex]K_A\subset Q_A\subset K_{A\sqrt{2}}[/tex], and since the integrand [tex]e^{-x^2-y^2}[/tex] is positive,
[tex]\iint_{K_A}e^{-x^2-y^2}dx\,dy\le\iint_{Q_A}e^{-x^2-y^2}dx\,dy\le\iint_{K_{A\sqrt{2}}}e^{-x^2-y^2}dx\,dy[/tex]
In Cartesian coordinates the integrand equals [tex]e^{-x^2}e^{-y^2}[/tex], so by Fubini's theorem the integral over the square is
[tex]\iint_{Q_A}e^{-x^2-y^2}dx\,dy=\int_0^Ae^{-x^2}dx\times \int_0^Ae^{-y^2}dy=\left(\int_0^Ae^{-x^2}dx\right)^2[/tex]
In polar coordinates [tex]x=r\cos\varphi,\ y=r\sin \varphi[/tex], [tex]dx\,dy=r\,dr\,d\varphi[/tex], the integral over the quarter disk is
[tex]\iint_{K_R}e^{-x^2-y^2}dx\,dy=\int_0^{\pi/2}d\varphi\int_0^Re^{-r^2}r\,dr=\frac{\pi}{2}\times \frac{1-e^{-R^2}}{2}=\frac{\pi}{4}\left(1-e^{-R^2}\right)[/tex]
Combining the three relations, we get
[tex]\frac{\pi}{4}\left(1-e^{-A^2}\right)\le\left(\int_0^Ae^{-x^2}dx\right)^2\le\frac{\pi}{4}\left(1-e^{-2A^2}\right)[/tex]
As [tex]A\to+\infty[/tex] both bounds tend to [tex]\frac{\pi}{4}[/tex], so [tex]I^2=\frac{\pi}{4}[/tex], and since [tex]I\gt 0[/tex], [tex]I=\frac{\sqrt{\pi}}{2}[/tex]. Thus the double integral over the first quadrant has the same value [tex]\frac{\pi}{4}=I^2[/tex] whether it is computed in Cartesian or in polar coordinates. Hence
[tex]\Gamma\left(\frac{1}{2}\right)=2I=\sqrt{\pi}[/tex]

Solution:
Let [tex]k=\sqrt{a^2+b^2+c^2}[/tex]. Both integrals exist because [tex]f[/tex] is continuous. If [tex]k=0[/tex], then [tex]a=b=c=0[/tex], the left-hand side equals [tex]4\pi f(0)[/tex] (the area of the unit sphere is [tex]4\pi[/tex]) and the right-hand side equals [tex]2\pi\times 2f(0)=4\pi f(0)[/tex].
Let [tex]k\gt 0[/tex] and [tex]\vec{e}_3=\frac{1}{k}(a, b, c)[/tex]. Complete [tex]\vec{e}_3[/tex] to an orthonormal basis [tex]\vec{e}_1,\ \vec{e}_2,\ \vec{e}_3[/tex] and introduce the new coordinates (scalar products with the position vector [tex]\vec{r}=(x, y, z)[/tex])
[tex]u=\vec{r}⋅\vec{e}_1,\qquad v=\vec{r}⋅\vec{e}_2,\qquad w=\vec{r}⋅\vec{e}_3[/tex]
This is an orthogonal change of coordinates: it preserves distances and areas, maps the unit sphere onto the unit sphere [tex]u^2+v^2+w^2=1[/tex] and therefore does not change a surface integral of the first kind. Since [tex]ax+by+cz=k\left(\vec{r}⋅\vec{e}_3\right)=kw[/tex], we get
[tex]\iint_S f(ax+by+cz)\,dS=\iint_{u^2+v^2+w^2=1}f(kw)\,dS[/tex]
Now parametrize this sphere with polar axis [tex]Ow[/tex]: [tex]u=\sin \theta\cos\varphi,\ v=\sin \theta\sin \varphi,\ w=\cos\theta[/tex], [tex]0\le\theta\le\pi,\ 0\le\varphi\le 2\pi[/tex]. For this parametrization the length of the cross product of the partial derivatives is [tex]\sin \theta[/tex], i.e. [tex]dS=\sin \theta\,d\theta\,d\varphi[/tex], and
[tex]\iint_{u^2+v^2+w^2=1}f(kw)\,dS=\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}f(k\cos\theta)\sin \theta\,d\theta=2\pi\int_{0}^{\pi}f(k\cos\theta)\sin \theta\,d\theta[/tex]
With [tex]t=\cos\theta[/tex], [tex]dt=-\sin \theta\,d\theta[/tex], where [tex]t[/tex] decreases from [tex]1[/tex] to [tex]-1[/tex], the last integral equals [tex]\int_{-1}^{1}f(kt)\,dt[/tex]. Therefore
[tex]\iint_S f(ax+by+cz)\,dS=2\pi\int_{-1}^{1}f\left(t\sqrt{a^2+b^2+c^2}\right)dt[/tex]
which proves Poisson's formula.