Joy of Mathematics

Is it a conservative field?

January 21, 2017January 21, 2017 Tolaso

(i) Let $\mathbb{D} \subset \mathbb{R}^2$ be the unit disk and $\partial \mathbb{D}$ be its positive oriented boundary. Evaluate the line integral

$\displaystyle \mathcal{J} = \ointctrclockwise \limits_{\partial \mathbb{D}} (x-y^3, x^3-y^2)\, {\rm d}(x, y)$

(ii) Can you deduce if the function

$f(x, y) =(x-y^3, x^3-y^2)$

is a conservative field using the above question?

Solution

(i) We are invoking Green’s theorem. Thus:

$\begin{aligned} \ointctrclockwise \limits_{\partial \mathbb{D}} \left ( x-y^3, x^3-y^2 \right ) {\rm d}\left ( x, y \right ) &=\iint \limits_{\mathbb{D}} \left ( \frac{\partial }{\partial x}Q(x, y) - \frac{\partial }{\partial y} P(x, y)\right )\, {\rm d}(x, y) \\ &= \iint \limits_{\mathbb{D}} \left ( 3x^2 + 3y^2 \right )\, {\rm d}(x, y)\\ &= 3\int_{0}^{1}\int_{0}^{2\pi} \, {\rm d} \theta \, {\rm d} \rho\\ &= 6\pi \neq 0 \end{aligned}$

(ii) If $f$ was a conservative field then then the line integral over all closed curves would have to be $0$ . But in the previous question we found one closed curve whose line integral is not $0$ . Thus $f$ is not a conservative field.

The exercise can also be found in the Jom Forum here.

The volumes are equal

January 18, 2017January 18, 2017 Tolaso

Prove that for every constant $c>0$ the set

$\mathcal{B}_{f, g} = \{ (x, y, z) \in \mathbb{R}^3 : (x-f(z))^2 + (y-g(z))^2 \leq c, \quad z \in [a, b] \}$

has the same volume for all continuous functions $f, g: [a, b] \rightarrow \mathbb{R}$ .

Solution

For every $z_0 \in [a, b]$ on the plane $z=z_0$ the set

$\{{(x,y,z_0)\in\mathbb{R}^3\,|\,(x-f(z_0))^2+(y-g(z_0))^2\leqslant c} \}\,, \quad c>0$

is a disk of constant radius $\sqrt{c}$ . Thus the set

$\mathcal{B}_{f,g}=\{{(x,y,z)\in\mathbb{R}^3\,|\,(x-f(z))^2+(y-g(z))^2\leqslant c\,,\;a\leqslant z\leqslant b}\}$

is a “cylinder” which axis is the curve

$\big\{{(f(z),g(z),z)\in\mathbb{R}\;|\;z\in[a,b]}\big\}$

and its radius is $\sqrt{c}$ .More specifically , the set $\mathcal{B}_{f, g}$ is bounded by the planes $z=a$ and $z=b$ and for every $t \in [a, b]$ the intersection of $\mathcal{B}_{f, g}$ with the plane $z=t$ is the disk

$\mathcal{D}_t=\big\{{(x,y,z)\in\mathbb{R}^3\,|\,(x-f(t))^2+(y-g(t))^2\leqslant c\,,\;z=t}\big\}$

The area of this disk is the same with the disk

$\mathcal{D}=\big\{{(x,y,z)\in\mathbb{R}^3\,|\,x^2+y^2\leqslant c\,,\;z=0}\big\}$

The latter one has an area of

$\begin{align*} \oiint \limits_{\mathcal{D}} \left \| \bar{N} \right \| \, {\rm d} \sigma &= \oiint \limits_{\mathcal{D}} \left \| \bar{e}_3 \right \|\, {\rm d}\sigma \\ &= \int_{-\sqrt{c}}^{\sqrt{c}} \int_{-\sqrt{c-x^2}}^{\sqrt{c-x^2}} \, {\rm d} (y, x)\\ &=\pi c \end{align*}$

It follows from Cavalieri’s Principal that $\mathcal{B}_{f, g}$ has the same volume and that is equal to

$\displaystyle \mathcal{V}\left ( \mathcal{B}_{f, g} \right ) = \int_{a}^{b} \pi c \, {\rm d}t = \pi c (b-a)$

which is the same for all continuous functions $f, g:[a, b] \rightarrow \mathbb{R}$ .

A somewhat visualization would be the following:

Rendered by QuickLaTeX.com

This was an exam’s question somewhere in Greece. The answer was migrated from mathematica.gr .

Work along an oriented field

January 18, 2017 Tolaso

Prove that the work

$\displaystyle \mathcal{W}=- \oint \limits_{\gamma} \frac{(x, y, z)}{\left ( x^2+y^2+z^2 \right )^{3/2}} \cdot \, {\rm d}(x, y, z)$

produced along a $\mathcal{C}^1$ oriented curve $\gamma$ of $\mathbb{R}^3 \setminus \{(0, 0, 0) \}$ depends only on the distances of starting and ending point of $\gamma$ about the origin.

Solution

Let us consider the function $f:\mathbb{R}^3 \setminus \{0 \} \rightarrow \mathbb{R}$ such that

$\displaystyle f\left ( x,y,z \right ) = - \frac{1}{\sqrt{x^2+y^2+z^2}}$

which is continuously differentiable in $\mathbb{R}^3 \setminus \{0\}$ and it holds that:

$\displaystyle {\rm grad }f\left ( x, y, z \right ) = \frac{(x, y, z)}{\left ( x^2+y^2+z^2 \right )^{3/2}}$

Hence the vector field $\displaystyle \frac{(x,y,z)}{( x^2+y^2+z^2)^{3/2}}$ is a conservative one. This , in return, means that $\mathcal{W}$ is actually independent of the road we choose , meaning that it only depends on distances of starting and ending point of $\gamma$ about the origin.

This was an exam’s question somewhere in Greece. The answer to this question was migrated from the Greek team of mathimatikoi.org forum.

Curves and line integrals

January 18, 2017 Tolaso

Let $\gamma$ be defined as

$\gamma(t) = e^{-t} (\cos t, \sin t )\quad , \; t \geq 0$

(i) Sketch the graph of $\gamma$ .

(ii) Evaluate the line integrals:

$\begin{matrix} & \displaystyle ({\rm i})\; \oint \limits_{\gamma}\left ( x^2 +y^2 \right )\, {\rm d}s & & ({\rm ii}) \displaystyle \oint \limits_{\gamma} (-y, x)\cdot {\rm d}(x, y) \end{matrix}$

Solution

(i) The graph of the curve $\gamma$ is depicted below:

Rendered by QuickLaTeX.com

(ii)

(i) For the first integral we have successively:

$\begin{align*} \oint \limits_{\gamma} \left(x^2+y^2\right)\,\mathrm{d}s&=\int_{0}^{\infty}\left(x^2(t)+y^2(t)\right)\,\||\gamma^\prime(t)||\,\mathrm{d}t\\ &=\int_{0}^{\infty}e^{-2\,t}\,\sqrt{2}\,e^{-t}\,\mathrm{d}t\\ &=\sqrt{2}\,\int_{0}^{\infty}e^{-3\,t}\,\mathrm{d}t\\ &=\left[-\dfrac{\sqrt{2}}{3}\,e^{-3\,t}\right]_{0}^{\infty}\\ &=\dfrac{\sqrt{2}}{3} \end{align*}$

(ii) For the second integral we have successively:

$\begin{align*} \oint \limits_{\gamma} \left(-y,x\right)\cdot d(x,y)&=\int_{0}^{\infty}\left(-y(t),x(t)\right)\cdot \gamma^\prime(t)\,\mathrm{d}t\\ &=\int_{0}^{\infty}-e^{-t}\cdot (-e^{-t})\,\mathrm{d}t\\ &=\int_{0}^{\infty}e^{-2\,t}\,\mathrm{d}t\\ &=\left[-\dfrac{e^{-2\,t}}{2}\right]_{0}^{\infty}\\ &=\dfrac{1}{2} \end{align*}$

This exercise was an exam’s question somewhere in Greece.