Line Integral MCQ Quiz in বাংলা - Objective Question with Answer for Line Integral - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

$\underset{c}{\oint }\left(ydx-xdy\right)$

where C is $x^2+y^2=\frac{1}{4}$ $\underset{2}{\overset{1}{\int }}gh=0$

By Green’s Theorem,

$\int \left(\varphi dx+\psi dy\right)=\int \int \left(\frac{\partial \psi }{\partial x}-\frac{\partial \varphi }{\partial y}\right)dxdy$

$\varphi =yand\psi =-x$

$\therefore \frac{\delta \varphi }{\delta y}=1and\frac{\delta \psi }{\delta x}=-1$

$I=\int \int \left(\frac{\partial \psi }{\partial x}-\frac{\partial \varphi }{\partial y}\right)dxdy=\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{0}{\overset{2\pi }{\int }}\left(-1-1\right)rd\theta dr$

$=-2\left(\underset{0}{\overset{\frac{1}{2}}{\int }}r.{\left(\theta \right)}_0^{2\pi }\right)dr$

$=-2\underset{0}{\overset{\frac{1}{2}}{\int }}r.\left(2\pi \right)dr$

$=-2\left(2\pi \right){\left(\frac{r^2}{2}\right)}_0^{\frac{1}{2}}=-2\pi \left(\frac{1}{2}\right)=\frac{-\pi }{2}$

$I=\underset{\left(0,2,1\right)}{\overset{\left(4,1,-1\right)}{\int }}\left(2zdx+2ydy+2xdz\right)$

$I=\underset{\left(0,2,1\right)}{\overset{\left(4,1,-1\right)}{\int }}\left(2zdx+2ydy+2xdz\right)$

$I=\underset{\left(0,2,1\right)}{\overset{\left(4,1,-1\right)}{\int }}\left(2d\left(xz\right)+2ydy\right)$

$=2{\left(xz\right)}_{\left(0,2,1\right)}^{\left(4,1,-1\right)}+2{\left(\frac{y^2}{2}\right)}_{\left(0,2,1\right)}^{\left(4,1,-1\right)}$

$={\left[2xz+y^2\right]}_{\left(0,2,1\right)}^{\left(4,1,-1\right)}$

$\left[2\left(4\right)\left(-1\right)+{\left(1\right)}^2\right]-{\left[2\right]}^2$

= [-8 + 1] – 4

= -7 - 4

= -11

Given:

$\overline{\mathrm{F}}={\mathrm{x}}^2\hat{\mathrm{i}}+{\mathrm{y}}^2\hat{\mathrm{j}}$

$\int \overline{\mathrm{F}}.\mathrm{d}\overline{\mathrm{r}}=\int \left({\mathrm{x}}^2\hat{\mathrm{i}}+{\mathrm{y}}^2\hat{\mathrm{j}}\right).\left(\mathrm{d}\mathrm{x}\hat{\mathrm{i}}+\mathrm{d}\mathrm{y}\hat{\mathrm{j}}\right)$

$\therefore \int \overline{\mathrm{F}}.\mathrm{d}\overline{\mathrm{r}}=\int {\mathrm{x}}^2\mathrm{d}\mathrm{x}+{\mathrm{y}}^2$

For the given region, straight-line joining (0, 0) to (1, 1)

x = y ⇒ dx = dy

$\int \overline{\mathrm{F}}.\mathrm{d}\mathrm{r}=\underset{0}{\overset{1}{\int }}{\mathrm{x}}^2\mathrm{d}\mathrm{x}+\underset{0}{\overset{1}{\int }}{\mathrm{x}}^2\mathrm{d}\mathrm{x}=\underset{0}{\overset{1}{\int }}2{\mathrm{x}}^2\mathrm{d}\mathrm{x}={2\left(\frac{{\mathrm{x}}^3}{3}\right)|}_0^1$

$\int \overline{\mathrm{F}}.\mathrm{d}\mathrm{r}=\frac{2}{3}=0.667$

$\frac{dy}{dx}=-tanx$

Length of the curve is given by:

$\begin{array}{l}=\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}\end{array}$

$\begin{array}{l}=\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\sqrt{1+{\left(-tanx\right)}^2}=\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}secxdx\\ {=\log \left|secx+tanx\right||}_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\\ =\log \left(\sqrt{2}+1\right)-\log \left(\sqrt{2}-1\right)=\log \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\\ =2\mathrm{l}\mathrm{o}\mathrm{g}\left(2.414\right)\end{array}$

$I=\underset{C}{\overset{}{\int }}\left[\left(y+z\right)dx+\left(x+z\right)dy+\left(x+y\right)dz\right]$

Arranging the terms of the integral given, we get:

$I=\underset{c}{\overset{}{\int }}\left[\left(ydx+xdy\right)+\left(zdx+zdz\right)+\left(zdy+ydz\right)\right]$

$=\underset{c}{\overset{}{\int }}d\left(xy\right)+d\left(xz\right)+d\left(yz\right)$

$={\left[xy+xz+yz\right]}_{\left(0,0,0\right)}^{\left(1,1,1\right)}$

= [(1 + 1 + 1) – (0)] = 3

Given:

∫ (x - 3y² + z) ds 

points: (0, 0, 0) and (1, 1, 1)

Analysis:

Equation of line:

 $\frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=\frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{y}}_2-{\mathrm{y}}_1}=\frac{\mathrm{z}-{\mathrm{z}}_1}{{\mathrm{z}}_2-{\mathrm{z}}_1}$

$\therefore \frac{\mathrm{x}-0}{1-0}-\frac{\mathrm{y}-0}{1-0}=\frac{\mathrm{z}-0}{1-0}=\mathrm{t}$

putting x = y = z = t, we get

∫ (t - 3t² + t) dt 

= ${\left[\frac{t^2}{2}-3\frac{t^3}{3}+\frac{t^2}{2}\right]}_0^1$

= (1 - 1) - (0 - 0) = 0

Explanation:

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{d}\mathrm{o}\mathrm{n}\mathrm{e}={\oint }_C\overline{F}\cdot d\overline{r}$

$\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{d}\mathrm{o}\mathrm{n}\mathrm{e}={\oint }_C\left(2x-y+z\right)dx+\left(x+y-z^2\right)dy+\left(3x-2y+4z\right)dz$

As z = 0, dz = 0

$\therefore W={\oint }_C\left(2x-y\right)dx+\left(x+y\right)dy$

Now,

Taking parametric equation of ellipse,

x = 5 cos θ, y = 4 sin θ

dx = -5 sin θ dθ , dy = 4 cos θ dθ

$W=\underset{0}{\overset{2\pi }{\int }}\left(10\cos \theta -4\sin \theta \right)\left(-5\sin \theta \right)d\theta +\left(5\cos \theta +4\sin \theta \right)\left(4\cos \theta \right)d\theta$

$W=\underset{0}{\overset{2\pi }{\int }}\left\{-50\sin \theta \cos \theta +20{\sin }^2\theta +20{\cos }^2\theta +16\sin \theta \cos \theta \right\}d\theta$

$W=\underset{0}{\overset{2\pi }{\int }}20\left({\sin }^2\theta +{\cos }^2\theta \right)d\theta -34\underset{0}{\overset{\pi }{\int }}\sin \theta \cos \theta d\theta$

$W=20{\left[\theta \right]}_0^{2\pi }$

First line segment: (0, 0) → (3, 4)

Second line segment: (3, 4) → (6, 0)

Now, the line integral becomes

${\int }_C\left(3dy-4dx\right)$ 

$=\underset{\left(0,0\right)}{\overset{\left(6,0\right)}{\int }}\left(-4dx+3dy\right)$ 

$={\left[\frac{-4x}{2}+3y\right]}_{\left(0,0\right)}^{\left(6,0\right)}$ 

= -4(6) + 3(0)

Concept:

${\int }_C\overline{F}\cdot d\overline{r}={\int }_C{F}_{1}dx+F_{2}dy$

${\int }_C\overline{F}\cdot d\overline{r}={\int }_C{x}^{2}dx+xydy$

For curve (line) y = x,

dy = dx

$\therefore {\int }_C{x}^{2}dx+xydy=\underset{0}{\overset{1}{\int }}{x}^{2}dx+x^{2}dx$

$=\underset{0}{\overset{1}{\int }}2x^{2}dx$

$=\frac{2}{3}{\left[x^3\right]}_0^1$

$\overrightarrow{F}=3xy\hat{i}-y^2\hat{j}$ and $\overrightarrow{R}=x\hat{i}+y\hat{j}$

$\underset{C}{\int }\overrightarrow{F}.d\overrightarrow{R}=\underset{C}{\int }\left(3xy\hat{i}-y^2\hat{j}\right).d\left(x\hat{i}+y\hat{j}\right)$

$=\underset{C}{\int }\left(3xydx-y^{2}dy\right)$

Substituting y = 2x² and x is varying from 0 to 1

dy = 4x dx

$\underset{C}{\int }\overrightarrow{F}.d\overrightarrow{R}=\underset{C}{\int }\left(3x\left(2x^2\right)dx-{\left(2x^2\right)}^2(4xdx\right)$

$=\underset{0}{\overset{1}{\int }}\left(6x^3-16x^5\right)dx$

$={\left[\frac{6}{4}{x}^4\right]}_0^1-{\left[\frac{16}{6}{x}^6\right]}_0^1$

$=\frac{6}{4}-\frac{8}{6}=\frac{18-32}{12}$

$=-\frac{7}{6}=-1.17$

Alternate Method:

$\overrightarrow{F}=3xy\hat{i}-y^2\hat{j}$ and

$\overrightarrow{R}=x\hat{i}+y\hat{j}$

$\underset{C}{\int }\overrightarrow{F}.d\overrightarrow{R}=\underset{C}{\int }\left(3xy\hat{i}-y^2\hat{j}\right).d\left(x\hat{i}+y\hat{j}\right)$

$=\underset{C}{\int }\left(3xydx-y^{2}dy\right)$

Let x = t

⇒ dx = dt

y = 2x² = 2t²

dy = 4tdt

t varies from 0 to 1.

$\underset{C}{\int }\overrightarrow{F}.d\overrightarrow{R}=\underset{C}{\int }\left(3\left(t\right)\left(2t^2\right)dt-{\left(2t^2\right)}^2(4tdt\right))$

$=\underset{0}{\overset{1}{\int }}\left(6t^3-16t^5\right)dt$

$={\left[\frac{6}{4}{t}^4\right]}_0^1-{\left[\frac{16}{6}{t}^6\right]}_0^1$

$=\frac{6}{4}-\frac{8}{6}=\frac{18-32}{12}$

$=-\frac{7}{6}=-1.17$