Multiple Integrals MCQ Quiz in मल्याळम - Objective Question with Answer for Multiple Integrals - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Area = 2 (Area in first quadrant)

$=2\left[\underset{0}{\overset{1}{\int }}\underset{-x^2+1}{\overset{8}{\int }}dydx+\underset{1}{\overset{3}{\int }}\underset{x^2-1}{\overset{8}{\int }}dydx\right]$ 

$=2\left[\underset{0}{\overset{1}{\int }}\left(7+x^2\right)dx+\underset{1}{\overset{3}{\int }}\left(9-x^2\right)dx\right]$ 

$=2\left[{\left[7x+\frac{x^3}{3}\right]}_0^1+{\left[9x-\frac{x^3}{3}\right]}_1^3\right]$ 

$=2\left[7+\frac{1}{3}+9\left(2\right)-\frac{1}{3}\left(26\right)\right]$ 

$=2\left[25-\frac{25}{3}\right]=\frac{100}{3}$ 

Concept:

Gauss divergence theorem

$\int \underset{S}{\overset{}{\int }}F\cdot nds=\underset{S}{\overset{}{\int }}div\cdot Fdv$ 

Calculation:

$\int \underset{S}{\overset{}{\int }}F\cdot nds=\underset{S}{\overset{}{\int }}div\cdot Fdv$

$=\underset{S}{\overset{}{\int }}\mathrm{\nabla }\left[\left(x+y\right)\hat{i}+\left(x+z\right)\hat{j}+\left(y+z\right)\hat{k}\right]dv$

$=\underset{S}{\overset{}{\int }}\left(1+0+1\right)dv$

$=2\int dv=2V$

$V=\frac{4}{3}\pi {\left(3\right)}^3=36\pi$

2 × 36 π = 72 π = 226.19

Concept:

Gauss divergence theorem:

It states that the surface integral of the normal component of a vector function $\overrightarrow{F}$ taken over a closed surface ‘S’ is equal to the volume integral of the divergence of that vector function $\overrightarrow{F}$ taken over a volume enclosed by the closed surface ‘S’.

$\underset{S}{\iint }\overrightarrow{F}.\hat{n}ds=\underset{V}{\iiint }\mathrm{\nabla }.\overrightarrow{F}dv$

Calculation:

Given:

F = xi + yj + zk

$\mathrm{\nabla }.\overrightarrow{F}=3$

$\frac{1}{4}\underset{S}{\iint }\overrightarrow{F}.\hat{n}ds=\frac{1}{4}\underset{V}{\iiint }\mathrm{\nabla }.\overrightarrow{F}dv$

$\frac{1}{4}\underset{S}{\iint }\overrightarrow{F}.\hat{n}ds=\frac{1}{4}\underset{V}{\iiint }\mathrm{\nabla }.\overrightarrow{F}dv=\frac{1}{4}\underset{V}{\iiint }3.dv=\frac{3}{4}\underset{V}{\iiint }dv$

$\frac{1}{4}\underset{S}{\iint }\overrightarrow{F}.\hat{n}ds=\frac{3}{4}\times volumeofthesphere$

$\frac{1}{4}\underset{S}{\iint }\overrightarrow{F}.\hat{n}ds=\frac{3}{4}\times \frac{4}{3}\times \pi \times {\left(1\right)}^3=\pi (\because radius=1)$

Stokes theorem:

It states that the line integral of a vector field $\overrightarrow{F}$ around any closed surface C is equal to the surface integral of the normal component of the curl of vector $\overrightarrow{F}$ over an unclosed surface ‘S’.

$\underset{C}{\oint }\overrightarrow{F}.\overrightarrow{dr}=\underset{S}{\iint }curl\overrightarrow{F}.\hat{n}ds$

Green's theorem:

$\oint \left(\mathrm{M}\mathrm{d}\mathrm{x}+\mathrm{N}\mathrm{d}\mathrm{y}\right)=\int \int \left(\frac{\partial \mathrm{N}}{\partial \mathrm{x}}-\frac{\partial \mathrm{M}}{\partial \mathrm{y}}\right)\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}$

Concept:

In multiple Integral Involving variables as limits, first draw the curves

y = x²

y = 2 - x

Calculation:

$\underset{0}{\stackrel{1}{\int }}x\left(\underset{y=x^2}{\stackrel{y=2-x}{\int }}ydy\right)dx$

$\underset{0}{\stackrel{1}{\int }}x{\left[\frac{y^2}{2}\right]}_{x^2}^{2-x}dx$

$\Rightarrow \frac{1}{2}\underset{0}{\stackrel{1}{\int }}x[{(2-x)}^2-{\left(x^2\right)}^2]dx$

$\Rightarrow \frac{1}{2}\underset{0}{\stackrel{1}{\int }}x(4+x^2-4x-x^4)dx$

$\Rightarrow \frac{1}{2}\underset{0}{\stackrel{1}{\int }}(-x^5+x^3-4x^2+4x)dx$

$\Rightarrow \frac{1}{2}{[-\left(\frac{x^6}{6}\right)+\left(\frac{x^4}{4}\right)-4\left(\frac{x^3}{3}\right)+4\left(\frac{x^2}{2}\right)]}_0^1$

$\Rightarrow \frac{1}{2}[\frac{-1}{6}(1-0)+\frac{1}{4}(1-0)\frac{4}{3}(1-0)+2(1-0)]$

$\Rightarrow \frac{1}{1}[\frac{-1}{6}+\frac{1}{4}-\frac{4}{3}+2]$

$\Rightarrow \frac{1}{2}\left[\frac{-2+3-16+24}{12}\right]$

$\Rightarrow \frac{1}{2}\left[\frac{9}{12}\right]=\frac{9}{24}=\frac{3}{8}$

Concept:

${\displaystyle {\int }_0^m{\int }_0^{x}f(x,y)dA}$

Since the inside limit is in terms of x, therefore we have to integrate first the 'y' terms and convert whole expression in terms of x.

Calculation:

Given:

${\displaystyle {\int }_0^1{\int }_0^x(x^2+y^2)dA}$

${\displaystyle {\int }_0^1[{\int }_0^x(x^2+y^2)dy]dx}$

${\displaystyle {\int }_0^1(x^2[y{]}_0^x+{\left[\frac{y^3}{3}\right]}_0^x)dx}$

${\displaystyle {\int }_0^1(x^3+\frac{x^3}{3})dx}$

${\displaystyle {\int }_0^1\left(\frac{4x^3}{3}\right)dx}$

${\displaystyle {\left(\frac{4x^4}{12}\right)}_0^1}$

${\displaystyle \frac{1}{3}}$

Concept:

Evaluation of Double integrals: It doesn’t matter which variable we integrate with respect to first, we will get the same answer regardless of the order of integration.

$\int \frac{dx}{x}=ln(x)+c$

Calculation:

Given:

I = ${\int }_1^a{\int }_1^b\frac{dxdy}{xy}$

= ${\int }_1^a\frac{dx}{x}\times {\int }_1^b\frac{dy}{y}$

= | ln (x) |_{1 to a} × | ln (y) |_{1 to b} 

= [ln (a) - ln (1)] × [ln(b) - ln(1)]

I = ln (a) ln (b) 

The region for the given constraint x > y² and y > x² is drawn as:

Limit of y: y = 0 to y = 1

Limit of x:

$x=y^{2}to{x}^2=y$

Or x = y to x = √y

Now, the volume under f(x, y) will be:

$V={\int }_{y=0}^{y=1}{\int }_{x=y^2}^{x=\sqrt{y}}f\left(x,y\right)dxdy$

The given limits are:

x = 0 to 3

y = x² to 9

The given limits are corresponding to a vertical strip ab as shown below.

Now, to change the order of integration consider the horizontal strip cd as shown.

Now, the limits are:

$x=0to\sqrt{y}$

y = 0 to 9

Now, the integration becomes

$\underset{y=0}{\overset{9}{\int }}\underset{x=0}{\overset{\sqrt{y}}{\int }}f\left(x,y\right)dxdy$

By comparing the above integral with the given integral, we get

p = 0, q = 9, r = 0, $s=\sqrt{y}$

Concept:

Let’s change the Cartesian coordinate limits into cylindrical planar polar coordinate,

⇒ x = r cosθ; y = r sinθ

Replace dx.dy = rdθ .dr

Calculation:

Given:

$y=\sqrt{r^2-x^2}$      ---(1)

From the graph of equation (1)

θ → 0 to π; r → 0 to 5

Then the limit of the double integration will be,

$I={\int }_0^5{\int }_0^{\pi }\left(r\times sin\theta \times r^2\right)rd\theta .dr$

$\Rightarrow I={\int }_0^5{\int }_0^{\pi }\left(r^{4}sin\theta \right)d\theta .dr$

$\Rightarrow I={\int }_0^{\pi }\left(\frac{5^5}{5}\times sin\theta \right)d\theta$

⇒ I = - [cos π – cos 0] × 625

⇒ I = 1250

Hint:

To Find $\int {}_0^{\frac{\pi }{2}}{\int }_0^{\pi }\cos (\mathrm{x}+\mathrm{y})\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}$, the limits of outermost integration is constant. so first integrate w.r.to x

When integrating with respect to x, we must think of y as a constant.

We know that:

$\mathrm{s}\mathrm{i}\mathrm{n}(\pi +\theta )=-\mathrm{s}\mathrm{i}\mathrm{n}\theta$ and

$\mathrm{s}\mathrm{i}\mathrm{n}(-\theta )=-\mathrm{s}\mathrm{i}\mathrm{n}\theta$

Calculation:

Consider, I = $\int {}_0^{\frac{\pi }{2}}{\int }_0^{\pi }\cos (\mathrm{x}+\mathrm{y})\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}$

To Find:

$\int {}_0^{\frac{\pi }{2}}{\int }_0^{\pi }\cos (\mathrm{x}+\mathrm{y})\mathrm{d}\mathrm{x}\mathrm{d}\mathrm{y}$
The limits of outermost integration are constant so first  integrate w.r.t x 

When integrating with respect to x, we must think of y as a constant.

$={\int }_0^{{\displaystyle \frac{\pi }{2}}}[\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{x}+\mathrm{y}){]}_0^{\pi }\mathrm{d}\mathrm{y}$

$={\int }_0^{{\displaystyle \frac{\pi }{2}}}(\mathrm{s}\mathrm{i}\mathrm{n}(\pi +\mathrm{y})-\mathrm{s}\mathrm{i}\mathrm{n}(0+\mathrm{y}))\mathrm{d}\mathrm{y}$

We know that:

$\mathrm{s}\mathrm{i}\mathrm{n}(\pi +\theta )=-\mathrm{s}\mathrm{i}\mathrm{n}\theta$ and $\mathrm{s}\mathrm{i}\mathrm{n}(-\theta )=-\mathrm{s}\mathrm{i}\mathrm{n}\theta$

$={\int }_0^{{\displaystyle \frac{\pi }{2}}}(-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{y}-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{y})\mathrm{d}\mathrm{y}$

$={\int }_0^{{\displaystyle \frac{\pi }{2}}}(-2\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{y})\mathrm{d}\mathrm{y}$

$=-2{\int }_0^{{\displaystyle \frac{\pi }{2}}}(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{y})\mathrm{d}\mathrm{y}$

= 2 [cos y]₀^π/2

= 2 [cos 90° - cos 0°]

= -2