Multiple Integrals MCQ Quiz - Objective Question with Answer for Multiple Integrals - Download Free PDF

Explanation:

 ${\int }_{x=3}^{x=4}{\int }_{y=1}^{y=2}\frac{dxdy}{(x+y{)}^2}$

= ${\int }_{x=3}^{x=4}-{\left[\frac{1}{(x+y)}dx\right]}_{y=1}^{y=2}$

$=-{\int }_{x=3}^{x=4}[\frac{1}{2+x}-\frac{1}{x+1}]$

= $-[\ln (2+x)-\ln (x+1){]}_{x=3}^{x=4}$

= $-{[\ln \left(\frac{2+x}{x+1}\right)]}_3^4$

$=-[\ln \left(\frac{6}{5}\right)-\ln \left(\frac{5}{4}\right)]$

= $-[\ln \left(\frac{6\times 4}{5\times 5}\right)]$

$=-\ln \left(\frac{24}{25}\right)$

$=\ln \left(\frac{25}{24}\right)$

Thus, ${\log }_e\left(\frac{a}{24}\right)=\ln \left(\frac{25}{24}\right)$

∴ a = 25

Explanation:

Intersection of the Surfaces:

The paraboloid $z=x^2+y^2$  intersects the sphere  $x^2+y^2+z^2=2$  where:

$z=x^2+y^2$  and  $x^2+y^2+z^2=2$

Substituting $z=x^2+y^2$  into the sphere equation:

$x^2+y^2+(x^2+y^2{)}^2=2$

Let $r^2=x^2+y^2$ :

$r^2+r^4=2$

Solving $r^2$ gives $r^2=1$ 

Thus, the intersection occurs at r = 1 

z ranges from the paraboloid ($z=x^2+y^2=r^2$) to the sphere ($z=\sqrt{2-r^2}$)

r ranges from 0 to 1

$\theta$  ranges from 0 to 2π 

In Cylindrical Coordinates: 

$x=r\cos \theta$ , $y=r\sin \theta$ , z = z , $dV=rdzdrd\theta$ 

The volume integral becomes:

$V={\int }_0^{2\pi }{\int }_0^1{\int }_{r^2}^{\sqrt{2-r^2}}rdzdrd\theta$ 

(Integrating with respect to z ) :

${\int }_{r^2}^{\sqrt{2-r^2}}rdz=r{\int }_{r^2}^{\sqrt{2-r^2}}dz=r[\sqrt{2-r^2}-r^2]$ 

(Integrating with respect to r ) :

${\int }_0^{1}r(\sqrt{2-r^2}-r^2)dr={\int }_0^{1}r\sqrt{2-r^2}dr-{\int }_0^1{r}^{3}dr$ 

For ${\int }_0^1{r}^{3}dr$  : ${\int }_0^1{r}^{3}dr={\left[\frac{r^4}{4}\right]}_0^1=\frac{1}{4}$ 

For ${\int }_0^{1}r\sqrt{2-r^2}dr$ , let $u=2-r^2$ , so $du=-2rdr$ 

${\int }_0^{1}r\sqrt{2-r^2}dr=-\frac{1}{2}{\int }_2^1{u}^{1/2}(-du)=\frac{1}{2}{\int }_1^2{u}^{1/2}du$ 

${\int }_1^2{u}^{1/2}du={\left[\frac{2}{3}{u}^{3/2}\right]}_1^2=\frac{2}{3}(2^{3/2}-1^{3/2})=\frac{2}{3}(2\sqrt{2}-1)$ 

So: ${\int }_0^{1}r\sqrt{2-r^2}dr=\frac{1}{2}\cdot \frac{2}{3}(2\sqrt{2}-1)=\frac{1}{3}(2\sqrt{2}-1)$

${\int }_0^{1}r(\sqrt{2-r^2}-r^2)dr=\frac{1}{3}(2\sqrt{2}-1)-\frac{1}{4}$ 

(with respect to $\theta$ ): Multiply by ${\int }_0^{2\pi }d\theta =2\pi$ :

$V=2\pi [\frac{1}{3}(2\sqrt{2}-1)-\frac{1}{4}]$

$V=2\pi (\frac{2\sqrt{2}}{3}-\frac{1}{3}-\frac{1}{4})$ 

$V=\frac{\pi }{6}(8\sqrt{2}-7)$ 

This is the volume of the region.

Hence Option(4) is the correct answer.

Explanation: 

$\frac{20}{7}{\iint }_{R}x{e}^{x^2-y^2}dxdy$ 

=  $\frac{20}{7}{\int }_0^1{\int }_y^{y+1}x{e}^{x^2-y^2}dxdy$

= $\frac{20}{7}\times \frac{1}{2}{\int }_0^1[e^{x^2}{]}_y^{y+1}{\int }_y^{y+1}x{e}^{x^2}dxdy$

= $\frac{20}{14}{\int }_0^1[e^{x^2}{]}_y^{y+1}{e}^{-y^2}dy$  

$=\frac{20}{14}{\int }_0^1(e^{2y+1}-1)du$ 

$=\frac{20}{7}\frac{1}{2}[e^u{]}_{y^2}^{(y+1{)}^2}$

$=\frac{10}{7}[e^3-e-2]$ 

Hence Option(1) is the correct answer.

Explanation:

Let I = ${\iiint }_{D}6xyzdxdydz$

The region D is the portion of the sphere in the first octant (where x, y, and z are all positive)

$0\le x\le a$

$0\le y\le \surd (a^2-x^2)$

$0\le z\le \surd (a^2-x^2-y^2)$

Now, the triple integral can be written as:

I = ${\int }_0^a{\int }_0^{\sqrt{a^2-x^2}}{\int }_0^{\sqrt{a^2-x^2-y^2}}6xyzdzdydx$

Integrate with respect to z:

I = ${\int }_0^a{\int }_0^{\sqrt{a^2-x^2}}{\left[\frac{6}{2}xy{z}^2\right]}_0^{\sqrt{a^2-x^2-y^2}}dydx$

Substitute the limits of integration for z:

I = ${\int }_0^a{\int }_0^{\sqrt{a^2-x^2}}3xy(a^2-x^2-y^2)dydx$

I = ${\int }_0^a{[3x(a^{2}y-x^{2}y-y^3)]}_0^{\sqrt{a^2-x^2}}dx$

Integrate with respect to y:

I = ${\int }_0^a{[3x(a^2\frac{y^2}{2}-\frac{x^2{y}^2}{2}-\frac{y^4}{4})]}_0^{\sqrt{a^2-x^2}}dx$

Substitute the limits of integration for y:

I = ${\int }_0^a\frac{3}{4}x(a^2-x^2{)}^{2}dx$

Integrate with respect to x:

I = ${[\frac{3}{4}(a^{4}x-2a^2{x}^3+x^5)]}_0^a$

Substitute the limits of integration for x:

$I=\frac{3}{4}\times \frac{a^6}{6}$

$I=\frac{a^6}{8}$

Hence Option(4) is the correct answer.

Explanation:

The region E is a tetrahedron in the first octant, bounded by the planes:
x = 0 ,
y = 0 ,
z = 0 , and x + y + z = 1  

$0\le z\le 1-x-y$,  $0\le y\le 1-x$,  $0\le x\le 1$

Now,

$I={\int }_0^1{\int }_0^{1-x}{\int }_0^{1-x-y}zdzdydx$

Integrate with respect to z :  

${\int }_0^{1-x-y}zdz={\left[\frac{z^2}{2}\right]}_0^{1-x-y}=\frac{(1-x-y{)}^2}{2}$

Integrate with respect to y :  

${\int }_0^{1-x}\frac{(1-x-y{)}^2}{2}dy=\frac{1}{2}{\int }_0^{1-x}(1-x-y{)}^{2}dy.$
 
Use substitution u = 1 - x - y , du = -dy , bounds: u = 1-x to u = 0 :

$\frac{1}{2}{\int }_{1-x}^0{u}^2(-du)=\frac{1}{2}{\int }_0^{1-x}{u}^{2}du$

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

= $\frac{1}{2}\cdot \frac{(1-x{)}^3}{3}$

= $\frac{(1-x{)}^3}{6}$

Integrate with respect to x :  

${\int }_0^1\frac{(1-x{)}^3}{6}dx=\frac{1}{6}{\int }_0^1(1-3x+3x^2-x^3)dx$

= $\frac{1}{6}(1-3\cdot \frac{1}{2}+3\cdot \frac{1}{3}-\frac{1}{4})$

= $\frac{1}{6}(1-\frac{3}{2}+1-\frac{1}{4})$

= $\frac{1}{6}(\frac{4}{4}-\frac{6}{4}+\frac{4}{4}-\frac{1}{4})$

= $\frac{1}{6}\cdot \frac{1}{4}$

=$\frac{1}{24}$

Hence Option(4) is the correct answer.

Given:

$I={\int }_{x=0}^1{\int }_{y=0}^{x^2}x{y}^{2}dydx$

0 ≤ y ≤ x² (this is represented by vertical strip)

And x varies from 0 to 1.

Now if we change the order of integration, we have to draw a horizon strip.

After changing the order of Integration

$\sqrt{y}\le x\le 1$

And, 0 ≤ y ≤ 1

∴ $I={\int }_{y=0}^1{\int }_{x=\sqrt{y}}^{1}x{y}^{2}dxdy$

Concept:

The triple integral can be evaluated as repeated integral:

${\int }_{{\text{x}}_1}^{{\text{x}}_2}{\int }_{{\text{y}}_1}^{{\text{y}}_2}{\int }_{{\text{z}}_1}^{{\text{z}}_2}\text{f}(\text{x},\text{y},\text{z})\text{ dz }\text{ dy }\text{ dx}$

First, f(x, y, z) is integrated w.r.t z between limits z₁ and z₂ keeping x and y constant, then integrated w.r.t y between limits y₁ and y₂ keeping x constant. The result is then integrated w.r.t x.

Calculation:

Given:

${\int }_0^1{\int }_{{\text{y}}^2}^{\text{1}}{\int }_0^{1-\text{x}}\text{x }\text{ dz }\text{ dx }\text{ dy}$

$={\int }_0^1{\int }_{{\text{y}}^2}^1\text{x }{\left|\text{z}\right|}_0^{1-\text{x}}\text{ dx }\text{ dy}$

$={\int }_0^1{\int }_{{\text{y}}^2}^1\text{x }(1-\text{x})\text{ dx }\text{ dy}$

$={\int }_0^1{|\frac{{\text{x}}^2}{2}-\frac{{\text{x}}^3}{3}|}_{{\text{y}}^2}^1\text{ dy}$

$={\int }_0^1(\frac{1}{2}-\frac{1}{3})-(\frac{{\text{y}}^4}{2}-\frac{{\text{y}}^6}{3})\text{ dy}$

$=\frac{1}{6}-{|\frac{{\text{y}}^5}{10}-\frac{{\text{y}}^7}{21}|}_0^1$

$=\frac{1}{6}-\frac{1}{10}+\frac{1}{21}=\frac{10}{60}-\frac{6}{60}+\frac{1}{21}=\frac{1}{15}+\frac{1}{21}$

$\underset{0}{\overset{2}{\int }}\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{x}+\mathrm{y}}\mathrm{d}\mathrm{y}\mathrm{d}\mathrm{x}=\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{\mathrm{x}}\left(\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{e}}^{\mathrm{y}}\mathrm{d}\mathrm{y}\right)\mathrm{d}\mathrm{x}$

$=\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{\mathrm{x}}{\left({\mathrm{e}}^{\mathrm{y}}\right)}_0^{\mathrm{x}}\mathrm{d}\mathrm{x}=\underset{0}{\overset{2}{\int }}{\mathrm{e}}^{\mathrm{x}}\left({\mathrm{e}}^{\mathrm{x}}-1\right)\mathrm{d}\mathrm{x}$

$=\underset{0}{\overset{2}{\int }}\left({\mathrm{e}}^{2\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\right)\mathrm{d}\mathrm{x}={\left(\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}-{\mathrm{e}}^{\mathrm{x}}\right)}_0^2$

$=\frac{{\mathrm{e}}^4}{2}-{\mathrm{e}}^2-\frac{1}{2}+1=\frac{{\mathrm{e}}^4}{2}-{\mathrm{e}}^2+\frac{1}{2}$

$=\frac{1}{2}\left({\mathrm{e}}^4-2{\mathrm{e}}^2+1\right)=\frac{1}{2}{\left({\mathrm{e}}^2-1\right)}^2$

The equation of line in intercept form is given by

$\frac{x}{2}+\frac{y}{1}=1$

⇒ $y=1-\frac{x}{2}\Rightarrow x=2-2y$

⇒ $\underset{0}{\overset{1}{\int }}\underset{0}{\overset{2-2y}{\int }}\left(xydx\right)dy=\underset{0}{\overset{1}{\int }}\left[\left(\frac{y{x}^2}{2}\right){|}_0^{2-2y}\right]dy$

$=\underset{0}{\overset{1}{\int }}\frac{y}{2}{\left(2-2y\right)}^{2}dy$

$=\underset{0}{\overset{1}{\int }}2y{\left(1-y\right)}^{2}dy=\underset{0}{\overset{1}{\int }}(2y-4y^2+2y^3)dy$

$={\left[y^2-\frac{4}{3}{y}^3+\frac{y^4}{2}\right]}_0^1=\frac{1}{6}$

Given,

$\frac{1}{2\pi }\int \underset{D}{\overset{}{\int }}\left(x+y+10\right)dxdy$

and x² + y² ≤ 4

Putting x = r.cosθ, y = r.sinθ and dx.dy = r.dr.dθ

$\mathrm{I}=\frac{1}{2\pi }\underset{\theta =0}{\overset{2\pi }{\int }}\underset{\mathrm{r}=0}{\overset{2}{\int }}\left(\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{s}\theta +\mathrm{r}\mathrm{s}\mathrm{i}\mathrm{n}\theta +10\right)\mathrm{r}\mathrm{d}\mathrm{r}\mathrm{d}\theta$

$=\frac{1}{2\pi }\underset{\theta =0}{\overset{2\pi }{\int }}{\left(\frac{{\mathrm{r}}^3}{3}\mathrm{c}\mathrm{o}\mathrm{s}\theta +\frac{{\mathrm{r}}^3}{3}\mathrm{s}\mathrm{i}\mathrm{n}\theta +5{\mathrm{r}}^2\right)}_0^2\mathrm{d}\theta$

$=\frac{1}{2\pi }\left[\underset{\theta =0}{\overset{2\pi }{\int }}\left(\frac{8}{3}\mathrm{c}\mathrm{o}\mathrm{s}\theta +\frac{8}{3}\mathrm{s}\mathrm{i}\mathrm{n}\theta +20\right)\mathrm{d}\theta \right]$

$=\frac{1}{2\pi }\left[0+0+20\left(2\pi \right)\right]=20$

The given curves are y = x and y = x²

Solving the equations, we get

x  = 0, x = 1

$Area=\underset{0}{\overset{1}{\int }}\left(x-x^2\right)dx$

$={\left(\frac{x^2}{2}-\frac{x^3}{3}\right)}_0^1=\frac{1}{2}-\frac{1}{3}$

$=\frac{1}{6}squnits$

Alternate Solution:

Concept:

Area of a region can be calculated by:

$\int \int dxdy$

Calculation:

Solving equation y = x² and we get y = x

We get intersection points i.e. (0,0) and (1,1)

Area of the region:

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

$\underset{x=0}{\overset{x=1}{\int }}{\left[y\right]}_{x^2}^x.dx$

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

${\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

$Given:\underset{y=0}{\overset{\pi }{\int }}\underset{x=y}{\overset{\pi }{\int }}\frac{\sin x}{x}dxdy$

Changing the order and take vertical strip, we get:

$I=\underset{x=0}{\overset{\pi }{\int }}\underset{y=0}{\overset{y=x}{\int }}\frac{\sin x}{x}dxdy$

$\underset{x=0}{\overset{\pi }{\int }}\frac{\sin x}{x}dx\underset{y=0}{\overset{y=x}{\int }}dy$

$\Rightarrow \underset{x=0}{\overset{\pi }{\int }}\frac{\sin x}{x}dx{\left[y\right]}_0^x$

$\Rightarrow \underset{x=0}{\overset{\pi }{\int }}\frac{\sin x}{x}dx\left[x-0\right]$

$\Rightarrow I=\underset{x=0}{\overset{\pi }{\int }}\sin xdx$

$I=(-cosx{)}_0^{\pi }$

I = (1 – (-1))

I = 2

Explanation:

We have the integration given as,

$I=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{x^2}{\int }}\left(x^2+y^2\right)dydx$

$I=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{x^2}{\int }}\left(x^2+y^2\right)\cdot dy\cdot dx$

$I=\underset{0}{\overset{1}{\int }}{\left[x^{2}y+\frac{y^3}{3}\right]}_0^{x^2}dx$

$I=\underset{0}{\overset{1}{\int }}\left[x^4+\frac{x^6}{3}\right]dx$

Placing the limits we get,

$I={\left[\frac{x^5}{5}+\frac{x^7}{7\times 3}\right]}_0^1$

$I=\frac{1}{5}+\frac{1}{21}=\frac{26}{105}$

Hence the required value of integration will be $\frac{26}{105}$.

Concept:

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Length major axis of ellipse = 2a

Length of minor axis of ellipse = 2b

Area $=\int \underset{R}{\overset{}{\int }}1\cdot dy\cdot dx$

Calculation:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow y=\sqrt{b\left(1-\frac{x^2}{a^2}\right)}$.

For the first quadrant, take a vertical strip as shown. Here, y coordinate varies from 0 to $\sqrt{b\left(1-\frac{x^2}{a^2}\right)}=\frac{b}{a}\sqrt{a^2-x^2}$.

Also, the x-coordinate varies from 0 to a

∴ Area $=\underset{0}{\overset{a}{\int }}\underset{0}{\overset{\frac{b}{a}\sqrt{a^2-x^2}}{\int }}\left(1\right)dydx$

$=\underset{0}{\overset{a}{\int }}\left(\frac{b}{a}\sqrt{a^2-x^2}\right)dx$

$=\frac{b}{a}{\left[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)\right]}_0^a=\frac{b}{a}\times \frac{a^2}{2}\times \frac{\pi }{2}=\frac{\pi ab}{4}$

∴ The total area of ellipse $=4\times \frac{\pi ab}{4}$ = πab units

Important Points

1) x² + y² = a² ; Represents a circle centred at (0, 0) and radius ‘a’ units.

2) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$; Represents Ellipse with major axis length 2a and minor axis length 2b and vertex at (0,0)

3) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$; Equation of Hyperbola .