Engineering Mathematics Test 5

y = x² – 4x + 4

dy = 2x dx – 4 dx = (2x - 4) dx

$\overrightarrow{dr}=dx\hat{i}+dy\hat{j}$ 

${\oint }_C\left(y\hat{i}-3x\hat{j}\right)\cdot \left(dx\hat{i}+dy\hat{j}\right)$ 

$={\oint }_{C}ydx-3xdy$ 

$={\oint }_C\left(x^2-4x+4\right)dx-3x\left(2x-4\right)dx$ 

$={\oint }_C\left(-5x^2+8x+4\right)dx$ 

$\underset{x=0}{\overset{2}{\int }}\left(-5x^2+8x+4\right)dx$ 

$={\left[\frac{-5x^3}{3}+4x^2+4x\right]}_0^2$ 

$=\frac{-5}{3}\left(8\right)+4\left(4\right)+4\left(2\right)$ 

$\frac{-40}{3}+24=\frac{32}{3}$

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:

According to Gauss Divergence theorem:

$\oint A.ds=\iiint (\mathrm{\nabla }.A)dv$

$\oint \overrightarrow{F.}\hat{n}ds=\iiint (\mathrm{\nabla }.F)dv$

Calculation:

Given that S is a closed surface:

$\oint \overrightarrow{r.}\hat{n}ds=\iiint (\mathrm{\nabla }.r)dv$

$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\mathrm{\nabla }.r=\frac{\partial }{\partial x}\left(x\right)+\frac{\partial }{\partial y}\left(y\right)+\frac{\partial }{\partial z}\left(z\right)=3$

$\oint \overrightarrow{r.}\hat{n}ds=\iiint (\mathrm{\nabla }.r)dv=\iiint 3dv=3V$

$\overrightarrow{F}\left(x,y,z\right)=\left(4z+5y\right)\hat{i}+\left(3z+5x\right)\hat{j}+\left(3y+4x\right)\hat{k}$

$\overrightarrow{dr}=dr\hat{i}+dy\hat{j}+dz\hat{k}$

$\overrightarrow{F}\cdot \overrightarrow{dr}=\left(4x+5y\right)dx+\left(3z+5x\right)dy+\left(3y+4x\right)dz$

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

⇒ x = y = z = t

⇒ dx = dy = dz = dt

Limits of t are: 0 to 1

$\overrightarrow{F}\cdot \overrightarrow{dr}=9tdt+8tdt+7tdt=24tdt$

${\int }_C\overrightarrow{F}\cdot \overrightarrow{dr}=\underset{t=0}{\overset{1}{\int }}24tdt={\left[24t^2\right]}_0^1=12$

Concept:

By stokes theorem,

$\underset{C}{\overset{}{\oint }}F.dr=\int \underset{S}{\overset{}{\int }}CurlF.Nds$

Calculation:

$\overrightarrow{F}\left(x,y,z\right)=x\hat{i}+y\hat{j}+3\left(x^2+y^2\right)\hat{k}$

$\mathrm{\nabla }\times \overrightarrow{F}=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& y& 3\left(x^2+y^2\right)\end{array}\right|$

= i (6y) – j(6x) = 6y î - 6x ĵ

S: x² + y² + z² – 64 = 0

$ds=+2x\hat{i}+2y\hat{j}+2z\hat{k}$

$\left(\mathrm{\nabla }\times \overrightarrow{F}\right)\cdot ds=\left(6y\hat{i}-6x\hat{j}\right)\cdot \left(2x\hat{i}+2y\hat{j}+2z\hat{k}\right)$

= 12xy – 12xy + 0 = 0

$\Rightarrow {\int }_C\overrightarrow{F}\cdot \overrightarrow{dr}=0$

$V=\underset{c}{\iiint }dv$

0 ≤ z ≤ 4 - xy

0 ≤ x ≤ 2

0 ≤ y ≤ 1

$V=\underset{c}{\iiint }dv=\underset{0}{\overset{2}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{4-xy}{\int }}dzdxdy$

$V=\underset{0}{\overset{2}{\int }}\underset{0}{\overset{1}{\int }}z{|}_0^{4-xy}dydx$

$V=\underset{0}{\overset{2}{\int }}\underset{0}{\overset{1}{\int }}\left(4-xy\right)dydx$

$V=\underset{0}{\overset{2}{\int }}{\left(4y-\frac{1}{2}x{y}^2\right)|}_0^{1}dx$

$=\underset{0}{\overset{2}{\int }}4-\frac{1}{2}xdx$

$\Rightarrow {\left(4x-\frac{1}{4}{x}^2\right)}_0^2=7$

Explanation:

Cut open the surface S by any plane and Let S₁, S₂ denotes its upper and lower portions.

Let C be the common curve bounding both these portions.

$\underset{S}{\overset{}{\int }}Curl\overrightarrow{F}.\overrightarrow{d}s=\underset{S_1}{\overset{}{\int }}Curl\overrightarrow{F}.\overrightarrow{d}s+\underset{S_2}{\overset{}{\int }}Curl\overrightarrow{F}.\overrightarrow{d}s$

By stokes theorem,

$\begin{array}{l}\underset{S}{\overset{}{\int }}Curl\overrightarrow{F}.\overrightarrow{d}s=\underset{C}{\overset{}{\int }}\overrightarrow{F}.\overrightarrow{d}r\\ \Rightarrow \underset{S}{\overset{}{\int }}Curl\overrightarrow{F}.\overrightarrow{d}s=\underset{C}{\overset{}{\int }}\overrightarrow{F}.\overrightarrow{d}r-\underset{C}{\overset{}{\int }}\overrightarrow{F}.\overrightarrow{d}r=0\end{array}$

The second integral is negative because it is traversed in a direction opposite to that of the first.

From Gauss divergence theorem,

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

$\overrightarrow{\mathrm{F}}=\left(x+z\right)i+\left(y+z\right)j+\left(x+y\right)k$

$\mathrm{\nabla }.\overrightarrow{\mathrm{F}}=1+1+0=2$

$\underset{V}{\iiint }\mathrm{\nabla }.\overrightarrow{F}dV=\iiint 2dV$

$2\iiint dV$

The volume of the sphere $=\frac{4}{3}\pi a^3$

$\Rightarrow \underset{V}{\iiint }\mathrm{\nabla }.\overrightarrow{F}dV=2\times \frac{4}{3}\pi a^3=\frac{8}{3}\pi a^3$