Engineering Mathematics Test 5

y = x² – 4x + 4

dy = 2x dx – 4 dx = (2x - 4) dx

\(\overrightarrow {dr} = dx\;\hat i + dy\;\hat j\) 

\(\mathop \oint \nolimits_C \left( {y\hat i - 3x\hat j} \right) \cdot \left( {dx\;\hat i + dy\;\hat j} \right)\) 

\( = \mathop \oint \nolimits_C ydx - 3x\;dy\) 

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

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

\(\mathop \smallint \limits_{x = 0}^2 \left( { - 5{x^2} + 8x + 4} \right)dx\) 

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

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

\(\frac{{ - 40}}{3} + 24 = \frac{{32}}{3}\)

Concept:

According to Gauss Divergence theorem:

\(\oint A.ds=\iiint{\left( \nabla .A \right)dv}\)

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

Calculation:

Given that S is a closed surface:

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

\(\vec r = x\hat i + y\hat j + z\hat k\)

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

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

C ≡ x² + y² = 1

Let x = r cos θ, y = r sin θ

⇒ dx = -sin θ dθ, dy = cos θ dθ

\(\begin{array}{l}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left[ {\left( {2\cos \theta + 3\sin \theta } \right)\left( { - \sin \theta } \right)d\theta - \left( {3\cos \theta - 4\sin \theta } \right)\left( {\cos \theta } \right)d\theta } \right]\\ = \mathop \smallint \limits_0^{2\pi } \left[ { - 2\sin \theta \cos \theta - 3{{\sin }^2}\theta - 3{{\cos }^2}\theta + 4\sin \theta \cos \theta } \right]d\theta \\ = \mathop \smallint \limits_0^{2\pi } \left( {2\sin \theta \cos \theta - 3} \right)d\theta \\ = \mathop \smallint \limits_0^{2\pi } \left( {\sin 2\theta - 3} \right)d\theta \end{array}\) 

= -3(2π) = -6π

Concept:

By stokes theorem,

\(\mathop \oint \limits_C^\; F.dr = \mathop \int\!\!\!\int \limits_S^\; Curl\;F.Nds\)

Calculation:

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

\(\nabla \times \vec F = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ x&y&{3\left( {{x^2} + {y^2}} \right)} \end{array}} \right|\)

= i (6y) – j(6x) = 6y î - 6x ĵ

S: x² + y² + z² – 64 = 0

\(ds = + 2x\hat i + 2y\hat j + 2z\hat k\)

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

= 12xy – 12xy + 0 = 0

\( \Rightarrow \mathop \smallint \nolimits_C \vec F \cdot \overrightarrow {dr} = 0\)

From Gauss divergence theorem,

\(\mathop \iint \limits_S F.ds = \iiint\limits_V {\nabla .\overrightarrow F dV}\)

\({\rm{\vec F}} = \left( {x + z} \right)i + \left( {y + z} \right)j + \left( {x + y} \right)k\)

\(\nabla .{\rm{\vec F}} = 1 + 1 + 0 = 2\)

\(\iiint\limits_{V}{\nabla .\overrightarrow{F}dV}=\iiint{2dV}\)

\(2\iiint{dV}\)

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

\(\Rightarrow \iiint\limits_{V}{\nabla .\overrightarrow{F}dV}=2\times \frac{4}{3}\pi {{a}^{3}}=\frac{8}{3}\pi {{a}^{3}}\)

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.

\(\mathop \smallint \limits_S^\; Curl\;\vec F.\vec ds = \mathop \smallint \limits_{{S_1}}^\; Curl\;\vec F. \vec ds + \mathop \smallint \limits_{{S_2}}^\; Curl\;\vec F.\vec ds\)

By stokes theorem,

\(\begin{array}{l} \mathop \smallint \limits_S^\; Curl\;\vec F.\vec ds = \mathop \smallint \limits_C^\; \vec F.\vec dr\\ \Rightarrow \mathop \smallint \limits_S^\; Curl\;\vec F.\vec ds = \mathop \smallint \limits_C^\; \vec F.\vec dr - \mathop \smallint \limits_C^\; \vec F.\vec dr = 0 \end{array}\)

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

\(\vec F = 3xy\hat i - {y^2}\hat j\) and \(\vec R = x\hat i + y\hat j\)

\(\mathop \smallint \limits_C \vec F.d\vec R = \mathop \smallint \limits_C \left( {3xy\hat i - {y^2}\hat j} \right).d\left( {x\hat i + y\hat j} \right) = \mathop \smallint \limits_C \left( {3xydx - {y^2}dy} \right)\)

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

⇒ dy = 4x dx

\(\mathop \smallint \limits_C \vec F.d\vec R = \mathop \smallint \limits_C \left( {3x\left( {2{x^2}} \right)dx - {{\left( {2{x^2}} \right)}^2}(4xdx} \right)\)

\(= \mathop \smallint \limits_0^1 \left( {6{x^3} - 16{x^5}} \right)dx\)

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

\(= \frac{6}{4} - \frac{8}{6} = \frac{{18 - 32}}{{12}} = - \frac{7}{6} = - 1.17\)

Alternate method:

\(\vec F = 3xy\hat i - {y^2}\hat j\) and \(\vec R = x\hat i + y\hat j\)

\(\mathop \smallint \limits_C \vec F.d\vec R = \mathop \smallint \limits_C \left( {3xy\hat i - {y^2}\hat j} \right).d\left( {x\hat i + y\hat j} \right) = \mathop \smallint \limits_C \left( {3xydx - {y^2}dy} \right)\)

Let x = t

⇒ dx = dt

y = 2x² = 2t²

⇒ dy = 4tdt

t varies from 0 to 1.

\(\mathop \smallint \limits_C \vec F.d\vec R = \mathop \smallint \limits_C \left( {3\left( t \right)\left( {2{t^2}} \right)dt - {{\left( {2{t^2}} \right)}^2}(4tdt} \right))\)

\(= \mathop \smallint \limits_0^1 \left( {6{t^3} - 16{t^5}} \right)dt\)

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

Concept:

Green’s Theorem:

If M(x, y), N(x, y), M_y and N_x be continuous in a region E of the xy-plane bounded by a closed curve C, then

\(\mathop \smallint \limits_C \left( {Mdx + Ndy} \right) = \mathop \int\!\!\!\int \limits_E \left( {\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}}} \right)dxdy\)

Calculation:

\(\frac{2}{\pi }\mathop \oint \limits_\gamma \left( { - {y^3}dx + {x^3}dy} \right)\)

It is in the form of \(\mathop \smallint \limits_C \left( {Mdx + Ndy} \right)\)

M = -y³ and N = x³

\(\frac{{\partial N}}{{\partial x}} = 3{x^2}\)

\(\frac{{\partial M}}{{\partial y}} = - 3{y^2}\)

\(\frac{2}{\pi }\mathop \oint \limits_\gamma \left( { - {y^3}dx + {x^3}dy} \right) = \frac{2}{\pi }\mathop \int\!\!\!\int \limits_E \left( {3{x^2} + 3{y^2}} \right)dxdy\)

Changing to polar coordinates (r, θ), r varies from 0 to 1 and θ varies from 0 to 2π.

\( = \frac{2}{\pi }\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^{2\pi } \left( {3{{\left( {r\cos \theta } \right)}^2} + 3{{\left( {r\sin \theta } \right)}^2}} \right)rd\theta dr\)

\( = \frac{2}{\pi }\mathop \smallint \limits_0^1 \mathop \smallint \limits_0^{2\pi } 3{r^3}d\theta dr\)

\( = \frac{2}{\pi }\mathop \smallint \limits_0^1 \left[ {3{r^3}\theta } \right]_0^{2\pi }dr\)

\( = \frac{6}{\pi }\mathop \smallint \limits_0^1 2\pi {r^3}dr\)