Engineering Mathematics Test 5

Green's theorem:

\(\oint \left( {{\rm{Mdx}} + {\rm{Ndy}}} \right) = \int\!\!\!\int \left( {\frac{{\partial {\rm{N}}}}{{\partial {\rm{x}}}} - \frac{{\partial {\rm{M}}}}{{\partial {\rm{y}}}}} \right){\rm{dxdy}}\)

Stokes theorem:

\(\oint \vec A \cdot d\vec l\) = \(\iint \left( {\vec \nabla \times \vec A} \right).d\vec s\)

Gauss Divergence theorem:

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

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

Explanation:

By Gauss-Divergence Theorem

\(\mathop \smallint \nolimits_s^{} \bar F.\bar nds = \mathop \smallint \nolimits_v^{} div\;\bar Fdv\) 

\(\mathop \smallint \nolimits_s^{} \bar F.\bar nds = \mathop \smallint \nolimits_v^{} \left( {1 + 1 + 0} \right)dv = \mathop \smallint \nolimits_v^{} 2dv\)

\(\mathop \smallint \nolimits_s^{} \bar F.\bar nds = 2v = 2\left( {\frac{4}{3}\pi {r^3}} \right) = \frac{{8\pi }}{3}{\left( 2 \right)^3}\)

\(\therefore \mathop \smallint \nolimits_s^{} \bar F.\bar nds = \frac{{64\pi }}{3}\)

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.

Explanation:

By Stokes theorem

\(\begin{array}{l} I = \mathop \int\!\!\!\int \limits_s^\; \left( {\nabla \times F} \right).ds = \mathop \oint \limits_C^\; F.dr\\ I = \mathop \oint \limits_C^\; F.dr = \mathop \oint \limits_C^\; \left( {3y\hat i - xz\hat j + y{z^2}\hat k} \right).\left( {dx\hat i + dy\hat j + dz\hat k} \right)\\ = \mathop \oint \limits_C^\; 3ydx - xzdy + y{z^2}dz \end{array}\)

S ≡ x² + y² = 16, z = 8

Let x = 4 cos θ, y = 4sin θ

C ≡ x² + y² = 16, θ = 0 to 2π

\(\begin{array}{l} I = \mathop \smallint \limits_0^{2\pi } 12\sin \theta \left( { - 4\sin \theta } \right)d\theta - 32\cos \theta \left( {4\cos \theta } \right)d\theta + 256\sin \theta \left( 0 \right)\\ = - 4\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {48{{\sin }^2}\theta + 128{{\cos }^2}\theta } \right)d\theta \\ = - 4\left( {48 \times \frac{1}{2} \times \frac{\pi }{2} + 128 \times \frac{1}{2} \times \frac{\pi }{2}} \right) \end{array}\)

= -176 π

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:

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

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

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

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

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

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

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

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

⇒ x = y = z = t

⇒ dx = dy = dz = dt

Limits of t are: 0 to 1

\(\vec F \cdot \overrightarrow {dr} = 9t\;dt + 8t\;dt + 7t\;dt = 24t\;dt\)

\(\mathop \smallint \nolimits_C \vec F \cdot \overrightarrow {dr} = \mathop \smallint \limits_{t = 0}^1 24t\;dt = \left[ {24{t^2}} \right]_0^1 = 12\)

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

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

\(\begin{array}{l} {\rm{V}} = \mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{{2\pi }}{3}} \mathop \smallint \limits_0^2 {{\rm{r}}^3}{\rm{co}}{{\rm{s}}^2}\phi {\rm{drd}}\phi {\rm{d\theta }}\\ {\rm{V}} = \mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{{2\pi }}{3}} \left[ {\frac{{{r^4}}}{4}} \right]_0^2{\rm{co}}{{\rm{s}}^2}\phi {\rm{d}}\phi {\rm{d\theta }}\\ {\rm{V}} = \mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{{2\pi }}{3}} 4.{\rm{co}}{{\rm{s}}^2}\phi {\rm{d}}\phi {\rm{d\theta }} \end{array}\)

\(\begin{array}{l} {\rm{Co}}{{\rm{s}}^2}\phi = \frac{{1 + {\rm{cos}}2\phi }}{2}\\ {\rm{V}} = 4\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{{2\pi }}{3}} \left( {\frac{{1 + Cos2\phi }}{2}} \right){\rm{d}}\phi {\rm{d\theta }}\\ {\rm{V}} = 2\mathop \smallint \limits_0^{2\pi } \mathop \smallint \limits_0^{\frac{{2\pi }}{3}} \left( {1 + Cos2\phi } \right){\rm{d}}\phi {\rm{d\theta }} \end{array}\)

\(\begin{array}{l} {\rm{V}} = 2\mathop \smallint \limits_0^{2\pi } \left[ {\phi + \frac{{sin2\phi }}{2}} \right]_0^{\frac{{2\pi }}{3}}d\theta \\ {\rm{V}} = 2\mathop \smallint \limits_0^{2\pi } \left[ {\frac{{2\pi }}{3} + \frac{{sin\left( {\frac{{2 \times 2\pi }}{3}} \right)}}{2}} \right]d\theta \\\end{array}\)

\(V = \left[ {\frac{{4\pi }}{3} + \;\sin \frac{{4\pi }}{3}} \right]\left[ \theta \right]_0^{2\pi }\)

\(V = \left[ {\frac{{4\pi }}{3} + \;\sin \frac{{4\pi }}{3}} \right]2\pi \)

\(V = \left[ {\frac{{8{\pi ^2}}}{3} + \;2{\rm{\pi \;}}\sin \frac{{4\pi }}{3}} \right]\)

\(V = \left[ {\frac{{8{\pi ^2}}}{3} + \;2\pi \left[ {\frac{{ - \sqrt 3 }}{2}} \right]} \right]\)

\(V = \left[ {\frac{{8{\pi ^2}}}{3} - \;\pi \sqrt 3 } \right]\)