Engineering Mathematics Test 6

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:

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

\(\nabla {\rm{\;}}.{\rm{\vec F\;}} = \frac{{\partial \left( {\rm{x}} \right)}}{{\partial {\rm{x}}}} + \frac{{\partial \left( {\rm{y}} \right)}}{{\partial {\rm{y}}}} + \frac{{\partial \left( {\rm{z}} \right)}}{{\partial {\rm{z}}}} = 3\)

x² + y² + z² = a²

∴ closed surface is a sphere of radius r

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:

The given integral can be written as

\(\mathop \oint \limits_c \left( {{e^x}\hat i + 2y\hat j - 1\hat k} \right).\overrightarrow {dr} = \mathop \oint \limits_c \vec F.\overrightarrow {dr}\)

Where,

\(\vec F = {e^x}\hat i + 2y\;\hat j - \hat k\)

Using stokes theorem

\(\mathop \oint \limits_c \vec F.\overrightarrow {dr} = \int\!\!\!\int \left( {\bar V \times \vec F} \right)ds\)

\(\bar V \times \vec F = \left| {\begin{array}{*{20}{c}} {\bar i}&{\hat j}&{\hat i}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{e^x}}&{2y}&{ - 1} \end{array}} \right| = 0\)

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.

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

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

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

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

Concept:

Eq. of plane through (a, 0, 0), (0, b, 0) and (0, 0, c) is

\(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\)

To evaluate the \(\mathop \smallint \nolimits_c \vec F \cdot d\vec r\) we use Stoke’s  theorem

\(\mathop \smallint \nolimits_C \vec F \cdot d\vec r = \int\!\!\!\int \left( {curl\;\vec F} \right) \cdot \hat n\;ds\)

Where,

\(d\vec r = dx\;\hat i + dy\;\hat j + dz\;\hat k\)

n̂ = unit normal to the plane.

\({\rm{Curl\;}}\vec F = \vec \nabla \times \vec F\)

Calculation:

Eq. of plane:

\(\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1 \Rightarrow 3x + 2y + z = 6\)

\(\vec n = 3\hat i + 2\hat j + \hat k\)

\(\hat n = \frac{{\vec n}}{{\left| {\vec n} \right|}}\)

\(\therefore \hat n = \frac{1}{{\sqrt {14} }}\left( {3\hat i + 2\hat j + \hat k} \right)\)

\({\rm{Curl\;}}\vec F = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {x + y}&{2x - z}&{y + z} \end{array}} \right|\)

= î (1 + 1) – ĵ (0) + k̂ (2 – 1)

= 2î + k̂ 

\(\therefore curl\;\vec F \cdot \hat n = \left( {2\hat i + \hat k} \right) \cdot \frac{1}{{\sqrt {14} }}\left( {3\hat i + 3\hat j + \hat k} \right)\)

\(\therefore curl\;\vec F \cdot \hat n = \frac{7}{{\sqrt {14} }}\)

Hence,

\(\mathop \smallint \nolimits_C \vec F \cdot d\vec r = \int\!\!\!\int \frac{7}{{\sqrt {14} }}ds = \frac{7}{{\sqrt {14} }}\int\!\!\!\int dx\;dy\)

\({\rm{Given\;area\;}} = 3\sqrt {14} = \int\!\!\!\int ds\)

\(\therefore \frac{7}{{\sqrt {14} }} \times 3\sqrt {14} = 21\)

Concept:

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

When z = 0

\(\vec r = x\hat i + y\hat j \Rightarrow d\vec r = dx\;\hat i + dy\;\hat j\) 

Circle x² + y² = a² ⇒ x = a cos θ

The line integral is given as:

\(\mathop \oint \nolimits_C \vec F \cdot d\vec r\)

Calculation:

\(\mathop \oint \nolimits_C \vec F \cdot d\vec r = \mathop \smallint \nolimits_C \left( {\sin y\hat i + x\left( {1 + \cos x} \right)\hat j} \right) \cdot \left( {dx\;\hat i + dy\;\hat j} \right)\)

\(\mathop \oint \nolimits_C \vec F \cdot d\vec r = \mathop \smallint \nolimits_C \left( {\sin y\;dx + x\left( {1 + \cos x} \right)dy} \right)\;\)

\(\mathop \oint \nolimits_C \vec F \cdot d\vec r = \mathop \smallint \nolimits_C d\left( {x\sin y} \right) + x\;dy\)

Now,

x = a cos θ, y = a sin θ

\(= \mathop \smallint \limits_0^{2\pi } \left[ {d\left( {a\cos \theta \sin \left( {a\sin \theta } \right)} \right) + {a^2}{{\cos }^2}\theta \;d\theta } \right]\)

\(= \left| {a\cos \theta \sin \left( {a\sin \theta } \right)} \right|_0^{2\pi } + \left[ {\frac{{{a^2}\theta }}{2} + \frac{{\sin 2\theta }}{2}} \right]_0^{2\pi }{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2}\)

\(\Rightarrow 0 + \frac{{{a^2}}}{2}\left( {2\pi } \right) + 0 - 0 = \pi {a^2}\)

Now,

a = 5 (Given)

Explanation:

Work done on semi-sphere \(= \mathop \smallint \nolimits \vec F.d\vec r\)

\(\begin{array}{l} \mathop \smallint \limits_c \bar F.d\bar r = \mathop \smallint \limits_s \left( {\bar \nabla \times \bar F} \right).\hat nds\\ \bar \nabla \times \bar F = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{x^2}}&{{y^2} + x}&{{z^2}} \end{array}} \right| = \hat i\left( {0 - 0} \right) - \hat j\left( {0 - 0} \right) + \hat k\left( {1 - 0} \right)\\ = \hat k \end{array}\)

now \(\hat n = \frac{{\nabla \emptyset }}{{\left| {\nabla \emptyset } \right|}}\) where \(\phi= {x^2} + {y^2} + {z^2} - 1\)

\(\begin{array}{l} \nabla \emptyset = \hat i\frac{{\partial \emptyset }}{{\partial x}} + \hat j\frac{{\partial \emptyset }}{{\partial y}} + \hat k\frac{{\partial \emptyset }}{{\partial z}}\\ = 2x\hat i + 2y\hat j + 2z\hat k\\ \left| {\nabla \emptyset } \right| = \sqrt {4{x^2} + 4{y^2} + 4{z^2}} = \sqrt {4\left( {{x^2} + {y^2} + {z^2}} \right)} = \sqrt {4 \times 4} = 4 \end{array}\)

now \(n = \frac{{2x\hat i + 2y\hat j + 2z\hat k}}{4} = \frac{1}{2}\left( {x\hat i + y\hat j + z\hat k} \right)\)

now \(\mathop \smallint \limits_c \bar F.d\bar r = \mathop \int\!\!\!\int \limits_s k.\frac{1}{2}\left( {x\hat l + y\hat j + z\hat k} \right)ds\)

\(= \mathop \int\!\!\!\int \limits_s \frac{z}{2}ds\)

where \(ds = \frac{{dx.dy}}{{\left| {\hat n.\hat k} \right|}} = \frac{{dx\ dy}}{{\left| {\frac{{x\hat l + y\hat j + z\hat k}}{2}.\hat k} \right|}} = \frac{{dx\ dy}}{{\left( {\frac{z}{2}} \right)}}\)

So \(\mathop \smallint \limits_c \bar F.d\bar r = \mathop \int\!\!\!\int \limits_s \frac{z}{2}.\frac{{dxdy}}{{\frac{z}{2}}} = \mathop \int\!\!\!\int \limits_s dxdy = area\ of\ plane\ of\ projection\)

\(= \pi \left( {{r^2}} \right) = \pi \left( 2^{2} \right) = 4 \pi\)