Engineering Mathematics Test 5

y = x² – 4x + 4

dy = 2x dx – 4 dx = (2x - 4) dx

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

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

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

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

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

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

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

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

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

First line segment: (0, 0) → (3, 4)

Second line segment: (3, 4) → (6, 0)

Now, the line integral becomes

\(\mathop \smallint \nolimits_C \left( {3dy - 4dx} \right)\;\) 

\( = \mathop \smallint \limits_{\left( {0,\;0} \right)}^{\left( {6,\;0} \right)} \left( { - 4dx + 3dy} \right)\) 

\( = \left[ {\frac{{ - 4x}}{2} + 3y} \right]_{\left( {0,\;0} \right)}^{\left( {6,\;0} \right)}\) 

= -4(6) + 3(0)

Concept:

According to Gauss Divergence theorem:

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

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

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

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

\(V = \iiint\limits_c {dv}\)

0 ≤ z ≤ 4 - xy

0 ≤ x ≤ 2

0 ≤ y ≤ 1

\(V = \iiint\limits_c {dv} = \mathop \smallint \limits_0^2 \mathop \smallint \limits_0^1 \mathop \smallint \limits_0^{4 - xy} dzdxdy\)

\(V = \mathop \smallint \limits_0^2 \mathop \smallint \limits_0^1 z\left. \right|_0^{4 - xy}dydx\)

\(V = \mathop \smallint \limits_0^2 \mathop \smallint \limits_0^1 \left( {4 - xy} \right)dydx\)

\(V = \mathop \smallint \limits_0^2 \left. {\left( {4y - \frac{1}{2}x{y^2}} \right)} \right|_0^1dx\)

\( = \mathop \smallint \limits_0^2 4 - \frac{1}{2}xdx\;\)

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

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.

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

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.

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 \(= \frac{4}{3}\pi {a^3}\)

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