Obtain the line integral

Wednesday, July 28, 2010

Obtain the line integral

Given $\displaystyle{f{=}}{{x}}^{{2}}{z}{d}{s}$

$\displaystyle{x}={\cos{{t}}},{y}={2}{t},{z}={\sin{{t}}}$ for $0<=t<=\pi$

We have to find the line integral i.e.

$\int_{c} f(x,y,z)ds=\int_{c} x^2z ds$

= $\int_{c} f(x(t),y(t),z(t)). ||r'(t)|| dt$

where ,

$||r'(t)||=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}$

= $\sqrt{sin^2t+2^2+cos^2t}$

= $\sqrt{1+4}$

= $\displaystyle\sqrt{{{5}}}$

Therefore we have,

$\int_{c}f(x,y,z)ds=\int_{0}^{\pi}cos^2tsint . \sqrt{5} dt$

= $\sqrt{5}\int_{0}^{\pi}cos^2tsint dt$

Take , $\displaystyle{\cos{{t}}}={u}$ , so $\displaystyle{{\cos}}^{{2}}{t}={{u}}^{{2}}$

Therefore, $\displaystyle-{\sin{{t}}}{\left.{d}{t}\right.}={d}{u}$

When t=0, then u=1 and when

t= $\pi$ , then u=-1

Hence we have,

$\int_{c}f(x,y,z)ds=\sqrt{5}\int_{1}^{-1}-u^2 du$

$= \sqrt{5}\int_{-1}^{1}u^2du$

= $\sqrt{5}[\frac{u^3}{3}]_{-1}^{1}$

= $\frac{2\sqrt{5}}{3}$

(b) Now we have the curve $\displaystyle{16}{y}={{x}}^{{4}}$ $\displaystyle{f{{\left({x},{y}\right)}}}={16}{y}-{{x}}^{{4}}$ parameterized by the curves

$\displaystyle{x}={2}{t},{y}={{t}}^{{4}}$ for $\displaystyle{0}\le{t}\le{1}$

We have to find the line integral :

$\int_{c} f(x,y) ds=\int_{c} xy ds$

$=\int_{c} f(x(t),y(t))||r'(t)|| dt$

where,

$||r'(t)||=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}$

$=\sqrt{2^2+(4t^3)^2}$

= $\sqrt{4+16t^6}$

= $2\sqrt{1+4t^6}$

Therefore we have,

$\int_{c} f(x,y) ds=\int_{0}^{1}2t^5. 2\sqrt{1+4t^6} dt$

$=4\int_{0}^{1}t^5\sqrt{1+4t^6} dt$

Now take,

$\sqrt{1+4t^6}=u$

Therefore,

$\frac{1}{2\sqrt{1+4t^6}}.24t^5 dt=du$

i.e. $\displaystyle{12}{{t}}^{{5}}{\left.{d}{t}\right.}={u}{d}{u}$

i.e. $t^5dt=\frac{u}{12} du$

When t=0, then u=1 and when

t=1, then u= $\sqrt{5}$

Therefore we have,

$\int_{c} f(x,y)ds=4\int_{1}^{\sqrt{5}}\frac{u^2}{12} du$

$=\int_{1}^{\sqrt{5}}\frac{u^2}{3} du$

$=[\frac{u^3}{9}]_{1}^{\sqrt{5}}$

= $\frac{5\sqrt{5}-1}{9}$

= 1.131

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