# Obtain the line integral

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Given f=x^2z ds

x=cost, y=2t, z=sint    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}

= 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 , cost=u, so cos^2t=u^2

Therefore,  -sint dt=du

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  16y=x^4 f(x,y)=16y-x^4 parameterized by  the curves

x=2t, y=t^4   for 0<=t<=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. 12t^5 dt=udu

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