Obtain the line integral

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             
 
http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsPtI.aspx