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             

 

Approved by eNotes Editorial Team
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