Respuesta :
a. We can parameterize [tex]C[/tex] by
[tex]x=t[/tex]
[tex]y=2t^2[/tex]
with [tex]0\le t\le1[/tex]. Then
[tex]\displaystyle\int_C5x\,\mathrm dS=\int_0^15t\sqrt{1+16t^2}\,\mathrm dt=\dfrac5{48}(17^{3/2}-1)[/tex]
b. We can parameterize the opposite direction by instead setting
[tex]x=1-t[/tex]
[tex]y=2(1-t)^2[/tex]
with [tex]0\le t\le 1[/tex]. Then
[tex]\displaystyle\int_C5x\,\mathrm dS=\int_0^15(1-t)\sqrt{1+16(1-t)^2}\,\mathrm dt=-\int_1^05u\sqrt{1+16u^2}\,\mathrm dt=\int_0^15u\sqrt{1+16u^2}\,\mathrm du[/tex]
which gives the same value as in part (a).
The parameterization of C in the direction from O to P can be determined by integrating the given function in terms of 't' nad in the same way parameterization of C in the direction from P to O can be determined by integrating the given function in terms of 'w'.
Given :
- [tex]f(x,y)=5x[/tex]
- C be the segment of the parabola [tex]y =2x^2[/tex] joining O(0,0) and P( 1,2).
A) The parameterization of C in the direction from O to P is given by:
[tex]x = t[/tex]
[tex]y = t^2[/tex]
where, [tex]0\leq t \leq 1[/tex].
[tex]\rm \int\limits^1_0 f(x,y)=\int\limits^1_0 {5x} \, dS[/tex]
      [tex]=\int\limits^1_0 {5t\sqrt{1+16t^2} } \, dt[/tex]
      [tex]= \dfrac{5}{48}(17^{\frac{3}{2}}-1)[/tex]
B). The parameterization of C in the direction from P to O is given by:
[tex]\rm x = 1-t[/tex]
[tex]\rm y = 2(1-t)^2[/tex]
where, [tex]0\leq t \leq 1[/tex].
[tex]\rm \int\limits^1_0 f(x,y)=\int\limits^1_0 {5x} \, dS[/tex]
      [tex]=\int\limits^1_0 {5(1-t)\sqrt{1+16(1-t)^2} } \, dt[/tex]
      [tex]=-\int\limits^0_1 {5(w)\sqrt{1+16(w)^2} } \, dt[/tex]
      [tex]=\int\limits^1_0 {5(w)\sqrt{1+16(w)^2} } \, dw[/tex]
      [tex]= \dfrac{5}{48}(17^{\frac{3}{2}}-1)[/tex]
For more information, refer to the link given below:
https://brainly.com/question/22008756
