Question 4 Evaluate the line integral:
(i) of π(π₯) = 4π₯^3 along the line segment from (β2,1) to (1,2).
(ii) where the curve πΆ is parameterized through π₯(π‘) = cos π‘, π¦(π‘) = sin π‘ and π§ = π‘^2 with π‘ β [0,2π] of β«(ππ π + ππ π + ππ π) πΆ
(iii) β« π (π₯, π¦, π§) β β π« πΆ , where πΉ(π₯, π¦, π§) = (5π§^2 , 2π₯, π₯ + 2π¦) and the curve πΆ is given by π₯ = π‘, π¦ = π‘^2, and π§ = π‘^2 with π‘ β [0,1]
(I)
Given"T(X)=4x^3"
Along line segment (-2,1) to (1,2(
Line integral is given by
"\\int T(X) dx=\\int_{X=-2}^{X=1}4x^3dx\\\\=\\frac{4x^4}{4}\\int_{X=-2}^{X=1}\\\\=1-(-2)^4\\\\=1-16\\\\Txdx=-15"
(ii)
Given X(t)=cos t,y(t)=sin t, z(t)=t2
dx=-sin ft, Dy=cos ft, dz=2tdt
Required lines integral is given by
"\\int ydx+xdy+zdz=\\int sin\\space tx-sin\\space tdt+cos \\space tx+2t^3dt"
"\\int_{t=0}^{2n}(cos^2t-sin^2t+2t^3)dt\\\\\n\n\\int_{t=0}^{2n}(cos\\space 2t+2t^3)dt\\\\=\\frac{1}{2}(2n)^4"
The required value of line integral is "2n^4"
(iii)
"Given\\\\x=t, y=t^2,z=t^2"
Such that
"dx=DT,Dy=2tdt, dz=2tdt"
Now "\\int f(X,y,z)dt=\\int_{t=0}^{t=1}(5t^2+4t^2+2t^2+4t^3)dt\\\\=\\frac{5t^3}{5}+\\frac{4t^3}{3}+\\frac{2t^3}{3}+\\frac{4t^4}{4}\\\\=1+2+1\\\\=4"
Required value of line integral is 4
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