Question

Question 4 Evaluate the line integral:

(i) of 𝑇(𝑥) = 4𝑥^3 along the line segment from (−2,1) to (1,2).

(ii) where the curve 𝐶 is parameterized through 𝑥(𝑡) = cos 𝑡, 𝑦(𝑡) = sin 𝑡 and 𝑧 = 𝑡^2 with 𝑡 ∈ [0,2𝜋] of ∫(𝒚𝒅𝒙 + 𝒙𝒅𝒚 + 𝒛𝒅𝒛) 𝐶

(iii) ∫ 𝐅(𝑥, 𝑦, 𝑧) ⋅ ⅆ𝐫 𝐶 , where 𝐹(𝑥, 𝑦, 𝑧) = (5𝑧^2 , 2𝑥, 𝑥 + 2𝑦) and the curve 𝐶 is given by 𝑥 = 𝑡, 𝑦 = 𝑡^2, and 𝑧 = 𝑡^2 with 𝑡 ∈ [0,1]

Answer 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

$\int ydx+xdy+zdz=\int sin\space tx-sin\space tdt+cos \space tx+2t^3dt$

$\int_{t=0}^{2n}(cos^2t-sin^2t+2t^3)dt\\ \int_{t=0}^{2n}(cos\space 2t+2t^3)dt\\=\frac{1}{2}(2n)^4$

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