Answer in Calculus for prezi #264744
Question 4 Evaluate the line integral:
(i) of ( ) = 4 ^3 along the line segment from (−2,1) to (1,2).
(ii) where the curve is parameterized through ( ) = cos , ( ) = sin and = ^2 with ∈ [0,2 ] of ∫( + + )
(iii) ∫ ( , , ) ⋅ ⅆ , where ( , , ) = (5 ^2 , 2 , + 2 ) and the curve is given by = , = ^2, and = ^2 with ∈ [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 sinspace tx-sinspace tdt+cos space tx+2t^3dt”
“int_{t=0}^{2n}(cos^2t-sin^2t+2t^3)dt\nnint_{t=0}^{2n}(cosspace 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
