Consider f(x, y, z) = x^3 - 3xyz + x^2z + y^3 + yz^2 + z^3 and that we are integrating โf = F
Thus, the integral is
F(3, 2, 1) - F(-1, 2, 3) = 3^3 - 3*3*2*1 + 3^2*1 + 2^3 + 2*1^2 + 1^3 - (-1^3-3*-(-1)*2*3+(-1)^2*3+2^3+2*(3)^2+3^3) =
29 - 73 =
-44
Mehran P.
asked 07/04/21How I can Evaluate (Line Integrals Of Vector Fields) โซc (๐๐^๐โ๐๐๐+๐๐๐) ๐ ๐ + (๐๐^๐โ๐๐๐+๐^๐) ๐ ๐ + (๐๐^๐โ๐๐๐+๐^๐+๐๐๐) ๐ ๐ along the line segment from (-1,2,3) to (3,2,-1)?
Consider f(x, y, z) = x^3 - 3xyz + x^2z + y^3 + yz^2 + z^3 and that we are integrating โf = F
Thus, the integral is
F(3, 2, 1) - F(-1, 2, 3) = 3^3 - 3*3*2*1 + 3^2*1 + 2^3 + 2*1^2 + 1^3 - (-1^3-3*-(-1)*2*3+(-1)^2*3+2^3+2*(3)^2+3^3) =
29 - 73 =
-44
Find the parametric equations of the line passing through the given points.
You should be getting
x = โ1 + t
y = 2
z = 3 โ t.
Obviously the interval of integration is 0 โค t โค 4.
then dx=dt , dy=0 and dz=โdt
Then substitute for x , y and z the corresponding values from the parametric equations
dx=dt , dy=0 and dz=โdt in the original integral then you will be getting an integral
in terms of t , with limits of integration from t = 0 to t= 4.
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Tom K.
07/05/21