.Evaluate the line integral ∫C F⋅dr where F= 〈−4sinx, 4cosy, 10xz〉 and C is the path given by r(t)=(2t3,−3t2,3t) for 0 ≤ t ≤ 1
∫C F⋅dr = ...........

Answers

Answer 1

The value of the line integral ∫C F⋅dr = 1.193.

To evaluate the line integral ∫C F⋅dr, we first need to calculate F⋅dr, where F= 〈−4sinx, 4cosy, 10xz〉 and dr is the differential of the vector function r(t)= (2t^3,-3t^2,3t) for 0 ≤ t ≤ 1.

We have dr= 〈6t^2,-6t,3〉dt.

Thus, F⋅dr= 〈−4sinx, 4cosy, 10xz〉⋅ 〈6t^2,-6t,3〉dt

= (-24t^2sin(2t^3))dt + (-24t^3cos(3t))dt + (30t^3x)dt

Now we integrate this expression over the limits 0 to 1 to get the value of the line integral:

∫C F⋅dr = ∫0^1 (-24t^2sin(2t^3))dt + ∫0^1 (-24t^3cos(3t))dt + ∫0^1 (30t^3x)dt

The first two integrals can be evaluated using substitution, while the third integral can be directly integrated.

After performing the integration, we get:

∫C F⋅dr = 2/3 - 1/9 + 3/5 = 1.193

Therefore, the value of the line integral ∫C F⋅dr is 1.193.

In conclusion, we evaluated the line integral by calculating the dot product of the vector function F and the differential of the given path r(t), and then integrating the resulting expression over the given limits.

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