The value of the line integral is 456.
To evaluate the integral ∫C[(2x+3y)I+(3x+2y)J+3z^2k]⋅dr where C is the line segment from (2,-1,3) to (4,2,-1), we can parameterize the line segment and use the line integral formula.
Parameterization:
Let's define a parameterization of the line segment C as follows:
r(t) = (2, -1, 3) + t*((4, 2, -1) - (2, -1, 3))
r(t) = (2 + 2t, -1 + 3t, 3 - 4t)
Line Integral Formula:
The line integral for a vector field F over a curve C parameterized by r(t) from a to b is:
∫C F(r(t)) ⋅ r'(t) dt
Evaluation:
Applying the formula with our parameterization:
∫C [(2x+3y)I+(3x+2y)J+3z^2k]⋅dr
= ∫_0^1 [(4 + 6t) + (-3 + 9t)I + (6 + 4t) + (2 + 6t)J + 3(3 - 4t)^2k] ⋅ (2, 3, -4) dt
= ∫_0^1 [10 + 15t + 8 + 10t + 27 - 36t + 144t^2] ⋅ (2, 3, -4) dt
= ∫_0^1 [184t^2 + 10t + 39] ⋅ (2, 3, -4) dt
= 2∫_0^1 (184t^2 + 10t + 39) dt
= 2 [61.33t^3 + 5t^2 + 39t] ∣_0^1
= 2(184 + 5 + 39) - 2(0 + 0 + 0)
= 2 × 228
= 456
Question:
Evaluate ∫C[(2x+3y)I+(3x+2y)J+3z2k]⋅Dr Where C Is The Line Segment From (2,−1,3) To (4,2,−1).