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Evaluate ᶴc x dx + xy dy + (x+z)dz
where C is the line segment from (3,2,1) to (2,0,2).

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The value of the line integral ∫_C x dx + xy dy + (x + z) dz along the curve C from (3,2,1) to (2,0,2) is 9/2.

To evaluate the line integral ∫_C x dx + xy dy + (x + z) dz along the curve C parameterized by r(t) = (x(t), y(t), z(t)), we need to find the line integral in terms of the parameter t and then integrate over the interval of parameterization.

Given the curve C parameterized by r(t) = (3 - t, 2t, 1 + t), we can find the line integral as follows:

∫_C x dx + xy dy + (x + z) dz = ∫_0^1 3 - t: -dt + (3 - t)(2t)dt + [(3 - t) + (1 + t)]dt

= ∫_0^1 -3t + t^2 + 6t - 2t^2 + 4t + t^2 dt

= ∫_0^1 9t dt

= [9t^2/2]_0^1

= 9/2 - 0

= 9/2

Therefore, the value of the line integral ∫_C x dx + xy dy + (x + z) dz along the curve C from (3,2,1) to (2,0,2) is 9/2.

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User Anjan Kant
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