Evaluate the line integral ∫c(−𝑥 + 𝑦 + 𝑧) 𝑑𝑥 + (𝑥 + 𝑦 − 𝑧) 𝑑𝑦 + (𝑥 − 𝑦 + 𝑧) 𝑑𝑧 where C is the line segment from the point 𝑃(−1, 2, 3) to 𝑄(3, −2, 1).

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karton · Accepted Answer

To evaluate the line integral ∫C(−x+y+z)dx+(x+y−z)dy+(x−y+z)dz, where C is the line segment from the point P(−1,2,3) to Q(3,−2,1), we can parameterize the line segment and then substitute the parameterization into the integral.Let&#039;s denote the parameter as t, where t varies from 0 to 1 along the line segment PQ. The position vector 𝐫(t) can be expressed as:𝐫(t)=𝐚+t(𝐛−𝐚)where 𝐚 is the position vector of P and 𝐛 is the position vector of Q.𝐚=(−123)𝐛=(3−21)Substituting the values, we have:𝐫(t)=(−123)+t((3−21)−(−123))Simplifying, we get:𝐫(t)=(−123)+t(4−4−2)Now, we can compute the line integral by substituting the parameterization into the integral:∫C(−x+y+z)dx+(x+y−z)dy+(x−y+z)dz=∫01[−(−1+4t)+(2−4t)+(3−2t)](4dt)+[(−1+4t)+(2−4t)−(3−2t)](−4dt)+[(−1+4t)−(2−4t)+(3−2t)](−2dt)Simplifying and combining like terms, we have:∫01(12t)dtEvaluating the integral, we get:[12t22]01=6(12)−6(02)=6Therefore, the value of the line integral ∫C(−x+y+z)dx+(x+y−z)dy+(x−y+z)dz along the line segment from P(−1,2,3) to Q(3,−2,1) is 6.

Evaluate the line integral ∫c(−𝑥 + 𝑦 +

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