Let C1 and C2 be two smooth parameterized curves that start at P0 and end at Q0 ≠ P0, but do not otherwise intersect. If the line integral of the function f(x, y, z) along C1 is equal to 43.1 and the line integral of f(x, y, z) along C2 is −15.9, what is the line integral around the closed loop formed by first following C1 from P0 to Q0, followed by the curve from Q0 to P0 along C2, but moving in the opposite direction? Incorrect: Your answer is incorrect.

Respuesta :

We're given that

[tex]\displaystyle \int_{C_1} f(x,y,z) \,ds = 43.1[/tex]

and

[tex]\displaystyle \int_{C_2} f(x,y,z) \,ds = -15.9[/tex]

The closed loop formed by joining C₁ and C₂ is then C₁ U (-C₂), which is to say we follow C₁ normally, then traverse C₂ in the opposite direction. Then

[tex]\displaystyle \int_{C_1 \cup (-C_2)} f(x,y,z) \,ds = \int_{C_1} f(x,y,z) \,ds + \int_{-C_2} f(x,y,z) \,ds[/tex]

[tex]\displaystyle \int_{C_1 \cup (-C_2)} f(x,y,z) \,ds = \int_{C_1} f(x,y,z) \,ds - \int_{C_2} f(x,y,z) \,ds[/tex]

[tex]\displaystyle \int_{C_1 \cup (-C_2)} f(x,y,z) \,ds = 43.1 - (-15.9) = \boxed{59}[/tex]

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