6) Compute int_{C} x y d s where C is line segment from (2,4) to (1,6).

Question

Accepted Answer

Step:1 ：Parameterize the line segment from (2,4) to (1,6) by setting (x(t) = 2 - t) and (y(t) = 4 + 2t) for (0 leq t leq 1).Step:2 ：Compute the differential (ds = sqrt{dx^2 + dy^2} = sqrt{(-dt)^2 + (2dt)^2} = sqrt{5} dt).Step:3 ：Substitute the parameterization and differential into the integral to get (int_{0}^{1} x(t) y(t) ds = int_{0}^{1} (2 - t)(4 + 2t) sqrt{5} dt).Step:4 ：Compute the integral to get the final answer (boxed{frac{22sqrt{5}}{3}}).

Answer

Step: 1 ：Parameterize the line segment from (2,4) to (1,6) by setting (x(t) = 2 - t) and (y(t) = 4 + 2t) for (0 leq t leq 1).Step: 2 ：Compute the differential (ds = sqrt{dx^2 + dy^2} = sqrt{(-dt)^2 + (2dt)^2} = sqrt{5} dt).Step: 3 ：Substitute the parameterization and differential into the integral to get (int_{0}^{1} x(t) y(t) ds = int_{0}^{1} (2 - t)(4 + 2t) sqrt{5} dt).Step: 4 ：Compute the integral to get the final answer (boxed{frac{22sqrt{5}}{3}}).

6) Compute $\int_{C} x y d s$ where $C$ is line segment from $(2,4)$ to $(1,6)$.

Answer

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