Question

If C is the line segment from (6,9) to 6,15() find the following line integral with respect to arc length. integral C of(xe^yds)

Answer 1

To evaluate the line integral, parametrize the line segment and calculate the arc length. Then, substitute the values into the line integral expression and evaluate the integral we get ∫01 (6e^(9+6t)) * 6 dt.

Step-by-step explanation:

To evaluate the line integral of (xe^yds) along the line segment from (6,9) to (6,15), we first need to parametrize the line segment.

Let t be a parameter that ranges from 0 to 1.

The coordinates of the line segment can be represented as (6,9) + t[(6,15)-(6,9)].

This simplifies to (6,9) + t(0,6) = (6,9+6t).

Now, we can calculate the arc length by integrating the magnitude of the derivative of this parametric curve, which is 6.

The line integral is then given by the integral from 0 to 1 of (x(t)e^y(t)) * ||r'(t)|| dt,

where x(t) = 6 and y(t) = 9+6t.

Substituting the values, the line integral becomes:

∫01 (6e^(9+6t)) * 6 dt.