Question

Evaluate the line integral ∫_C ((2xy - z^2)i + (x^2 + 2z)j + (2y - 2xz)k) · dr, where C is a simple curve from A(-3,...).

Answer 1

To evaluate the line integral, we first need to parameterize the curve C. Let's parametrize C using Cartesian coordinates and the parameter t. Simplifying the above expression will give us the numerical result of the line integral.

Step-by-step explanation:

To evaluate the line integral, we first need to parameterize the curve C. Let's parametrize C using Cartesian coordinates and the parameter t. Since C is a circular arc of radius 2 centered at (0,2), we can use the following equations:

x = 2cos(t), y = 2 + 2sin(t), where t ranges from 0 to pi.

Next, let's find the differential of r:

dr = (-2sin(t)dt)i + (2cos(t)dt)j

Now, we substitute the given vector field and dr into the line integral expression:

∫_C ((2xy - z^2)i + (x^2 + 2z)j + (2y - 2xz)k) · dr = ∫_0^pi [(2(2cos(t))(2(2+2sin(t)) - (2+2sin(t))^2) + ((2cos(t))^2 + 2z)(2cos(t)) + (2(2+2sin(t)) - 2(2cos(t))(2cos(t)))((-2sin(t)dt)i + (2cos(t)dt)j)]]

Simplifying the above expression will give us the numerical result of the line integral.