Calculus 3 Line integral

Question

Find , where  is the line segment from (0,-5,5) to (2,-11,1)

Answer 1

Let's parametrize the line segment that the two given A (0,-5,5) and B (2,-11,1)points define.

The directing vector AB = < 2, -6 ,-4 > or any scalar multiple of AB , say conviniettly

u =< 1 ,- 3 , -2 >

Then the parametric equations of the line segment are

x = x(t) = t

y = y(t) = -5 - 3t

z = z(t) = 5 - 2t.

Then ∫_C ƒ (x ,y ,z ) d s = ∫^t=1_t=0 [ t² ( 5 -2 t ) ] [ (dx/dt)² + (dy/dt)² + (dz/dt)² ]^1/2dt =

∫^t=1_t=0 [ t² ( 5 -2 t ) ] [ ( 1)² + (-3)² + (-2)² ]^1/2dt

∫^t=1_t=0 [ 5t² -2 t³ ] [ ( 1)² + (-3)² + (-2)² ]^1/2dt

∫^t=1_t=0 [ 5t² - 2t³] [ ( 14)]^1/2dt

√14 ∫^t=1_t=0 [ 5t² - 2t³]dt

√14 [ 5/3 t³ - 1/2 t⁴ ]|¹₀

√14 [ 10/6 - 3/6 ] = 7√14/6