Question

Evaluate the line integral ∫CF⋅dr, where F(x,y,z)=−xi−2yj−zk and C is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

Answer 1

Answer:

$\mathbf { \int _ c Fdr= - \dfrac{9 \pi ^2}{8} - \dfrac{3}{2} }$

Step-by-step explanation:

GIven that:

$r(t) = (sin \ t, cos \ t, t)$

$r' (t) = (cos \ t, -sin \ \ t, 1)$

$F(x, y, z) = ( -x, -2y , - z)$

$F(r(t)) = ( sin \ t , - 2 cos \ t, - 1 t)$

$F(r(t)) \times r'(t) = (sin \ t, - 2 \ cos \ t , -1 \ t)( cos \ t , - sin \ t , 1 )$

$= sin t \ cos t + 2 \ sin t \ cos t - 1t$

$= 3sint \ cost - 1 t$

$\int _ c Fdr = \int ^b_a f(r(t)) \times r'(t) \ dt$

$= \int^{3 \pi/2}_0 \ [3 sin \ t \ cos \ t - 1 \ t ] \ dt$

$= 3 \int ^{3 \pi/2} _0 \ cos \ t ( sin \ t \ dt ) - 1 \int ^{3 \pi/2}_0 \ (t) \ dt$

Let   cos t = u   &

sint  dt = du

$= 3 \int ^{3 \pi/2_} } _0 \ udu - 1 \int ^{3 \pi/2}_0 \ (t) \ dt$

$= 3 [ \dfrac{u^2}{2}]^{3 \pi/2}_0 - 1 [ \dfrac{t^2}{2}]^{3 \pi/2}_0$

$= \dfrac{3}{2} [ cos ^2 \ t ] ^{3 \pi/2} _0- \dfrac{1}{2} ( \dfrac{ 3 \pi }{2 })^2 - 0^2$

$= \dfrac{3}{2} [ cos ^2 \ \dfrac{3 \pi}{2} - cos ^2 \ 0 ] - \dfrac{1}{2}( \dfrac{9 \pi^2}{4})$

$= \dfrac{3}{2} ( 0 -1 ) - \dfrac{9 \pi^2}{8}$

$= - \dfrac{3}{2} - \dfrac{9 \pi ^2}{8}$

$\mathbf { \int _ c Fdr= - \dfrac{9 \pi ^2}{8} - \dfrac{3}{2} }$

Answer 2

To answer this question, we apply:

∫CF×dr = ∫ c F (r(t)) × dr

Solution is:

( 1/2)  - (9/8)×π

We know  r(t) =  sint i + cost j + t k

Then  dr = ( cost i - sint j + k ) dt

And  F ( x , y , z ) = -x i - 2y j - z k

Then  F ( r(t)) =  - sint i - 2 × cost j - t × k

And F ( r(t)) × dr  = (- sint×cost + 2 ×sint ×cost - t ) dt

∫F (r(t)) × dr = ∫ (- sint×cost + 2 ×sint ×cost - t ) dt  

Integration limits 0≤  t  ≤ ( 3/2 ) π

∫ (- sint×cost + 2 ×sint ×cost - t ) dt  = ∫ ( sint ×cost - t ) dt

∫ sint ×cost × dt  - ∫ t ×  dt

∫F (r(t)) × dr = (1/2) sin²t  - ( 1/2) × t² | 0  y  (3/2) π

∫F (r(t)) × dr = (1/2)× ( -1)² - 0 - ( 9/8 ) × π - 0

∫F (r(t)) × dr = ( 1/2)  - (9/8)×π

Related Link :https://brainly.com/question/3645828