Evaluate the given integrals - line integrals

[chwala]

Homework Statement

See attached (Refreshing on this)

Relevant Equations

Line integrals

My interest is on question $37$. Highlighted in Red.

For part (a) I have the following lines;

$\int_c A. dr = 4t(2t+3) +2t^5 + 3t^2(t^4-2t^2) dt$

$\left[\dfrac {8t^3}{3}+ 6t^2+\dfrac{t^6}{3} + \dfrac{3t^7}{7} - \dfrac{6t^5}{5}\right]_0^1$


$=\dfrac{288}{35}$

For part (b) for $(0,0,0)$ to $(0,0,1)$ i was having $dx=0, dy=0$.

$\int_{ z=0}^1 (0×z-0) dz=0$

also for $(0,0,0)$ to $(0,1,1)$ where $dx=0$ and $dz=0$

$\int_{y=0}^1 (0 . 0) dy=0$

from $(0,1,1)$ to $(2,1,1)$


$\int_{x=0}^2 (2+3)dx=[5x]_0^2 = 10$ not sure about this approach... need to recheck or give direction.

lastly,

for part (c),

from $(0,0,0)$ t0 $(2,1,1)$ we have the parametric form,

$x=2t, y=t, z=t$

Therefore,

$\int_0^1 [(2t+3)2 + 2t^2 +t^2-2t]dt$

$=\int_0^1 [4t+6 + 2t^2 +t^2-2t]dt$

$=[2t^2+6t+\dfrac{2t^3}{3}+\dfrac{t^3}{3}- t^2 ]_0^1= [t^2+t^3+6t]_0^1=8$

there could be a better approach.

cheers.

 

[slider142]

These are all correct methodologies. Good job!

 

[Gavran]

Excellent job except some typos.