- #1

chwala

Gold Member

- 2,662

- 352

- 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.

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.

Last edited by a moderator: