- #1
chwala
Gold Member
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- 369
- 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.
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