Help solving line integral question

  • Thread starter Jaqsan
  • Start date
  • #1
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h

Homework Statement



Evaluate ∫xy|dr| over the path given by x=t^3, y=t^2, t=0...2

Homework Equations



x=t^3, y=t^2, t=0...2

The Attempt at a Solution



x=t^3, y=t^2
y^(3/2) =x, y=t, x=t^(3/2), t=0...4
∫0to4 t^5/2 [Sqrt((3t^(1/2))/2)^2 +(1)^2]
=∫0to4 t^5/2 [Sqrt(9t/4 + 1) dt

HELP PLEASE, I'm not sure this is right. Help or point me in the right direction, would you? :-)
 

Answers and Replies

  • #2
pasmith
Homework Helper
2,036
661
h

Homework Statement



Evaluate ∫xy|dr| over the path given by x=t^3, y=t^2, t=0...2

Homework Equations



x=t^3, y=t^2, t=0...2

The Attempt at a Solution



x=t^3, y=t^2
y^(3/2) =x, y=t, x=t^(3/2), t=0...4
∫0to4 t^5/2 [Sqrt((3t^(1/2))/2)^2 +(1)^2]
=∫0to4 t^5/2 [Sqrt(9t/4 + 1) dt

HELP PLEASE, I'm not sure this is right. Help or point me in the right direction, would you? :-)

The usual notation for [itex]\|\mathrm{d}\mathbf{r}\|[/itex] is [itex]\mathrm{d}s[/itex].

Your starting point is
[tex]
\int_C xy\,\mathrm{d}s =
\int_0^2 x(t)y(t)\|\mathbf{r}'(t)\|\,\mathrm{d}t =
\int_0^2 x(t)y(t)\sqrt{(x'(t))^2 + (y'(t))^2}\,\mathrm{d}t.
[/tex]

Now substitute [itex]x(t) = t^3[/itex] and [itex]y(t) = t^2[/itex].
 
  • #3
17
0
The usual notation for [itex]\|\mathrm{d}\mathbf{r}\|[/itex] is [itex]\mathrm{d}s[/itex].

Your starting point is
[tex]
\int_C xy\,\mathrm{d}s =
\int_0^2 x(t)y(t)\|\mathbf{r}'(t)\|\,\mathrm{d}t =
\int_0^2 x(t)y(t)\sqrt{(x'(t))^2 + (y'(t))^2}\,\mathrm{d}t.
[/tex]

Now substitute [itex]x(t) = t^3[/itex] and [itex]y(t) = t^2[/itex].

My integral comes out to the same answer of 102.842 but your method seems a whole lot easier. I think I was just thinking too much about it. Thanks.
 

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