Help solving line integral question

[Jaqsan]

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? :-)

 

[pasmith]

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 \|\mathrm{d}\mathbf{r}\| is \mathrm{d}s.

Your starting point is
<br /> \int_C xy\,\mathrm{d}s =<br /> \int_0^2 x(t)y(t)\|\mathbf{r}&#039;(t)\|\,\mathrm{d}t = <br /> \int_0^2 x(t)y(t)\sqrt{(x&#039;(t))^2 + (y&#039;(t))^2}\,\mathrm{d}t.<br />

Now substitute x(t) = t^3 and y(t) = t^2.

 

[Jaqsan]

The usual notation for \|\mathrm{d}\mathbf{r}\| is \mathrm{d}s.

Your starting point is
<br /> \int_C xy\,\mathrm{d}s =<br /> \int_0^2 x(t)y(t)\|\mathbf{r}&#039;(t)\|\,\mathrm{d}t = <br /> \int_0^2 x(t)y(t)\sqrt{(x&#039;(t))^2 + (y&#039;(t))^2}\,\mathrm{d}t.<br />

Now substitute x(t) = t^3 and y(t) = t^2.

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.