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
$$\int_C xy\,\mathrm{d}s = \int_0^2 x(t)y(t)\|\mathbf{r}'(t)\|\,\mathrm{d}t = \int_0^2 x(t)y(t)\sqrt{(x'(t))^2 + (y'(t))^2}\,\mathrm{d}t.$$

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
$$\int_C xy\,\mathrm{d}s = \int_0^2 x(t)y(t)\|\mathbf{r}'(t)\|\,\mathrm{d}t = \int_0^2 x(t)y(t)\sqrt{(x'(t))^2 + (y'(t))^2}\,\mathrm{d}t.$$

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.