# Misunderstanding Line (Curve) integral of first and second kind

Can some one explain to me what is a curve integral of first kind and what is a curve integral of 2nd kind? I have to understand this for my calculus exam.

I know how to compute the area of the wall between the graph of f(x,y) and curve $\gamma$ on XoY plane in this case: The area is $\int f(x,y) \,ds$, where $\,ds = \sqrt {(\,dx)^2 + (\,dy)^2}$ and $x$ and $y$ get parametrized like this: $x = x(t)$ and $y = y(t)$

So my integral becomes (maybe not written 100% correctly): $$\int f(x,y) \sqrt{(\,dx)^2 + (\,dy)^2}$$

But we know that $x = x(t)$ and $y=y(t)$ the integral becomes: $$\int f(x(t),y(t)) \sqrt{(\,dx)^2 + (\,dy)^2}$$

Then: $$\sqrt{(\,dx)^2 + (\,dy)^2} = \frac {\,dt}{\,dt} \sqrt{(\,dx)^2 + (\,dy)^2} = \frac {1}{\,dt}\sqrt{(\,dx)^2 + (\,dy)^2}\,dt = \sqrt{(\frac{\,dx}{\,dt})^2 + (\frac{\,dy}{\,dt})^2}\,dt$$

And finaly the formula for the area: $$\int f(x(t),y(t)) \sqrt{(\frac{\,dx(t)}{\,dt})^2 + (\frac{\,dy(t)}{\,dt})^2}\,dt$$

Following this I have another question is this the first kind line integral? Is it the second kind line integral? Or it has nothing in common with first and second kind line integrals?

I played around with $f: R^2 \rightarrow R$ and $\gamma$ a parametrized path on XoY plane so a discussion with an $f: R^n \rightarrow R^m$ would be appreciated.

• A line integral of the 1st kind, is the integral of a scalar ($\mathbb{R}^2 \to \mathbb{R}$, say) function, such as the one used in your question to calculate the area of a fence. A line integral of the 2nd kind, is the integral of a vector function ($\mathbb{R}^2 \to \mathbb{R}^2$), which you'd use to calculate e.g. the work done by a force acting along the line. – Teddy Sep 3 '12 at 12:01
• Ok now I see I understand integral of 1st kind, but the 2nd kind is still strange to me. – Dr.Optix Sep 3 '12 at 12:22
• If you want a general explanation about the 2nd-type line integral, you should consult a textbook. It would do a better job. If you'd like something more specific explained, then please ask specifically. – Teddy Sep 3 '12 at 13:08
• In my teacher text book it is explained in chinese (romanian actually but still very strange). When someone tells me this is actually and area, this is actually a volume then I understand it. I will research terms like work done by a force so I can understand it. i really appreciate that now I understand what first integral of first kind is. – Dr.Optix Sep 3 '12 at 13:44
• @Teddy Let's assume I parametrize my curve $\gamma$ with a position vector in 2D to keep it easy: $r(t) = (2t)i + (t^2)j$ where t is time and I do $r'(t) = 2i + (2t)j$ I get the speed of a particle. That's what you mean by work done by a vector field acting on the line? Actually a particle traveling on that line? – Dr.Optix Sep 3 '12 at 17:18

A line integral of the 1st kind, is the integral of a scalar ($\mathbb{R}^2 \to \mathbb{R}$, say) function, such as the one used in your question to calculate the area of a fence. Another interpretation is the mass of a wire with variable density $f(x,y,z)$.
A line integral of the 2nd kind, is the integral of a vector function ($\mathbb{R}^2 \to \mathbb{R}^2$), which you'd use to calculate e.g. the work done by a force acting along the line.
The way you calculate a 2nd kind integral, say of a (vector) function $\vec{F}(x,y)$ along a curve $\vec{r}(t)$, where $a\le t\le b$ is this: $\int_a^b \vec{F}(\vec{r}(t)) \cdot \overrightarrow{r'}(t)\, dt$. What we did is translate a 2nd-kind line integral into an ordinary integral. Here $\vec{r'}(t)$ is the particle's velocity. It can also be viewed as a tangent vector to the path, at $\vec{r}(t)$.