Calc3.8

The Line Integral

Definition

We've integrated single variable functions along lines, and two variable functions over regions of the plane, and even three variable functions over regions of space. But lines, and curves, can be contained in the plane, and surfaces can be contained in space. Why have we been restricting ourselves to integrating functions over regions which have the same dimensions as their domains?

Well, because it is simpler, and useful to have under our belts before we expand our repetoire. But there is no reason we can't integrate a function over a region in its domain which is a lower dimension than the domain itself.

We'll start with the line integral. Technically, it should be called the curve integral, but let's let that slide for now and get to the definition. We're trying to integrate over a curve, say, C. At first, we'll let this curve lie in the xy plane - it's not difficult to generalize to three dimensions. So we proceed as before: we divide this region into n pieces $C_{k},1\leq k\leq n$, and we pick a point $(x_{k},y_{k})\in C_{k}$ for each piece. Letting the length of $C_{k}$ be $L_{k}$. The definition follows as expected:

$\int _{C}f(x,y)dL=\lim _{L_{k}\to 0,n\to \infty }\sum _{k=1}^{n}f(x_{k},y_{k})L_{k}$

Computation

As the length goes to zero, the Pythagorean theorem provides a more and more precise result for the length element dL.

As before, we're not going to be using the Riemann sum definition to compute line integrals. Let $x=x(t),y=y(t)$ be a parametric description of $C$, so that C starts $x(t_{0}),y(t_{0})$ and ends at $x(t_{1}),y(t_{1})$. This allows us to compute the length element, dL:

$dL={\sqrt {dx^{2}+dy^{2}}}$

But we don't have any formulas for dx or dy! The closest we've got are formulas for $dx/dt,dy/dt$; we'll have to plug these in and see if we can patch things up:

$={\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}{\sqrt {dt^{2}}}$

Now we can just plug in our formula for dL into the integral to get an easy-to-use formula for a line integral:

$\int _{C}f(x,y)dL=\int _{t_{0}}^{t_{1}}f(x,y){\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}dt$

The Surface Integral

Definition

Now we turn to the idea of integrating a three-variable function over a surface, S. This is called a surface integral. As we always do, we divide this region into n pieces $S_{k},1\leq k\leq n$, and we pick a point $(x_{k},y_{k},z_{k})\in S_{k}$ for each piece. Letting the area of $S_{k}$ be $A_{k}$. The definition follows as expected:

$\iint _{S}f(x,y,z)dL=\lim _{A_{k}\to 0,n\to \infty }\sum _{k=1}^{n}f(x_{k},y_{k},z_{k})A_{k}$

Computation

Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=x(u,v),y=y(u,v),z=z(u,v) } be a parametric description of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S } , so that S will be defined as the image under these functions these three functions x,y,z of some region R of the uv plane.

The surface area element and the vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u},\vec{v}} .

As before, the trick to finding a convenient formula for a surface integral is finding a formula for that surface area element, dA. What does such a surface element even look like?

You'll remember that in our second lecture, we defined the cross product of two vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}\times\vec{v}} as having length equal to the area of the parallelogram bordered on two sides by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v}} , and direction perpendicular to either vector, with a right handed orientation. We can use this to our advantage now - all we need to do is figure out two vectors which will border a surface area element, say, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}, \vec{v}} , and we'll be able to compute the surface integral as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \iint_S f(x,y,z)dA = \iint_R f(x(u,v),y(u,v),z(u,v)) |\vec{u}\times\vec{v}| dudv }

It's easy enough to do this: each border vector will have some small x component, some small y component, and some small z component. By choosing our surface elements Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_k \ } to be in line with the curves of constant u and curves of constant v on the surface, these small components will then be

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}=\begin{pmatrix}\partial_u x\\\partial_u y\\\partial_u z\end{pmatrix}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v}=\begin{pmatrix}\partial_v x\\\partial_v y\\\partial_v z\end{pmatrix}}