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Line 18: − =Generalized Integration= − − =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 <math>C_k, 1\leq k \leq n</math>, and we pick a point <math>(x_k,y_k)\in C_k</math> for each piece. Letting the length of <math>C_k</math> be <math>L_k</math>. The definition follows as expected: − − <math>\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 </math> − − ==Computation== − [[Image:Arc length.jpg|right|197px|thumb|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 <math>x=x(t),y=y(t) </math> be a parametric description of <math>C </math>, so that ''C'' starts <math>x(t_0),y(t_0)</math> and ends at <math>x(t_1),y(t_1)</math>. This allows us to compute the length element, ''dL'': − − <math>dL = \sqrt{dx^2+dy^2} </math> − − But we don't have any formulas for ''dx'' or ''dy''! The closest we've got are formulas for <math>dx/dt, dy/dt</math>; we'll have to plug these in and see if we can patch things up: − − <math>= \sqrt{ \left({ \frac{dx}{dt} }\right)^2 + \left({ \frac{dy}{dt} }\right)^2 } \sqrt{dt^2} </math> − − Now we can just plug in our formula for ''dL'' into the integral to get an easy-to-use formula for a line integral: − − <math>\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 </math> − − =Green's Theorem in the Plane= − − =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 <math>S_k, 1\leq k \leq n</math>, and we pick a point <math>(x_k,y_k,z_k)\in S_k</math> for each piece. Letting the area of <math>S_k</math> be <math>A_k</math>. The definition follows as expected: − − <math>\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 </math> − − ==Computation== − − Let <math>x=x(u,v),y=y(u,v),z=z(u,v) </math> be a parametric description of <math>S </math>, 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. − [[Image:Surface element.jpg|right|thumb|200px|The surface area element and the vectors <math>\vec{u},\vec{v}</math>.]] − 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 <math>\vec{u}\times\vec{v}</math> as having length equal to the area of the parallelogram bordered on two sides by <math>\vec{u}</math> and <math>\vec{v}</math>, 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, <math>\vec{u}, \vec{v}</math>, and we'll be able to compute the surface integral as − − <math>\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 </math> − − 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 <math>S_k \ </math> to be in line with the curves of constant ''u'' and curves of constant ''v'' on the surface, these small components will then be − − <math>\vec{u}=\begin{pmatrix}\partial_u x\\\partial_u y\\\partial_u z\end{pmatrix}</math>, <math>\vec{v}=\begin{pmatrix}\partial_v x\\\partial_v y\\\partial_v z\end{pmatrix}</math>
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