Math 275 Notes Topic 5.1: Line Element and Scalar Line Integrals Textbook Section: 16.2 More Details on Line Elements (vector dr, and scalar ds): http://www.math.oregonstate.edu/bridgebook/book/math/drvec From the Toolbox (what you need from previous classes): alc I/II: Evaluating integrals b a f (t) dt. omputing differentials du. Recall: this is done as a part of u- substitution: if u = f (t), then du = f (t) dt. alc III: Parameterizing a curve using either coordinate functions (aka parametric equations) or a vector function: x = x(t), y = y(t), z = z(t) or r(t) = x(t) î + y(t) ĵ + z(t) k omputing the derivative r (t) of a vector function r(t), and computing its magnitude r (t). Recall: if r(t) represents a curve, then r (t) is a tangent vector to the curve. Learning Objectives (New Skills) & Important oncepts Learning Objectives (New Skills): ompute the vector line element dr and the scalar line element ds along a curve. Set up and evaluate scalar line integrals f ds. Use scalar line integrals to compute the length of a curve, and the mass of 1-d objects.
Important oncepts: The vector differential dr = dx î +dy ĵ +dz k is an infinitesimal displacement vector. Its magnitude is ds = dr = dx 2 + dy 2 + dz 2 is an infinitesimal version of the Pythagorean theorem. Hence, dr represents an infinitesimal change in position, and ds provides way to measure distance over infinitesimal increments. Given a parameterization r(t) for a curve : The vector line element is the vector differential dr = r (t) dt. Since dr is a multiple of the tangent vector r (t), dr is tangent to the curve. The vector line element represents an infinitesimal change in position over a curve (or: the distance traveled along the curve in an infinitesimal time increment). The scalar line element (or element of arc length) is the magnitude of the vector line element ds = dr = r (t) dt). This gives a way of measuring distance along a curve. A scalar line integral f ds is a one-dimensional integral. The region of integration is a curve. The integrand f of a scalar line integral is a scalar-valued function, for example: z = f (x, y) or w = f (x, y, z). The Big Picture The vector line element dr is the differential of the position vector r. It represents an infinitesimal change in position along the curve parameterized by r. If r(t) is a vector function parameterizing a curve, then dr = r(t) dt. Since this is a multiple of the tangent vector r (t), dr is also tangent to the curve parameterized by r(t). The scalar line element ds = dr is the infinitesimal yardstick. It gives a way of measuring distance along along a curve. A scalar line integral f ds is a one-dimensional integral where the region of integration is a curve, the integrand is a scalar-valued function f, and the differential is the scalar line element ds. 2
How to Read a Scalar Line Integral Scalar line integrals: f ds is the region (or domain) of integration. is a curve. In this class, is a curve in R 2 or R 3. f is a scalar-valued function. In this class, f will be a function of two or three variables: z = f (x, y) or w = f (x, y, z). ds is the scalar line element. ds represents an infinitesimal distance. In R 3 : ds = dx 2 + dy 2 + dz 2 Evaluating Scalar Line Integrals Using a Parametrization A common method for evaluating scalar line integrals involves using a parameterization for the curve: Find a Parameterization for the urve: Describe the curve using either coordinate functions (aka parametric equations) or a vector function: x = x(t), y = y(t), z = z(t) or r(t) = x(t) î + y(t) ĵ + z(t) k Find Limits of Integration: These are the values of the parameter t that mark the endpoints of the curve : a t b ompute ds: ds = r (t) dt Restrict the Integrand f to the urve : Replace x, y, and z in the integrand with the coordinate functions x(t), y(t), z(t): f (x, y, z) = f ( x(t), y(t), z(t) ). 3
Set Up the Scalar Line Integral: b f (x, y, z) ds = f ( x(t), y(t), z(t) ) r (t) dt a The integral on the right is an integral of a single variable (the parameter t), which can be evaluated using methods learned in alc I/II. More Details If ds = dx 2 + dy 2 + dz 2, where does the computational formula ds = r (t) dt come from? Suppose a curve is described by the vector function r(t) = x(t) î + y(t) ĵ + z(t) k. Then, along this curve : (d [ ]) 2 ( [ ]) 2 ( [ ]) 2 ds = x(t) + d y(t) + d z(t) From using u-substitution to compute integrals, we know that, if u = f (t), then du = f (t) dt, so: (d [ ]) 2 ( [ ]) 2 ( [ ]) 2 ds = x(t) + d y(t) + d z(t) [x = (t) dt ] 2 [ + y (t) dt ] 2 [ + z (t) dt ] 2 = [x (t) ] 2 [ + y (t) ] 2 [ + z (t) ] 2 dt = r (t) dt Some Applications of Scalar Line Integrals All applications of integrals begin with the idea that an integral works by chopping up and adding. For scalar line integrals, the region of integration (the curve ) is chopped up into infinitely many infinitesimal (very very small) segments of length ds. The adding up is accomplished by integrating f ds over the curve. Scalar Line Integrals and Arc Length hop up the curve into infinitesimal segments of length ds. To find the total length of, add up the lengths of all these segments: s = ds 4
Scalar Line Integrals and Mass hop up the curve into infinitesimal segments of length ds. If δ 0 represents the line density (density per unit length) at each point along, then the mass of an infinitesimal segment is δ ds. To find the total mass m of, add up the masses of all the segments by integrating δ ds over : m = δ ds Technical Details: Scalar Line Integrals as Limits of Riemann Sums Integration is a process of chopping and adding. integral f ds: For a scalar line The curve is chopped into small segments. The length of a segment is s. The function f is evaluated at a point P i in each segment, and multiplied by the length of the segment: f (P i ) s. These products are added in a Riemann sum: f ds n f (P 1 ) s The value of the line integral is given by the limit (if it exists) as n, in such a way that s 0: i=1 f ds = lim n n f (P 1 ) s Working under the infinite magnifying glass, we add up (integrate) infinitely many of the infinitesimal quantities f ds. i=1 5