Engineering Calculus Notes 220

Engineering Calculus Notes 220 - 208 CHAPTER 2. CURVES then...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 208 CHAPTER 2. CURVES then the element of arclength is given by ˙ (x)2 + (y )2 dx ˙ ds = = 1 + (2x)2 dx = 1 + 4x2 dx so 1 (x)( x ds = 1 + 4x2 dx) 0 C which we can do using the substitution u = 1 + 4x2 du = 8x dx 1 x dx = du 8 x=0↔u=1 x=1↔u=5 and 1 x 1 5 1/2 u du 81 5 1 = u3/2 12 1 √ 5 5−1 = . 12 1 + 4x2 dx = 0 If we try to use the same parametrization to find the second integral, we have x2 ds y ds = C C = 0 1 x2 1 + 4x2 dx ...
View Full Document

This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online