Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.1 - Line Integrals - Exercises 16.1 - Page 944: 19


Part a: $4\sqrt5$ Part b: $\frac{1}{12}(17\sqrt{17}-1)$

Work Step by Step

The question asks us to find the integral $\int_{C}xds$, where C is two different curves. For part a, curve $C$ is a straight line segment given by $x=t$ and $y=\frac{t}{2}$ from $(0,0)$ to $(4,2)$. Thus, the parametrization of $C$ is $r(t)=ti+\frac{t}{2}j$. We then must find $|v(t)|$, which is the derivative of r above. $|v(t)|=|1i+\frac{1}{2}|=\frac{\sqrt{5}}{2}$ Thus, combining everything: $\int_{C}xds=\int_{0}^{4}{t \cdot \frac{\sqrt{5}}{2}}dt=4\sqrt5$ For part b, a similar approach is taken for the new curve given: $(x,y)=(t,t^2)$ from $(0,0)$ to $(2,4)$ $f(x,y)=x$ $|v(t)|=\sqrt{(1)^2+(2t)^2}=\sqrt{4t^2+1}$ Combining all that together: $\int_C x ds= \int_{0}^{2}t\sqrt{4t^2+1}dt=\frac{1}{12}(17\sqrt{17}-1)$ Where we used u-substitution ($u=4t^2+1$) to integrate.
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