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: 33

Answer

$$2\sqrt2 - 1$$

Work Step by Step

We set up the integral to find the mass: $$M=\int_C \delta (t) |\mathbf{v}(t)|dt$$ where C is given by the curve $$\mathbf{r}(t)=t^2\mathbf{i}+2t\mathbf{j}$$ and the $t$ range is $$0 \leq t \leq 1$$ Thus, we have: $$M=\int_{0}^{1}\frac{3t}{2}\sqrt{(2t-0)^2+(2)^2}dt=3\int_{0}^{1}t\sqrt{t^2+1}dt=\frac{3}{2}\int_{1}^{2}u^{1/2}=\frac{3}{2}\frac{2}{3}(2\sqrt{2}-1)=2\sqrt{2}-1$$
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