Evaluate fc 2x ds, where consists of the arc of the parabola y= x2 from to followed by the vertical line segment from (1, 1) to(1, 2)

The given line integral is ∫C2xdS where C is the arc of the parabola y=x2 from (0,0) to (1,1) and…

Answered: Evaluate fc 2x ds, where consists of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Evaluate f_c 2x ds, where consists of the arc of the parabola y= x² from to followed by the vertical line segment from (1, 1) to
(1, 2)

Expert Solution

Step 1

The given line integral is $\int_{C} 2 x d S$ where C is the arc of the parabola $y = x^{2}$ from (0,0) to (1,1) and followed by vertical line segment (1,1) to (1,2)

Now, let $C_{1} : y = x^{2}$ can be parametrize by $x = t a n d y = t^{2}$ where $0 \leq t \leq 1$

So, we get

$\begin{array}{rcl} \int_{C_{1}} 2 x d S & = & \int_{t = 0}^{1} 2 t \sqrt{1 + {(2 t)}^{2}} d t \\ = & \int_{t = 0}^{1} 2 t \sqrt{1 + 4 t^{2}} d t \end{array}$

we put $1 + 4 t^{2} = u \Rightarrow 8 t d t = d u$

$\begin{array}{rcl} \int_{t = 0}^{1} 2 t \sqrt{1 + 4 t^{2}} d t & = & \frac{1}{4} \int_{u = 1}^{5} u^{\frac{1}{2}} d u \\ = & = \frac{1}{4} {(\frac{u^{\frac{3}{2}}}{\frac{3}{2}})}_{1}^{5} \\ = & \frac{1}{6} (5 \sqrt{5} - 1) \end{array}$

Finally we can write

$\begin{array}{rcl} \int_{C_{1}} 2 x d S & = & \frac{1}{6} (5 \sqrt{5} - 1) \end{array}$

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