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I recently came across an AP Calculus question:

The length of the arc of the parabola $4x=y2$ cut off by the line $x=2$ is given by the integral

a)$\int_{- 1}^1 \sqrt( x^2 +1 ) dx $

b) $\int_{- 1}^1 \sqrt( 1+x ) dx$

c)$ 1/2 \int _0^2\sqrt( 4+y^2 ) dy $

d) $\int _{-1}^1\sqrt( 4+y^2 ) dy $

My answer:

The parabola is symmetric to the $x$-axis and for$ x=2, y = +-\sqrt(8)$

Which means the integral should be:

$2\int_{-2\sqrt(2)}^0 {\sqrt( 4 + y^2 ) dy}$] $ But that isn't even an option. Where am I going wrong?

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    use $$\LaTeX$$ please! – 2017-01-13
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    You are finding the area cut off, not the arc length cut off. – 2017-01-13

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The answer will be option (C). To find arc lengths, you should use Pythagoras theorem to differential triangles with sides dx and dy, and then substitute dy/dx from given function.

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    That is what I did. Also the answer at the back of the book is option (d). – 2017-01-29