Find the arc-length of a circle with radius a.From the equation of a circle, I found out the equation for the one quadrant, which is: y = a 2 − x 2 I tried solving the problem, and here's what I did: A = 4 ∫ 0 a 1 + x 2 a 2 − x 2 d x A = 4 ∫ 0 a a a 2 − x 2 d xAnd this is where I need help. How can solve this integral now?

Question

Find the arc-length of a circle with radius a.From the equation of a circle, I found out the equation for the one quadrant, which is:  y  =            a      2        −          x      2      I tried solving the problem, and here&#039;s what I did:  A  =  4      ∫    0    a        1    +                  x        2                              a          2                −                  x          2                      d  x  A  =  4      ∫    0    a        a                            a          2                −                  x          2                      d  xAnd this is where I need help. How can solve this integral now?

Kaiden Porter · Accepted Answer

Fix the problem, it should be   y  =            a      2        −          x      2       as i edited it. Then use the substitution   x  =  a  sin  ⁡  θ

Find the arc-length of a circle with radius a. From the equation of a circle, I found out the equat

Answered question

Answer & Explanation

New Questions in High school geometry