## Related questions with answers

Question

In this exercise, find a definite integral that represents the arc length of the curve over the indicated interval and observe that the integral cannot be evaluated with the techniques studied so far

$x=\sqrt{36-y^2},\quad0\leq y\leq3$

Solution

VerifiedAnswered 2 years ago

Answered 2 years ago

Step 1

1 of 3To find the arc length of the function we first need to derivate the function one time:

$f'(y)=\frac{-y}{\sqrt{36-y^2}}$

The area of integration will be: $0<y<3$

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