Question

In Exercises \(15-22,\) (a) graph the function, highlighting the part indicated by the given interval, (b) 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, and (c) use the integration capabilities of a graphing utility to approximate the are length. $$ x=\sqrt{36-y^{2}}, \quad 0 \leq y \leq 3 $$

Short answer

The function is a semicircle with radius 6 and the arc length over the interval \([0, 3]\) is represented by the integral \(L = \int_0^3 \sqrt{1 + \left(\frac{-x}{\sqrt{36 - x^2}}\right)^2} \, dx\). Although this integral cannot be solved by standard techniques, a graphing utility can be used to approximate its value.

Step-by-step solution

Rearrange and Identify Function to be Graphed

Rearrange \(x=\sqrt{36-y^{2}}\) for \(y\). This results in \(y = \sqrt{36 - x^2}\). This function represents a semi-circle with a radius of 6, which will be drawn with the relevant part shaded in.

Derive Integral for Arc Length

The formula for the arc length of a curve \(y = f(x)\) from \(x=a\) to \(x=b\) is \(L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx\). First take the derivative of the function \(f'(x)= -x/ \sqrt{36 - x^2}\), to substitute into the formula. This results in \(L = \int_0^3 \sqrt{1 + \left(\frac{-x}{\sqrt{36 - x^2}}\right)^2} \, dx\). As noted in the exercise, this integral cannot be computed using commonly known techniques, but it correctly represents the arc length of the curve on the interval \([0, 3]\).

Estimate Arc Length with Graphing Utility

You will now use a graphing utility to approximate the arc length. Plot the function \(y = \sqrt{36 - x^2}\) and use the utility’s integration feature to compute the definite integral obtained in Step 2, over the interval \([0, 3]\). This will give you an estimate of the arc length.