The following problems involve the computation of arc length of differentiable functions on closed intervals. Let's first begin by finding a general formula for computing arc length. Consider a graph of a function of unknown length $L$ which can be represented as $ y=f(x) $ for $ a \le x \le b $ or $ x=g(y) $ for $ c \le y \le d $. We will derive the arc length formula using the differential of arc length, $ ds $, a small change in arc length $s$, and write $ds$ in terms of $dx$, the differential of $x$, and $dy$, the differential of $y$ (See the graph below.).


Using the Pythagorean Theorem we will assume that $$ (ds)^2 = (dx)^2 + (dy)^2 $$ so that $$ ds = \sqrt{ (dx)^2 + (dy)^2 } $$ It then follows that the total arc length $L$ from $x=a$ to $x=b$ is $$ ARC = \displaystyle{ \int_{s=0}^{s=L} 1 \ ds } = \displaystyle{ \int_{x=a}^{x=b} \sqrt{ (dx)^2 + (dy)^2 } } $$ $$ = \displaystyle{ \int_{a}^{b} \sqrt{ \Bigg(1 + {(dy)^2 \over (dx)^2}\Bigg) (dx)^2} } $$ $$ = \displaystyle{ \int_{a}^{b} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } dx } $$ i.e., $$ ARC = \displaystyle{ \int_{a}^{b} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } dx } $$ Similarly, it follows that the total arc length $L$ from $y=c$ to $y=d$ is $$ ARC = \displaystyle{ \int_{s=0}^{s=L} 1 \ ds } = \displaystyle{ \int_{y=c}^{y=d} \sqrt{ (dx)^2 + (dy)^2 } } $$ $$ = \displaystyle{ \int_{c}^{d} \sqrt{ \Bigg({(dx)^2 \over (dy)^2} + 1 \Bigg) (dy)^2} } $$ $$ = \displaystyle{ \int_{c}^{d} \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } dy } $$ i.e., $$ ARC = \displaystyle{ \int_{c}^{d} \sqrt{ 1 + \Big({dx \over dy}\Big)^2 } dy } $$

Most of the following problems are average. But because the arc length formula includes a square root, most problems will require relatively intense and very careful algebraic simplification, including manipulation of fractions and creation of perfect squares.

Compute the area of the region enclosed by the graphs of the given equations. Use vertical cross-sections on Problems 1-16.

Click HERE to return to the original list of various types of calculus problems.

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A heartfelt "Thank you" goes to The MathJax Consortium and the online Desmos Grapher for making the construction of graphs and this webpage fun and easy.

Duane Kouba ... May 3, 2017