Science:Math Exam Resources/Courses/MATH101/April 2005/Question 02 (d)

MATH101 April 2005

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[hide] Question 02 (d)
Let $\displaystyle R$ be the finite region bounded by the curve $\displaystyle y=4-x^{2}$ and bounded below by $\displaystyle y=2-x$ . Express the length of the upper curve that bounds $\displaystyle R$ as a definite integral, and using an appropriate substitution express your answer as an integral involving trigonometric functions. You do not need to evaluate this trigonometric integral.

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[show] Hint
What is the formula for computing the length of a curve? (see arclength in your textbook/notes if you do not remember).

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[show] Solution
From part c), we found that the integral needed to be taken from $x=-1$ to $x=2$ . Thus we need to evaluate the arclength of the curve $y=4-x^{2}$ in between those two points. We recall the formula for arclength is given by $l=\int _{a}^{b}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx.$ Making the appropriate substitutions for our situation we get: $l=\int _{-1}^{2}{\sqrt {1+(-2x)^{2}}}\,dx=\int _{-1}^{2}{\sqrt {1+4x^{2}}}\,dx$ Now we need to choose the appropriate trigonometric substitution. Making the substitution $\tan(\theta )=2x$ , allows the integral to be more easily evaluated in terms of trig functions: ${\begin{aligned}l&=\int _{-1}^{2}{\sqrt {1+4x^{2}}}\,dx\\&=\int _{\arctan(-2)}^{\arctan(4)}{\sqrt {1+\tan ^{2}(\theta )}}\cdot {\frac {1}{2}}\sec ^{2}(\theta )\,d\theta \\&=\int _{\arctan(-2)}^{\arctan(4)}\sec(\theta )\cdot {\frac {1}{2}}\sec ^{2}(\theta )\,d\theta \\&={\frac {1}{2}}\int _{\arctan(-2)}^{\arctan(4)}\sec ^{3}(\theta )\,d\theta .\end{aligned}}$