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

MATH101 April 2005

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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.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: ${\begin{aligned}l&=\int _{-1}^{2}{\sqrt {1+4x^{2}}}\,dx\\&=\int _{\arctan(-2)}^{\arctan(4)}{\sqrt {1+\tan ^{2}(\theta )}}\cdot {\frac {1}{2}}\sec ^{2}(\theta )\,d\theta \\&=\int _{\arctan(-2)}^{\arctan(4)}\sec(\theta )\cdot {\frac {1}{2}}\sec ^{2}(\theta )\,d\theta \\&={\frac {1}{2}}\int _{\arctan(-2)}^{\arctan(4)}\sec ^{3}(\theta )\,d\theta .\end{aligned}}$

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