Science:Math Exam Resources/Courses/MATH105/April 2010/Question 03/Solution 1

First, we sketch each of the three curves to see what is the region that we are talking about.

We will need to find the coordinates of points A and B which are the intersection of curves. For A we have

${\begin{aligned}-x&=-{\frac {2}{x}}\\x^{2}&=2\\x&=\pm {\sqrt {2}}\end{aligned}}$

We can see from the picture that we are interested in the point with a positive x coordinate, so

$\displaystyle A=({\sqrt {2}},-{\sqrt {2}})$

For B we have

${\begin{aligned}-4{\sqrt {x}}&=-{\frac {2}{x}}\\x^{3/2}&={\frac {1}{2}}\\x&=\left({\frac {1}{2}}\right)^{2/3}={\frac {1}{\sqrt[{3}]{4}}}\end{aligned}}$

And so since y = -2/x we have

$\displaystyle B=\left({\frac {1}{\sqrt[{3}]{4}}},-2{\sqrt[{3}]{4}}\right)$

This allows us to write the desired area as the sum of 2 integrals

${\begin{aligned}{\text{area}}&=\int _{0}^{\frac {1}{\sqrt[{3}]{4}}}\left((-x)-(-4{\sqrt {x}})\right)\,dx+\int _{\frac {1}{\sqrt[{3}]{4}}}^{\sqrt {2}}\left((-x)-(-{\frac {2}{x}})\right)\,dx\\&=\int _{0}^{\frac {1}{\sqrt[{3}]{4}}}\left(-x+4{\sqrt {x}}\right)\,dx+\int _{\frac {1}{\sqrt[{3}]{4}}}^{\sqrt {2}}\left(-x+{\frac {2}{x}}\right)\,dx\\&=\left[-{\frac {x^{2}}{2}}+{\frac {4\cdot 2}{3}}x^{3/2}\right]_{0}^{\frac {1}{\sqrt[{3}]{4}}}+\left[-{\frac {x^{2}}{2}}+2\ln(x)\right]_{\frac {1}{\sqrt[{3}]{4}}}^{\sqrt {2}}\\&=-{\frac {(2^{-2/3})^{2}}{2}}+{\frac {8}{3}}(2^{-2/3})^{3/2}+0-0-{\frac {({\sqrt {2}})^{2}}{2}}+2\ln({\sqrt {2}})+{\frac {(2^{-2/3})^{2}}{2}}-2\ln(2^{-2/3})\\&={\frac {4}{3}}-1+\ln(2)+{\frac {4}{3}}\ln(2)\\&={\frac {1}{3}}+{\frac {7}{3}}\ln(2)\end{aligned}}$