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

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

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

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

For B we have

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

And so since y = -2/x we have

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

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

$\begin{array}{rl}\text{area}& =\underset{0}{\overset{\frac{1}{\sqrt[3]{4}}}{\int }}((-x)-(-4\sqrt{x}))dx+\underset{\frac{1}{\sqrt[3]{4}}}{\overset{\sqrt{2}}{\int }}((-x)-(-\frac{2}{x}))dx\\ & =\underset{0}{\overset{\frac{1}{\sqrt[3]{4}}}{\int }}(-x+4\sqrt{x})dx+\underset{\frac{1}{\sqrt[3]{4}}}{\overset{\sqrt{2}}{\int }}(-x+\frac{2}{x})dx\\ & ={[-\frac{x^2}{2}+\frac{4\cdot 2}{3}{x}^{3/2}]}_0^{\frac{1}{\sqrt[3]{4}}}+{[-\frac{x^2}{2}+2\ln (x)]}_{\frac{1}{\sqrt[3]{4}}}^{\sqrt{2}}\\ & =-\frac{(2^{-2/3}{)}^2}{2}+\frac{8}{3}(2^{-2/3}{)}^{3/2}+0-0-\frac{(\sqrt{2}{)}^2}{2}+2\ln (\sqrt{2})+\frac{(2^{-2/3}{)}^2}{2}-2\ln (2^{-2/3})\\ & =\frac{4}{3}-1+\ln (2)+\frac{4}{3}\ln (2)\\ & =\frac{1}{3}+\frac{7}{3}\ln (2)\end{array}$