Science:Math Exam Resources/Courses/MATH101/April 2014/Question 03/Solution 1

The given region is the following:

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We can find the bounds of the region by finding the intersection points of two graphs. By substituting $y={\frac {1}{x}}$ into $3x+3y=10$, we have

${\begin{aligned}3x+{\frac {3}{x}}=&10\\3x^{2}+3=&10x\\3x^{2}-10x+3=&0\\(3x-1)(x-3)=&0.\end{aligned}}$

Thus, at $x={\frac {1}{3}}$ and $x=3$, the two functions intersect and therefore the region to generate volume is bounded on $1/3\leq x\leq 3$. We will compute the volume of the region by determining the volume generated by the wider function ($3x+3y=10$) on its own and subtracting the volume generated by the narrower function ($y=1/x$). For a general function given by $y=f(x)$ on $a\leq x\leq b$, we think of the volume generated around the x-axis by integrating small cylinders with height $dx$ and radius $r=f(x)$. The volume generated by this is

$V=\int _{a}^{b}\pi f(x)^{2}{\textrm {d}}x.$

For the function $3x+3y=10$, we need to write this as $y=f(x)$ to get $y=10/3-x$. The volume generated by this first function is therefore

$V_{1}=\pi \int _{\frac {1}{3}}^{3}\left({\frac {10}{3}}-x\right)^{2}{\textrm {d}}x=\pi \int _{\frac {1}{3}}^{3}{\frac {100}{9}}-{\frac {20}{3}}x+x^{2}{\textrm {d}}x=\pi \left[{\frac {100}{9}}x-{\frac {10}{3}}x^{2}+{\frac {1}{3}}x^{3}\right]_{\frac {1}{3}}^{3}={\frac {728}{81}}\pi .$

For the second function $y=1/x$ the volume generated is

$V_{2}=\pi \int _{\frac {1}{3}}^{3}\left({\frac {1}{x}}\right)^{2}{\textrm {d}}x=-\pi \left[{\frac {1}{x}}\right]_{\frac {1}{3}}^{3}={\frac {8}{3}}\pi .$

Therefore, the volume generated by the region between these two functions is

$V=V_{1}-V_{2}={\frac {512}{81}}\pi .$