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

MATH101 April 2014

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Question 03
Long Problem. Show your work. No credit will be given for the answer without the correct accompanying work. Find the volume of the solid generated by rotating the finite region bounded by $\displaystyle y=1/x$ and $\displaystyle 3x+3y=10$ about the x-axis. It will be useful to sketch the region first.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
How can we express the volume generated by rotating a single function, $f(x)$, around the x-axis?

Hint 2
If we think of the volume generated by each function separately, how can we isolate the volume between the functions?

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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. The given region is the following: . 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 $V=V_{1}-V_{2}={\frac {512}{81}}\pi .$

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