Science:Math Exam Resources/Courses/MATH103/April 2010/Question 08

MATH103 April 2010

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Question 08
The graph of the function $y=f(x)=x(1-x),\quad 0\leq x\leq 0.5$ is rotated about the y axis to form a trumpet shape whose base is at (0,0). Determine the volume of fluid that can be contained inside this shape.

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
(Note: This hint was given on the exam question itself.) One way to solve this problem is to find the function $x=g(y)$ that describes the same curve and use it to set up the appropriate integral.)

Hint 2
The formula for the disc method, for a function y = f(x) rotated about the y-axis, is $V=\int _{y_{\text{min}}}^{y_{\text{max}}}\pi \left(f^{-1}(y)\right)^{2}\,dy.$

Hint 3
The formula for the shell method, for a function y = f(x) rotated about the y-axis, is $V=\int _{x_{\text{min}}}^{x_{\text{max}}}2\pi xh(x)\,dx,$ where h(x) is the height of the shell at x.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We want to use the disk method and so we need to take the inverse of the function $\ f(x)=x(1-x).$ $f(x)=y=x-x^{2}\$ $x^{2}-x+y=0\$ $x_{1/2}={\frac {1\pm {\sqrt {1-4y}}}{2}}$ We get two possible inverse functions $f_{1}^{-1}(y)={\frac {1}{2}}+{\frac {1}{2}}{\sqrt {1-4y}}\$ and $f_{2}^{-1}(y)={\frac {1}{2}}-{\frac {1}{2}}{\sqrt {1-4y}}.\$ On the picture we see that for our purpose we need the function $f_{2}^{-1}$. To find the boundaries for the integral we calculate $f(x=0)=0,\qquad f\left(x={\frac {1}{2}}\right)={\frac {1}{4}}.$ Now we use the formula for the disk method $V=\int \pi r(z)^{2}dz\$. Here $\ V$ is the volume and $\ r$ is the function of the radius. We need the function $f_{2}^{-1}$ for the radius. ${\begin{aligned}V&=\int _{0}^{\frac {1}{4}}\pi \left({\frac {1}{2}}-{\frac {\sqrt {1-4y}}{2}}\right)^{2}\,dy\\&=\pi \int _{0}^{\frac {1}{4}}\left({\frac {1}{4}}-{\frac {1}{2}}{\sqrt {1-4y}}+{\frac {1-4y}{4}}\right)\,dy\\&=\pi \int _{0}^{\frac {1}{4}}\left({\frac {1}{2}}-{\frac {1}{2}}{\sqrt {1-4y}}-y\right)\,dy\\&={\frac {\pi }{2}}\int _{0}^{\frac {1}{4}}\left(1-(1-4y)^{\frac {1}{2}}-2y\right)\,dy\\&={\frac {\pi }{2}}\left[y-{\frac {2}{3}}(1-4y)^{\frac {3}{2}}\left(-{\frac {1}{4}}\right)-y^{2}\right]_{0}^{\frac {1}{4}}\\&={\frac {\pi }{2}}\left({\frac {1}{4}}+{\frac {1}{6}}\left(1-4(1/4)\right)^{\frac {3}{2}}-(1/4)^{2}-0-{\frac {1}{6}}(1-4(0))^{\frac {3}{2}}+0^{2}\right)\\&={\frac {\pi }{2}}\left({\frac {1}{4}}-{\frac {1}{16}}-{\frac {1}{6}}\right)\\&={\frac {\pi }{96}}\end{aligned}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For this problem it is much easier to use the shell method. For using the shell method we need to know the height $\ h$ of the trumpet for every $\ x\in [0,{\frac {1}{2}}].$ This is $\ h(x)=f(1/2)-f(x)={\frac {1}{4}}-x(1-x).$ Now we can easily integrate ${\begin{aligned}V&=\int _{0}^{\frac {1}{2}}2\pi xh(x)dx\\&=2\pi \int _{0}^{\frac {1}{2}}x\left({\frac {1}{4}}-x+x^{2}\right)dx\\&=2\pi \left[{\frac {1}{4}}{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}\right]_{0}^{\frac {1}{2}}\\&=2\pi \left[{\frac {1}{4}}{\frac {1}{8}}-{\frac {1}{8}}{\frac {1}{3}}+{\frac {1}{16}}{\frac {1}{4}}-0\right]\\&={\frac {\pi }{96}}\end{aligned}}$

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