Weekly Challenge 28: The right volume

6495 /www/nrich/html/content/id/6495/ 1 2011-03-15T10:10:20 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/898&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/6426&part=">Try this next </a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html?1274115048"> Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/1408&part=">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7066&part=solution">Last week's solution</a> </li> </ul> <div> A segment of the curve $y=f(x)$ starts at $(0, 0)$ and ends at $(1, 1)$. It is rotated about the $x$ axis to form a volume. Can you find a curve which will result in a volume of $1$? </div> <div class="framework"> Did you know ... ? That volumes of revolution are often used in mechanics calculations. They were used by Archimedes around 250 BC to calculate volumes of solids created by rotating conic sections about an axis.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> If we suppose that the curve $y=f(x)$ is integrable then the volume so created will be $$ V = \int^1_0 \pi y^2 dx $$ To get a feel for the sort of curve we might need, first consider the special case $y=x$, which clearly passes through the two points. Then, $$ V = \int^1_0 \pi x^2 = \pi\left[\frac{x^3}{3}\right]^1_0 = \frac{\pi}{3} $$ This is slightly larger than $1$, so we could consider a family of curves which at beneath $y=x$ with enough flexibility to all us to vary the final value of the volume. A simple choice is parabolas $y=Ax(x-1)$ for some multiplicative constant. These give $$ V = \int^1_0 \pi Ax^2(x-1)^2dx $$ Now that I see it, I'm not too keen on doing this integral, so I'm going to consider instead $y=A\sqrt{|x(x-1)|}$ This gives a volume $$ V = \int^1_0 \pi A x(1-x) dx = \pi A\left[\frac{x^2}{2}-\frac{x^3}{3}\right]^1_0 = \pi A \left[\frac{1}{2}-\frac{1}{3}\right] = \frac{\pi A}{6} $$ Thus, a curve which has the required properties is $$ y = \frac{6}{\pi}\sqrt{|x(x-1)|} $$ There are, of course, others! </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Curves such as $y=x, y=x^{3/2}, y=x^{5.4}$ pass through the required points. Can you think of a more general curve to try to revolve? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> If we suppose that the curve $y=f(x)$ is integrable then the volume so created will be $$ V = \int^1_0 \pi y^2 dx $$ To get a feel for the sort of curve we might need, first consider the special case $y=x$, which clearly passes through the two points. Then, $$ V = \int^1_0 \pi x^2 = \pi\left[\frac{x^3}{3}\right]^1_0 = \frac{\pi}{3} $$ This is slightly larger than $1$, so we could consider a family of curves which at beneath $y=x$ with enough flexibility to all us to vary the final value of the volume. A simple choice is parabolas $y=Ax(x-1)$ for some multiplicative constant. These give $$ V = \int^1_0 \pi Ax^2(x-1)^2dx $$ Now that I see it, I'm not too keen on doing this integral, so I'm going to consider instead $y=A\sqrt{|x(x-1)|}$ This gives a volume $$ V = \int^1_0 \pi A x(1-x) dx = \pi A\left[\frac{x^2}{2}-\frac{x^3}{3}\right]^1_0 = \pi A \left[\frac{1}{2}-\frac{1}{3}\right] = \frac{\pi A}{6} $$ Thus, a curve which has the required properties is $$ y = \frac{6}{\pi}\sqrt{|x(x-1)|} $$ There are, of course, others! </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 28: The right volume Can you rotate a curve to make a volume of 1? Collections Weekly Challenge Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Differentiation and Integration A2 Secondary Mapping Document DisplayCabinet