How do you find the volume of the solid of revolution formed by revolving this function around the $x$-axis: $y=2.33 \cos\left(\frac{25\pi}{119}\left(x-2.47\right)\right)$ between the limits $x=2.47$ and $x=4.85$?

Full working would be greatly appreciated!

  • $\begingroup$ Are you familiar with any standard methods for this? You want to use what is called "disc method" (en.wikipedia.org/wiki/Solid_of_revolution#Disc_method). $\endgroup$ – Carser Jul 30 '16 at 3:52
  • $\begingroup$ Yes I am familiar with that method. That is what I am trying to use but I am just unsure with how to expand the brackets when you have to square the entire function: (2.33cos(25π119(x−2.47)))^2. Could you please show me how to expand the bracket? Does the 2.33 become squared and the cos become squared or does something happen to inside the cos bracket as well $\endgroup$ – Emma Jul 30 '16 at 4:11

The volume is given by $$ V = \pi \int_{2.47}^{4.85} \left( 2.33 \cos \left( \frac{25 \pi}{119}(x-2.47) \right) \right)^2 \ dx $$ $$ = \pi (2.33)^2 \int_{2.47}^{4.85} \cos^2 \left(\frac{25 \pi}{119} (x-2.47) \right) \ dx $$ Letting $u=(x-2.47)$, then $du= dx$ and we can substitute to get $$ = \pi (2.33)^2 \int_{2.47}^{4.85} \cos^2 \left(\frac{25 \pi}{119}u\right) \ du $$ and if we can use $$ \int \cos^2(a x) dx = \frac{2ax+\sin(2ax)}{4a} $$ then we have $$ = \pi (2.33)^2 \left[ \frac{2\frac{25 \pi}{119}u+\sin(2\frac{25 \pi}{119}u)}{4 \frac{25 \pi}{119}} \right]_{2.47}^{4.85} $$ which I'll happily leave for you to simplify.


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