Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about x = 6.

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Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about x = 6.



y = x, y = 0, x = 3.
Revolving around x = 6, what is the volume of the shape formed using shell method?



I've tried this:



http://www3.wolframalpha.com/Calculate/MSP/MSP447521i89ce0d69c7cb5000040fc75g3e5g51hca?MSPStoreType=image/gif&s=44



From what I've researched online and from my notes in class, this should be working but I can't find what I'm doing wrong.
I've also tried integrating from 3 to 6. Not sure what else I could be doing wrong. Thank you.





please include your thoughts about questions or what You have treid
– Deepesh Meena
4 hours ago




2 Answers
2



At any $x$, you have a shell whose radius is $6-x$ and whose height is $x$, so its volume is $2pi x(6-x)Delta x$.



$2piint_0^3(6-x)(x-0)mathbb dx=2piint_0^3(6x-x^2)mathbb dx=2pi [3x^2-fracx^33]_0^3=2pi[18-0]=36pi$





Thank you so much. I wasn't thinking about it in the sense of subtracting but rather adding the extra length so I was stuck with adding 3.
– Ivan Ahumada
4 hours ago





Sure. The radius is $(6-x)$, when rotating about $x=6$...
– Chris Custer
3 hours ago






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