Someone please check! :) (the answer says it's coming out wrong, but I can't find out why.)
solids of revolution, yesssssss you're finally starting the fun stuff! lol
hehe that's one way to look at it ;) Do you want a screen shot of the graph too? or can you envision it fine?
Oh they gave you the diagram? Nahh it's prolly fine
Yep yep. Well, as part of the question I had to find the matching graph.
\[\pi \int\limits_{0}^{1}(4-\frac{ 3 }{ 2 }x)^2\] is my set up. I'm unsure about the 3/2x part though.
1/2, ya?
Looks good besides that :o
In what sense are you "unsure about the (3/2)x part?"
Ahh, how do I get that part? Like I was unsure about if it was correct. Which it isn't xD
Grr, I'm not on my computer :P can't draw it out..
...want the graph now? :}
Two suggestions for you: Graph this situation, if you haven't already, and share your graph. Show your work where that (4-[x/2])^2 is concerned. Note that it may be well to expand that, but not necessarily essential.
zepdrix and I can give you much better feedback if you'd please share every step of the process with us.
Ok Imma get on my puter :P brb i love doing these types of problems lol
Love your enthusiasm, zepdrix! But let's make a point of guiding Babynini as she attempts to find her own solution to this problem.
@Babynini : What method have you chosenf or finding the volume of this 3D solid?
\[\pi \int\limits_{0}^{1}(\frac{ x^2 }{ 4 }-4x+16)\]\[\pi (\frac{ x^3 }{ 12 }-2x^2+16x)_0^1\] \[\pi(\frac{ 169 }{ 12 })=\frac{ 169\pi }{ 12 }\]
Mind explaining why you chose to square (4-x/2)? where's your "dx?"
Whoops, I know the dx is meant to be there, I just didn't..write it haha. I squared it because I was simplifying (4-1/2x)^2 from the previous step :)
Are you going to draw it now, zep? xD I want to see!!
What was "the previous step?" I'd still like for you to share the name of the method you're using here and why you chose to use that method.
There are 3 primary methods to choose from when one wants to find the volume of a solid of revolution. Which have your chosen?
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