Calculus II - Volume of Solids Hello, having some problems with this volume of a solid question. I cannot find a similar example, so I come to you for assistance! Find the volume of the solid generated by revolving the shaded region about the y-axis. x = 3tan((PI*y)/3)

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Calculus II - Volume of Solids Hello, having some problems with this volume of a solid question. I cannot find a similar example, so I come to you for assistance! Find the volume of the solid generated by revolving the shaded region about the y-axis. x = 3tan((PI*y)/3)

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\[x = 3\tan ((\pi*y)/3)\]
From y = 0 to 1 |dw:1327692518967:dw|

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I thought the setup would be; \[\int\limits_{0}^{1} \pi(3\tan((\pi/3)y)^2\] But this does not seem to be right. Any help would be appreciated.
the bounds is from 0 to 1 right? up the y axis
Yes, 0 to 1 up the Y axis.
\[\pi\int_{0}^{1}(3\ tan(\frac{\pi}{3}y))^2dy\]does seem to be appropriate so where is the hiccup your having in this?
Interesting... I guess I am having some problems with the arithmetic and integration. Trying to work out the answer now.
\[\pi\int_{0}^{1}3^2\ tan^2(\frac{\pi}{3}y)dy\] \[9\pi\int_{0}^{1} tan^2(\frac{\pi}{3}y)dy\] can you think of a simpler way to write tan^2 that would be more doable?
Ahh, sec^2x - 1 I assume?
\[9\pi\int_{0}^{1} sec^2(\frac{\pi}{3}y)-1\ dy\] yes
I forgot my trig identities! Ahhh
now for practice and to see whats missing lets see that htis comes from the derivative of tan :) tan(ny) = n sec^2(y) so we are missing the n part which is just a constant
pi/3 is missing meaning we had a 3/pi to get rid of it to begin with am i seeing that right or you wanna go for the so called usub?
Probably u-sub, much more familiar with that.
u = pi/3 y du/dy = pi/3 dy = du/(pi/3) = 3du/pi
same results, we need to include a 3/pi
\[9\pi\int_{0}^{1} \frac{3}{\pi}(sec^2(u)-1)\ du\] \[\frac{3*9\pi}{\pi}\int_{0}^{1}sec^2(u)-1\ du\] \[27\int_{0}^{1}sec^2(u)-1\ du\]
now, do you want to go ahead with this? or do we dare change the limits to match u instead of y :)
\[u=\frac{\pi}{3}*0=0\] \[u=\frac{\pi}{3}*1=\frac{\pi}{3}\] \[27\int_{0}^{pi/3}sec^2(u)-1\ du\] fancy!!!
Sorry for the slow replies. At work trying to go through your work step by step while taking the odd call. This has been great... my biggest downfall was forgetting tan^(x) = sec^(x) -1.. The tan was just making my integral the nastiest thing ever. Now that you have changed the interval it is possible to sub those values in for U as opposed to subbing U back in and using the original interval?
yep, now the whole thing is speaking in us and not partial tongues lol
Haha, great stuff man. Thanks so much. I find this volume of solid stuff to just be some of the nastiest stuff imaginable. Do you have any recommended resources for learning the stuff or just practice, practice, practice?
\[27(tan(60^o)-\frac{pi}{3})\cancel{-27(tan(0^0)-0)}^{(\ 0\ )}\]
practice practice practice; after a bit it starts to sink in and make sense :)
i started with the older textbooks out of the college libraries and learnt myself up til now
Great, well thanks again for all of the help. Im going to try working through this start to finish again and see about getting the right answer :)
if im right about this: \[Volume=27\sqrt{3}-9pi\]
I'll let you know. I use Mymathlab for HW, so it only allows the right answer. Can be a bit of a pain (ex .01 off)
mathlab costs money; i use interactmath.com. same setup but free; and you can choose from alot of different subjects
http://www.wolframalpha.com/input/?i=integrate+pi%289tan%5E2%28y*pi%2F3%29%29+from+0+to+1 wolfram helps too :)
Yep! We got it! Oh, I'm doing a paid online Calculus II class.. don't have a choice.

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