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anonymous
 4 years ago
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 yaxis.
x = 3tan((PI*y)/3)
anonymous
 4 years ago
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 yaxis. x = 3tan((PI*y)/3)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0More information coming 

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x = 3\tan ((\pi*y)/3)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0From y = 0 to 1 dw:1327692518967:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the bounds is from 0 to 1 right? up the y axis

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, 0 to 1 up the Y axis.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\pi\int_{0}^{1}(3\ tan(\frac{\pi}{3}y))^2dy\]does seem to be appropriate so where is the hiccup your having in this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Interesting... I guess I am having some problems with the arithmetic and integration. Trying to work out the answer now.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[\pi\int_{0}^{1}3^2\ tan^2(\frac{\pi}{3}y)dy\] \[9\pi\int_{0}^{1} tan^2(\frac{\pi}{3}y)dy\] can you think of a simpler way to write tan^2 that would be more doable?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ahh, sec^2x  1 I assume?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[9\pi\int_{0}^{1} sec^2(\frac{\pi}{3}y)1\ dy\] yes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I forgot my trig identities! Ahhh

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1now 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1pi/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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Probably usub, much more familiar with that.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1u = pi/3 y du/dy = pi/3 dy = du/(pi/3) = 3du/pi

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1same results, we need to include a 3/pi

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[9\pi\int_{0}^{1} \frac{3}{\pi}(sec^2(u)1)\ du\] \[\frac{3*9\pi}{\pi}\int_{0}^{1}sec^2(u)1\ du\] \[27\int_{0}^{1}sec^2(u)1\ du\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1now, do you want to go ahead with this? or do we dare change the limits to match u instead of y :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[u=\frac{\pi}{3}*0=0\] \[u=\frac{\pi}{3}*1=\frac{\pi}{3}\] \[27\int_{0}^{pi/3}sec^2(u)1\ du\] fancy!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry 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?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yep, now the whole thing is speaking in us and not partial tongues lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, 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?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[27(tan(60^o)\frac{pi}{3})\cancel{27(tan(0^0)0)}^{(\ 0\ )}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1practice practice practice; after a bit it starts to sink in and make sense :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i started with the older textbooks out of the college libraries and learnt myself up til now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Great, 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 :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if im right about this: \[Volume=27\sqrt{3}9pi\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'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)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1mathlab costs money; i use interactmath.com. same setup but free; and you can choose from alot of different subjects

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=integrate+pi%289tan%5E2%28y*pi%2F3%29%29+from+0+to+1 wolfram helps too :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep! We got it! Oh, I'm doing a paid online Calculus II class.. don't have a choice.
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