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 y-axis. x = 3tan((PI*y)/3)

1. anonymous

2. anonymous

$x = 3\tan ((\pi*y)/3)$

3. anonymous

From y = 0 to 1 |dw:1327692518967:dw|

4. anonymous

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.

5. amistre64

the bounds is from 0 to 1 right? up the y axis

6. anonymous

Yes, 0 to 1 up the Y axis.

7. amistre64

$\pi\int_{0}^{1}(3\ tan(\frac{\pi}{3}y))^2dy$does seem to be appropriate so where is the hiccup your having in this?

8. anonymous

Interesting... I guess I am having some problems with the arithmetic and integration. Trying to work out the answer now.

9. amistre64

$\pi\int_{0}^{1}3^2\ tan^2(\frac{\pi}{3}y)dy$ $9\pi\int_{0}^{1} tan^2(\frac{\pi}{3}y)dy$ can you think of a simpler way to write tan^2 that would be more doable?

10. anonymous

Ahh, sec^2x - 1 I assume?

11. amistre64

$9\pi\int_{0}^{1} sec^2(\frac{\pi}{3}y)-1\ dy$ yes

12. anonymous

I forgot my trig identities! Ahhh

13. amistre64

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

14. amistre64

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?

15. anonymous

Probably u-sub, much more familiar with that.

16. amistre64

u = pi/3 y du/dy = pi/3 dy = du/(pi/3) = 3du/pi

17. amistre64

same results, we need to include a 3/pi

18. amistre64

$9\pi\int_{0}^{1} \frac{3}{\pi}(sec^2(u)-1)\ du$ $\frac{3*9\pi}{\pi}\int_{0}^{1}sec^2(u)-1\ du$ $27\int_{0}^{1}sec^2(u)-1\ du$

19. amistre64

now, do you want to go ahead with this? or do we dare change the limits to match u instead of y :)

20. amistre64

$u=\frac{\pi}{3}*0=0$ $u=\frac{\pi}{3}*1=\frac{\pi}{3}$ $27\int_{0}^{pi/3}sec^2(u)-1\ du$ fancy!!!

21. anonymous

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?

22. amistre64

yep, now the whole thing is speaking in us and not partial tongues lol

23. anonymous

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?

24. amistre64

$27(tan(60^o)-\frac{pi}{3})\cancel{-27(tan(0^0)-0)}^{(\ 0\ )}$

25. amistre64

practice practice practice; after a bit it starts to sink in and make sense :)

26. amistre64

i started with the older textbooks out of the college libraries and learnt myself up til now

27. anonymous

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 :)

28. amistre64

if im right about this: $Volume=27\sqrt{3}-9pi$

29. anonymous

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)

30. amistre64

mathlab costs money; i use interactmath.com. same setup but free; and you can choose from alot of different subjects

31. amistre64
32. anonymous

Yep! We got it! Oh, I'm doing a paid online Calculus II class.. don't have a choice.