Question

Math practice for volumes of solids of revolutions. Please can you check my answers.

Answer 1

@SithsAndGiggles

Answer 2

I'd be happy to check any answers if you share all of the work you did towards obtaining that answer.

Answer 3

Thanks for posting the problems. What have you done so far? If you work on paper, how about sharing a digital image of your work?

Answer 4

I have 11 and 12 but am having trouble with figuring out 13.
eleven I have \[v = \int\limits_{-2}^{-5} \pi[6^{2} - (-2x - 4)^{2}] dx\]
12. I have \[v=\int\limits_{-2}^{-5} [\pi(5^{2} - 2^{2})] dy\]

Answer 5

I am stuck on thirteen

Answer 6

REAL 13 sorry I messed up the equation.
Area=\[(3)^{2}\pi \]
volume =\[\int\limits_{-5}^{1}[\pi(3)^{2}]dy\]

Answer 7

14.
area = \[(6)^{2}\pi - (3)^{2}\pi \]
volume = \[\int\limits_{6}^{-9} [\pi(6)^{2} - (3)^{2}]dx\]

Answer 8

15. area= \[(3)^{2}\pi - (1.5)^{2}\pi \]
volume = \[\int\limits_{-5}^{4}[\pi(3)^{2}-(1.5)^2] dy\]

Answer 9

@whpalmer4 could you take a look and check if these are right for me? I want to make sure I am doing this correctly

Answer 10

May I assume you still want feedback on #13? Thank you for posting your work.

Answer 11

Yes, I would like feedback on 11-15 please. I fixed 13 and deleted the post I messed up on.

Answer 12

Will help you, 'though can't promise to discuss all 5 problems with you. Which would you like to concentrate on first?

Answer 13

are all of them wrong? and I guess lets start with 11

Answer 14

In no way did I mean to imply that any, let alone all, are wrong.
Let's start with 11.

Answer 15

Regarding #11, what are you certain about and what are you questioning? How may I help?

Answer 16

I know you have to use the disk method and I found the area first then the volume. Honestly this is the only one I feel is completely correct.

Answer 17

yes I have this graph written down on my paper.