Question

Find the volume of the solid generated by revolving the region bounded by the graphs of y=3sqrt(x) and y=(x^2/8) around the x-axis in cubic units rounded to the nearest hundredth.

Answer 1

The answer is 587.2un^3, but no matter what I try, I cannot get to that value. Help with figuring out the steps would be much appreciated.

Answer 2

a usual mistake of people is limits of the integral...or the method used (i.e. shell, washer, disk, etc)

Answer 3

are you doing double integral or washer disk...?

Answer 4

I have tried disk, washer, and shell methods but can't come up with that number. I think it should be the washer method, but I'm doing something wrong in the math.

Answer 5

First off, what does that shape look like? Now what are your limits of integration?

Answer 6

it seems the washer is the most preferrable according to @Kainui 's graph

Answer 7

do you mean the values that define the integral? I was using 0 and the point where they graphs intersect at 8,6535

Answer 8

But now that I've seen a bigger picture than my calculator, it looks like 0 is wrong.

Answer 9

^^that was 8.6535

Answer 10

Wait, is that \[\frac{ x^2 }{ 8 } =or =x^\frac{ 2 }{ 8 }\]

Answer 11

Oh I see ok lol

Answer 12

The first equation.

Answer 13

Alright, so you know the limits of integration, good. Now what shapes are you making with those two graphs and how?

Answer 14

I like to think of it this way. You're making two solid shapes but you're subtracting one of the solid shapes from the other.

Answer 15

Okay, so we're still using 0 and 8.6535 for the region correct? And the way I would set it up is V=pi\[\int\limits_{0}^{8.6535}\]((3\[\sqrt{x)}\]-(x^2)/8

Answer 16

Sorry I couldn't really figure out the equation machine. so integrand would be (3sqrt(x))^2-((x^2)/8)^2

Answer 17

Well I just checked and I found that they intersect at about 8.320335

Answer 18

Darn it I was reading the wrong value. 8.6535 is the y value, not the x. Let me try plugging in 8.32035

Answer 19

With the region (0,8.32035) I got the answer 186.9155

Answer 20

\[\pi[ \int\limits_{0}^{8.32}3\sqrt{x}dx-\int\limits_{0}^{8.32}\frac{ x^2dx }{ 8 }]\] that's the integration I found, I might have made a mistake I didn't really think about it too much

Answer 21

Yeah I forgot to square the function lol

Answer 22

Does that look like what you were doing? I couldn't really tell, but I can explain what I did and why.

Answer 23

Yes, but I squared both of the functions. So after squaring and integrating I got pi[((9/2)x^2)-((x^5)/320)]

Answer 24

Yeah that's what I got, I'm about to plug in the answer myself.

Answer 25

I'm also getting 186.915. How do you know that's not the right answer lol?

Answer 26

Haha it is a practice version of the final exam which came with an answer key - so hopefully the key would be correct...

Answer 27

Oh wait.

Answer 28

What's 186.92 times pi?

Answer 29

;)

Answer 30

Oh jeeze. Lol now I feel really silly. Thanks for walking the whole thing through with me though!

Answer 31

Haha no problem, I made the same mistake too rofl.