Question

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Answer 1

You didn't subtract the two radii. The integral needs to be like this

int[2^2 - x^2]

Answer 2

Wait what, I'm confused

Answer 3

The area of a circle is pi * r^2. Here you have to consider only the shaded region, so the area of the shaded region will be pi * (outer radius)^2 - pi * (inner radius)^2

Answer 4

Then integrate the area under limits 0 to 2 to find the volume

Answer 5

\[\pi \int\limits_{0}^{2}4-x^2 dx ?\]

Answer 6

Yes

Answer 7

Are you 100% sure?

Answer 8

Why is it from 0 to 2 and not from 0 to 4?

Answer 9

Since you are integrating with respect to x, so the limits are of x values not y values from the graph. 0 to 2 are the x values while 0 to 4 are the y values of the shaded region. And yes it is correct

Answer 10

Is it better to integrate with respect to y or x in this case? Because I wasnt sure

Answer 11

when you have the equation in terms of x integrate in x; when you have the equation in y integrate in y. it also varies according to the equations given. in this question, however, integrating in x is simpler.

Answer 12

What is the axis of revolution in this case? I see the specifications of the region only, and I am led to believe that "\(a\) solid of revolution" means we have a choice in any that we want; although I don't see what made the outer radius 2 and inner radius x^2.

Answer 13

I just went along with the boundaries of how it looks like when I graphed it

Answer 14

@jennisicle what we did above was assuming that the bounded area is revolved around the y-axis to make the solid. However it is not stated clearly in the question so can you post the axis around which the solid is revolved?

Answer 15

I don't know what you're asking me to post, because that's the entire question I had.

Answer 16

it didn't tell me much else

Answer 17

Like in the very first picture I posted where it has the question is the exact same way the question is worded

Answer 18

Then I assume you should go with the simpler solid generated by revolving the bounded area around the x axis.

Answer 19

But that answer I got in the second picture; does my work look correct and everything?

Answer 20

It will then give the integral as follows:
\[\int\limits_{0}^{2} \pi x^2 dx\]
Note that again we are taking the limits from 0 to 2 since the integral is with respect to x. This is for the first approach (revolving around x axis)

Answer 21

I thought you said it was 4-x^2 from 0 to 2?

Answer 22

Yes that is for the second approach when you form the solid by revolving the bounded area around the y axis.

Since the question states that you form a solid by revolving it, but it does not state the axis around which you have to revolve, so you can either revolve it around y axis or x axis.

Answer 23

Oh, I understand. But both of these approaches will give me 64/3 pi in the end?

Answer 24

No. Both form different solids. The first one gives you a volume of 8pi/3

Answer 25

The second one gives you a volume of 8pi

Answer 26

Go with the first one since it is a simpler solid than that formed in the second one. Because when you revolve around the x axis in this figure it gives you a non hollow solid while if you revolve around y axis it gives you a hollow solid

Answer 27

How did you get 8pi/3 though? I keep getting 16pi/3

Answer 28

\[\int\limits_{0}^{2}\pi x^2 dx\]
then you get
pi/3 x^3 from 0 to 2
putting the values,
pi/3 (2^3 - 0^3)
= pi/3(8)
= 8pi/3

Answer 29

wait.. that is incorrect let me do it again sorry

Answer 30

I thought we were doing the 4-x^2 one?

Answer 31

Oh its confusing! Do the pi x^2 because it is more probable that the question meant this! Sorry! Its correct work will be:
\[\int\limits_{0}^{2} \pi * (x^2)^2 dx = 32 \pi/5\]

Answer 32

(we forgot to square the already x^2, which, by the way, you did on your first answer) The only error you did was to take the incorrect limits in your first answer.

Answer 33

Oh okay, I just did it again and got 32/5 pi.

Answer 34

So I'm sure that's correct. Thanks :)

Answer 35

Glad to help!