Integration and volume problem: Find volume generated by area between y=x^2+2, y=6, rotated abt x-axis

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Aminta_1900

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"Find the volume generated when the area between the curve y = x^2 + 2 and the line y = 6 is rotated about the x-axis."

AAA.jpg

I understand that this would form a ring-shape of sorts, but am I right in using this method below to obtain the given answer? Only the answer in my text book gives simply "295 (3 s. f.)" without using Pi.

CCC.jpg
 
How do you think we can check whether the 2 answers are the same?
Of course, the answers being the same does not mean that the method is applied correctly. So, what is it that you are not sure about?
 
Aminta_1900 said:
"Find the volume generated when the area between the curve y = x^2 + 2 and the line y = 6 is rotated about the x-axis."

View attachment 36751

I understand that this would form a ring-shape of sorts, but am I right in using this method below to obtain the given answer? Only the answer in my text book gives simply "295 (3 s. f.)" without using Pi.

View attachment 36753
Click to expand...
How did you decide on the limits of integration (-2 to +2)?
 
The value of [imath](x^2+2)^2 - 6^2[/imath] is negative in the [imath](-2,2)[/imath] interval, so the result of the integration should be negative too. But the absolute value of the result seems correct.
 
Aminta_1900 said:
"Find the volume generated when the area between the curve y = x^2 + 2 and the line y = 6 is rotated about the x-axis."


View attachment 36753
Click to expand...

Without commenting either way on whether you used the correct set-up for your first line, I wanted to make certain comments.

In your first line, supposedly the integrand does not need any more grouping symbols, but it looks more organized if it is written as
\(\displaystyle [(x^2 + 2)^2 - (6)^2] dx. \)

In your second line, you are missing "dx" in two places, and so I would have your integrands written as
\(\displaystyle (x^4 + 4x^2 + 4 - 36) dx \ \ and \ \ (x^4 + 4x^2 - 32) dx, \ \) respectively.

To ease your computation, because of the y-axis symmetry, the integral can be rewritten as \(\displaystyle \ 2\pi \ \) times the integral
from 0 to 2 of \(\displaystyle \ (x^4 + 4x^2 - 32)dx.\)

Supposing your numerical answer is correct at the end, I would write the pi symbol after the fraction:

\(\displaystyle \dfrac{1,408}{15}\pi \ \ units^3 \ \ \ \) However, the integral in your third line produces a negative value.

Edit: \(\displaystyle \ \ \)The correct stream-lined integral can be rewritten as \(\displaystyle \ 2\pi \ \) times the integral from 0 to 2 of \(\displaystyle \ [(6)^2 - (x^2 + 2)^2]dx\)
if you want to save on calculations.
 
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Aminta_1900 said:
Only the answer in my text book gives simply "295 (3 s. f.)" without using Pi.
Click to expand...
What does "3 s.f." mean? Ah - three significant figures! I was thinking "square feet" at first ...

A calculator shows that your answer, [imath]\frac{1408\pi}{15}[/imath], comes to 294.89, which rounds to 295, so it appears to be correct. Did you not do this?

The correct graph is:

1699915104947.png​

This confirms your limits of integration, which you presumably found by solving x^2+2 = 6.

The method, as has been pointed out, is correct except that you subtracted the square of the outer radius from the square of the inner radius, resulting in a negative answer that you didn't notice.

On the whole, good work, but be more careful!
 
lookagain said:
Without commenting either way on whether you used the correct set-up for your first line, I wanted to make certain comments.

In your first line, supposedly the integrand does not need any more grouping symbols, but it looks more organized if it is written as
\(\displaystyle [(x^2 + 2)^2 - (6)^2] dx. \)

In your second line, you are missing "dx" in two places, and so I would have your integrands written as
\(\displaystyle (x^4 + 4x^2 + 4 - 36) dx \ \ and \ \ (x^4 + 4x^2 - 32) dx, \ \) respectively.

To ease your computation, because of the y-axis symmetry, the integral can be rewritten as \(\displaystyle \ 2\pi \ \) times the integral
from 0 to 2 of \(\displaystyle \ (x^4 + 4x^2 - 32)dx.\)

Supposing your numerical answer is correct at the end, I would write the pi symbol after the fraction:

\(\displaystyle \dfrac{1,408}{15}\pi \ \ units^3 \ \ \ \) However, the integral in your third line produces a negative value.

Edit: \(\displaystyle \ \ \)The correct stream-lined integral can be rewritten as \(\displaystyle \ 2\pi \ \) times the integral from 0 to 2 of \(\displaystyle \ [(6)^2 - (x^2 + 2)^2]dx\)
if you want to save on calculations.
Click to expand...
OK, thanks a lot for the clarity. I'm reading over what you said carefully to better understand this.
 
Dr.Peterson said:
What does "3 s.f." mean? Ah - three significant figures! I was thinking "square feet" at first ...

A calculator shows that your answer, [imath]\frac{1408\pi}{15}[/imath], comes to 294.89, which rounds to 295, so it appears to be correct. Did you not do this?

The correct graph is:


This confirms your limits of integration, which you presumably found by solving x^2+2 = 6.

The method, as has been pointed out, is correct except that you subtracted the square of the outer radius from the square of the inner radius, resulting in a negative answer that you didn't notice.

On the whole, good work, but be more careful!
Click to expand...
OK, thanks again for the help. No, I didn't try dividing 1408pi by 15, I just thought it was odd that the answer was in this format. All others had involved pi.

In future I'll be aware of subtracting the inner from the outer. Thanks.
 
Aminta_1900 said:
x^2 + 2 = 6 since y = y

x^2 = 6 - 2

x^2 = 4

x = sqr(4)

x = 2, -2
Click to expand...
Correct -

but "your" sketch did not match that.
Now you have been shown the "correct" method to calculate a volume of "revolution" . Generally, there are two ways to calculate volume of revolution.
Method of disks &
Method of washers
Which method are you using here?
 
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khansaheb said:
Correct -

but "your" sketch did not match that.
Now you have been shown the "correct" method to calculate a volume of "revolution" . Generally, there are two ways to calculate volume of revolution.
Method of disks &
Method of washers
Which method are you using here?
Click to expand...
I've never heard these terms before so I have no idea what you're talking about.
 
Aminta_1900 said:
I've never heard these terms before so I have no idea what you're talking about.
Click to expand...

Okay; what methods *have* you been given for answering this sort of exercise? What was the source of your equations in your initial post?
 
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