Triple integration - find volume

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yecko
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Homework Statement


螢幕快照 2017-11-12 上午11.34.45.png
螢幕快照 2017-11-12 上午11.34.57.png


Homework Equations


in the pic

The Attempt at a Solution


why is there an extra "r" in the highlighted line?

my attempt:
## \int_0^{2\pi}\int_0^2\int_{r^2}^41\ dy\ r\ dr\ d\theta ##
= ##\int_0^{2\pi}\int_0^2\left(4-r^2\right)\ r\ dr\ d\theta##

thanks
 

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  • 螢幕快照 2017-11-12 上午11.34.57.png
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Answers and Replies

  • #2
why is there an extra "r" in the highlighted line?
Because it is part of the correct form of the differential area in polar coordinates.
$$dA=r\;dr\;d\theta$$
 
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  • #3
yecko
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##\int_0^{2\pi}\int_0^2\int_{r^2}^41\ dy\ r\ dr\ d\theta##
I have considered r dr dθ already in my calculation... but the answer is r *r *dr*dθ
 
  • #4
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I have considered r dr dθ already in my calculation... but the answer is r *r *dr*dθ
It comes from the substitutions for polar coordinates in the x-z plane, for the expression ##\sqrt{x^2 + z^2}##. This is explained in the example where they use ##x = r\cos(\theta)## and ##z = r\sin(\theta)##.
 
  • #5
yecko
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√x^2+z
but why is there exists this expression in the first line of the second pic?
isn't that always 1 in the middle and integrate by looking at the boundaries for the triple integration?

my attempt : ##\int_0^{2\pi}\int_0^2\int_{r^2}^41\ dy\ r\ dr\ dθ##
mechanism: I would like to get the height first, then get the area of the cylinder

(my concepts are a little bit messy... sorry for that... and thank you very much for your patience)
 
  • #6
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but why is there exists this expression in the first line of the second pic?
isn't that always 1 in the middle and integrate by looking at the boundaries for the triple integration?
my attempt : ##\int_0^{2\pi}\int_0^2\int_{r^2}^41\ dy\ r\ dr\ dθ##

(my concepts are a little bit messy... sorry for that... and thank you very much for your patience)
If the goal is to find the volume using a triple integral, then the integrand will be 1, something like this ##\iiint_E 1~dV##.
However, in this problem you are asked to evaluate an integral that doesn't represent a volume.
 
  • #7
yecko
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the solution: ## \int_0^{2\pi}\int_0^2\int_{r^2}^4r\ dy\ r\ dr\ dθ ##
so why the solution put r in the middle?
 
  • #8
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the solution: ## \int_0^{2\pi}\int_0^2\int_{r^2}^4r\ dy\ r\ dr\ dθ ##
so why the solution put r in the middle?
I don't know what you're talking about. The solution in the example you posted doesn't have y in it. They converted the integral to polar form - no x, y, or z.
 
  • #9
yecko
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I don't know what you're talking about. The solution in the example you posted doesn't have y in it. They converted the integral to polar form - no x, y, or z.
its the same...
##\int_0^{2\pi}\int_0^2\int_{r^2}^4r\ dy\ r\ dr\ dθ## =## \int_0^{2\pi}\int_0^2\left(4-r^2\right)r\ \ r\ dr\ dθ##

so why the solution put r in the middle?
 
  • #10
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its the same...
##\int_0^{2\pi}\int_0^2\int_{r^2}^4r\ dy\ r\ dr\ dθ## =## \int_0^{2\pi}\int_0^2\left(4-r^2\right)r\ \ r\ dr\ dθ##

so why the solution put r in the middle?
I don't know. It seems that you put it in the middle, since the textbook image you posted doesn't have this integral in it - ##\int_0^{2\pi}\int_0^2\int_{r^2}^4r\ dy\ r\ dr\ dθ##.

You could rewrite that integral like so: ##\int_0^{2\pi}\int_0^2\int_{r^2}^4\ dy\ r^2\ dr\ dθ##, and it wouldn't make any difference - you get the same integral as shown on the right in what I quoted from you.
 
  • #11
yecko
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But why not## \int_0^{2\pi}\int_0^2\int_{r^2}^41\ dy\ r\ dr\ dθ## which is 1 in the middle?
why is solution is ##\int_0^{2\pi}\int_0^2\int_{r^2}^4r\ dy\ r\ dr\ dθ## = ##\int_0^{2\pi}\int_0^2\left(4-r^2\right)r\ \ r\ dr\ dθ## ?

However, in this problem you are asked to evaluate an integral that doesn't represent a volume
it is asking for volume in the question...
 
  • #12
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But why not## \int_0^{2\pi}\int_0^2\int_{r^2}^41\ dy\ r\ dr\ dθ## which is 1 in the middle?
You keep asking this question, which was answered in posts #2 and #4. What parts of those posts do you not understand?
yecko said:
why is solution is ##\int_0^{2\pi}\int_0^2\int_{r^2}^4r\ dy\ r\ dr\ dθ## = ##\int_0^{2\pi}\int_0^2\left(4-r^2\right)r\ \ r\ dr\ dθ## ?


it is asking for volume in the question...
NO, IT IS NOT!
As it says in the image you posted,
Evaluate ##\iiint_E \sqrt{x^2 + z^2}~dV##, where E is the region bounded by the paraboloid ##y = x^2 + z^2## and the plane ##y = 4##.
This is NOT a volume integral!

Please take more care in your reading.
 
  • #13
yecko
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If the goal is to find the volume using a triple integral, then the integrand will be 1, something like this ##\iiint_E 1~dV##.
However, in this problem you are asked to evaluate an integral that doesn't represent a volume.
it is not a volume, so it is just a physically-meaning mathematics expression?
i see. thanks for telling me.
 

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