/sci/ - Science & Math » Thread #11938930

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Triple Integral

Anonymous Sun 26 Jul 05:04:57 2020 No.11938930 View Reply Original Report

Quoted By: >>11938946 >>11939236

Alright /sci/. I tried, I really did, but I don't get this one at all.

Anonymous

Anonymous Sun 26 Jul 2020 05:09:34 No.11938946 Report

Quoted By: >>11938949

>>11938930
Is this for your PHD dissertation?

Anonymous

Anonymous Sun 26 Jul 2020 05:10:35 No.11938949 Report

Quoted By: >>11938950

>>11938946
Calc 3

Anonymous

Anonymous Sun 26 Jul 2020 05:11:42 No.11938950 Report

Quoted By: >>11938958 >>11939177

>>11938949
Ah, I was assuming this was for your PHD in triple integrals.

Anonymous

Anonymous Sun 26 Jul 2020 05:15:43 No.11938958 Report

Quoted By:

>>11938950
This is a quiz question. It feels like a dissertation.

Anonymous

Anonymous Sun 26 Jul 2020 05:22:23 No.11938969 Report

Quoted By:

bump

Anonymous

Anonymous Sun 26 Jul 2020 05:30:43 No.11938982 Report

Quoted By: >>11938986 >>11939007 >>11939018

just fucking translate into Spherical coordinates, its super easy
and since its an average, just divide by the total volume of the space after the integral

Anonymous

Anonymous Sun 26 Jul 2020 05:31:59 No.11938986 Report

Quoted By:

>>11938982
this.

Anonymous

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Anonymous Sun 26 Jul 2020 05:36:12 No.11938995 Report

Quoted By: >>11938999 >>11939007 >>11939016 >>11939018 >>11939020 >>11939024 >>11939241

This is how you set up the integral

Anonymous

Anonymous Sun 26 Jul 2020 05:37:57 No.11938999 Report

Quoted By: >>11939041

>>11938995
you are very bad at drawing.

Anonymous

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Anonymous Sun 26 Jul 2020 05:42:05 No.11939007 Report

Quoted By: >>11939019

>>11938982
>>11938995
Alright, check this out. I did exactly as you suggested, and I got zero for my answer. Does this make any sense?

Anonymous

Anonymous Sun 26 Jul 2020 05:46:45 No.11939016 Report

Quoted By:

>>11938995
you fucked it up lmao

Anonymous

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Anonymous Sun 26 Jul 2020 05:47:05 No.11939018 Report

Quoted By: >>11939024

>>11938982
>>11938995
Alright, check this out. I did exactly as you suggested, and I got zero for my answer. Does this make any sense?

Anonymous

Anonymous Sun 26 Jul 2020 05:47:11 No.11939019 Report

Quoted By: >>11939041

>>11939007
No it does not. Since we are in the first octant, x, y and z are all positive (and 0 on the boundaries). So you are integrating a bunch of positive stuff, and should get a positive answer.

Anonymous

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Anonymous Sun 26 Jul 2020 05:48:00 No.11939020 Report

Quoted By:

>>11938995
>using theta as the azimuthal angle instead of phi

Anonymous

Anonymous Sun 26 Jul 2020 05:49:27 No.11939024 Report

Quoted By: >>11939145

>>11938995
>>11939018
youre only integrating over the first octant, the angles are only between 0 and pi/4

Anonymous

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Anonymous Sun 26 Jul 2020 06:02:23 No.11939041 Report

Quoted By: >>11939092 >>11939233

>>11938999
>>11939019
Like this? What say you anons?

Anonymous

Anonymous Sun 26 Jul 2020 06:21:35 No.11939068 Report

Quoted By:

bump

Anonymous

Anonymous Sun 26 Jul 2020 06:35:19 No.11939092 Report

Quoted By: >>11939130 >>11939142

>>11939041
yes, thats right, but pull the integrals out from one another $\displaystyle\int \int ab\ \text{d}a\ \text{d}b = \int a\ \text{d}a* \int b\ \text{d}b$
shits gross to do inside

Anonymous

Anonymous Sun 26 Jul 2020 06:49:21 No.11939130 Report

Quoted By:

>>11939092
They never thought us that. My professor kinda sucks at online teaching.

Anonymous

Anonymous Sun 26 Jul 2020 06:53:00 No.11939142 Report

Quoted By: >>11939235

>>11939092
They never taught us that. My professor kinda sucks at online teaching.

Anonymous

Anonymous Sun 26 Jul 2020 06:53:33 No.11939145 Report

Quoted By: >>11939222

>>11939024
>first octant
QUESTION 6 • 8 POINTS
Find the average value of the function f (x, y, z) = 4x3yz on the region bounded by the sphere of radius 2 in the first octant. Express the average value as an exact answer.
Provide your answer below:

Anonymous

Anonymous Sun 26 Jul 2020 06:58:59 No.11939159 Report

Quoted By:

its an unsupervised online course and you are already cheating, why not just use wolfram alpha?

Anonymous

Anonymous Sun 26 Jul 2020 07:07:10 No.11939177 Report

Quoted By:

>>11938950
kek

Anonymous

Anonymous Sun 26 Jul 2020 07:47:38 No.11939222 Report

Quoted By:

>>11939145
thanks anon

Anonymous

Anonymous Sun 26 Jul 2020 07:54:18 No.11939233 Report

Quoted By: >>11939235

>>11939041
$ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{2} 4 \left ( \rho^3 \: \sin^3 \phi \: \cos^3 \theta \right ) (\rho \: \sin \phi \: \cos \theta) ( \rho \: \cos \theta) ( \rho^2 \: \sin \phi) ~ \text{d} \rho \, \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{2} 4 \, \rho^7 \: \sin^5 \phi \: \cos \phi \: \sin \theta \: \cos^3 \theta ~ \text{d} \rho \, \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} 128 \: \sin^5 \phi \: \cos \phi \: \sin \theta \: \cos^3 \theta ~ \text{d} \rho \, \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \dfrac{64 \: \cos^3 \theta \: \sin \theta}{3} ~ \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \dfrac{64 \: \cos^3 \theta \: \sin \theta}{3} ~ \text{d} \theta = \dfrac{\cancel{3}}{\cancel{4} \pi} \cdot \dfrac{\cancelto{4}{16}}{\cancel{3}} = \boxed{ \dfrac{4}{ \pi}} $

Anonymous

Anonymous Sun 26 Jul 2020 07:56:13 No.11939235 Report

Quoted By: >>11939251 >>11939255

>>11939142
you are really stoopid if you didn't notice this
>>11939233
you should be permabanned for contributing to hw threads

Anonymous

Anonymous Sun 26 Jul 2020 07:56:34 No.11939236 Report

Quoted By: >>11939243

>>11938930
Integrate the function from 0 to 2 on x,y,z, then divide by the volume of the sphere in that octant.

duh

Anonymous

Anonymous Sun 26 Jul 2020 07:57:51 No.11939241 Report

Quoted By: >>11939243

>>11938995
pretty sure you can (and should) just do it in cartesian coords

Anonymous

Anonymous Sun 26 Jul 2020 07:58:53 No.11939243 Report

Quoted By:

>>11939241
nevermind... ya convert to spherical and do>>11939236

Anonymous

Anonymous Sun 26 Jul 2020 08:00:43 No.11939245 Report

Quoted By: >>11939265

$ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{2} 4 \left ( \rho^3 \: \sin^3 \phi \: \cos^3 \theta \right ) (\rho \: \sin \phi \: \cos \theta) ( \rho \: \cos \theta) ( \rho^2 \: \sin \phi) ~ \text{d} \rho \, \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{2} 4 \, \rho^7 \: \sin^5 \phi \: \cos \phi \: \sin \theta \: \cos^3 \theta ~ \text{d} \rho \, \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} 128 \: \sin^5 \phi \: \cos \phi \: \sin \theta \: \cos^3 \theta ~ \text{d} \rho \, \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \dfrac{64 \: \cos^3 \theta \: \sin \theta}{3} ~ \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \dfrac{64 \: \cos^3 \theta \: \sin \theta}{3} ~ \text{d} \theta = \dfrac{3}{4 \pi} \cdot \dfrac{16}{3} = \boxed{ \dfrac{4}{ \pi}} $

Anonymous

Anonymous Sun 26 Jul 2020 08:04:03 No.11939251 Report

Quoted By: >>11939256

>>11939235
>shitty whining
When you were born your mom started crying upon seeing you, and insisted that the doctor keep on slapping.

Anonymous

Anonymous Sun 26 Jul 2020 08:06:44 No.11939255 Report

Quoted By:

>>11939235
anons like this whiny cunt make me want to be pro-choice

Anonymous

Anonymous Sun 26 Jul 2020 08:07:15 No.11939256 Report

Quoted By:

>>11939251
She tried to claim that they got their patients' procedures switched in the chart and that she was actually here for an abortion.

Anonymous

Anonymous Sun 26 Jul 2020 08:10:23 No.11939265 Report

Quoted By:

>>11939245
thanks for being a bro and formatting it like this anon

Anonymous

Anonymous Sun 26 Jul 2020 08:14:48 No.11939274 Report

Quoted By:

$ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{2} 4 \left ( \rho^3 \: \sin^3 \phi \: \cos^3 \theta \right ) (\rho \: \sin \phi \: \cos \theta) ( \rho \: \cos \theta) ( \rho^2 \: \sin \phi) ~ \text{d} \rho \, \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{2} 4 \, \rho^7 \: \sin^5 \phi \: \cos \phi \: \sin \theta \: \cos^3 \theta ~ \text{d} \rho \, \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \int_{0}^{ \frac{ \pi}{2}} 128 \: \sin^5 \phi \: \cos \phi \: \sin \theta \: \cos^3 \theta ~ \text{d} \phi \, \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \dfrac{64 \: \cos^3 \theta \: \sin \theta}{3} ~ \text{d} \theta \\ \displaystyle \dfrac{3}{4 \pi} \int_{0}^{ \frac{ \pi}{2}} \dfrac{64 \: \cos^3 \theta \: \sin \theta}{3} ~ \text{d} \theta = \dfrac{3}{4 \pi} \cdot \dfrac{16}{3} = \boxed{ \dfrac{4}{ \pi}} $

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