Dgdrxrt

Weaponising the Grasp-at-a-Distance spell

How can I list files in reverse time order by a command and pass them as arguments to another command?

The Nth Gryphon Number

Did John Wesley plagiarize Matthew Henry...?

French equivalents of おしゃれは足元から (Every good outfit starts with the shoes)

"Destructive power" carried by a B-52?

What is the proper term for etching or digging of wall to hide conduit of cables

Why not use the yoke to control yaw, as well as pitch and roll?

How does the body cool itself in a stillsuit?

Why does BitLocker not use RSA?

Why are current probes so expensive?

Why can't fire hurt Daenerys but it did to Jon Snow in season 1?

Should man-made satellites feature an intelligent inverted "cow catcher"?

Flight departed from the gate 5 min before scheduled departure time. Refund options

Found this skink in my tomato plant bucket. Is he trapped? Or could he leave if he wanted?

How does TikZ render an arc?

What did Turing mean when saying that "machines cannot give rise to surprises" is due to a fallacy?

What is "Lambda" in Heston's original paper on stochastic volatility models?

Any stored/leased 737s that could substitute for grounded MAXs?

Did any compiler fully use 80-bit floating point?

Where and when has Thucydides been studied?

Why is there so little support for joining EFTA in the British parliament?

In musical terms, what properties are varied by the human voice to produce different words / syllables?

Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?

I still can't find this average value

0

$begingroup$

I am trying to find the average value of the function $f(x,y,z) = y^2+2(x+1)+z$ over the region in the cylinder $x^2+y^2=4$ that is bounded above by the plane $z=4x$ and below by the $xy$-plane. I am stuck trying to find the volume of the region because I cannot come up with the bounds.

integration multivariable-calculus

share|cite|improve this question

edited Apr 2 at 15:47

Sriram Gopalakrishnan

2245

asked Apr 2 at 15:18

UchuukoUchuuko

468

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given that you're working with a cylinder, have you tried changing to cylindrical coordinates?
  $endgroup$
  – Sriram Gopalakrishnan
  Apr 2 at 15:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried, but I don't think I am doing it correctly at all (I am also considering the shadow in order to get bounds for my last two integrals). I don't think I really know how to convert because I wasn't given clear steps on how to do so.
  $endgroup$
  – Uchuuko
  Apr 2 at 15:44
  

  
  
  
  

add a comment | 

0

$begingroup$

I am trying to find the average value of the function $f(x,y,z) = y^2+2(x+1)+z$ over the region in the cylinder $x^2+y^2=4$ that is bounded above by the plane $z=4x$ and below by the $xy$-plane. I am stuck trying to find the volume of the region because I cannot come up with the bounds.

integration multivariable-calculus

share|cite|improve this question

edited Apr 2 at 15:47

Sriram Gopalakrishnan

2245

asked Apr 2 at 15:18

UchuukoUchuuko

468

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given that you're working with a cylinder, have you tried changing to cylindrical coordinates?
  $endgroup$
  – Sriram Gopalakrishnan
  Apr 2 at 15:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I tried, but I don't think I am doing it correctly at all (I am also considering the shadow in order to get bounds for my last two integrals). I don't think I really know how to convert because I wasn't given clear steps on how to do so.
  $endgroup$
  – Uchuuko
  Apr 2 at 15:44
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to find the average value of the function $f(x,y,z) = y^2+2(x+1)+z$ over the region in the cylinder $x^2+y^2=4$ that is bounded above by the plane $z=4x$ and below by the $xy$-plane. I am stuck trying to find the volume of the region because I cannot come up with the bounds.

integration multivariable-calculus

share|cite|improve this question

edited Apr 2 at 15:47

Sriram Gopalakrishnan

2245

asked Apr 2 at 15:18

UchuukoUchuuko

468

$endgroup$

share|cite|improve this question

edited Apr 2 at 15:47

Sriram Gopalakrishnan

2245

asked Apr 2 at 15:18

UchuukoUchuuko

468

share|cite|improve this question

edited Apr 2 at 15:47

Sriram Gopalakrishnan

2245

asked Apr 2 at 15:18

UchuukoUchuuko

468

edited Apr 2 at 15:47

Sriram Gopalakrishnan

2245

edited Apr 2 at 15:47

Sriram Gopalakrishnan

2245

asked Apr 2 at 15:18

UchuukoUchuuko

468

asked Apr 2 at 15:18

UchuukoUchuuko

468

1 Answer
1

active

oldest

votes

2

$begingroup$

Cylindrical coordinates are useful here. Recall how we change to cylindrical coordinates:
$$xmapsto rcostheta$$
$$ymapsto rsintheta$$
$$zmapsto z$$
So, we have $x^2+y^2=r^2(cos^2theta+sin^2theta)=r^2$. Using the equation for the cylinder given, we know that $x^2+y^2=4=r^2$, so $r=2$ (the negative square root corresponds to the portion of the cylinder below the $xy$-plane, which we don't care about). The bounds on our integral for $z$ are then $0$ and $4rcostheta=8costheta$. To find our "lower" integral bound for $r$, we need to solve $4rcostheta=0$ for $r$. We find that $r=0$ is the lower bound for the $r$ portion of our triple integral. Likewise, to find the bounds on $theta$, we need to solve $4rcostheta=0$ for $theta$. That is, we need $costheta=0$. This happens when $theta=pmpi$. So the bounds for the $theta$ portion of our integral are $-pi$ and $pi$. Lastly, we need to change the function we're integrating to cylindrical coordinates. Using the above substitutions, we can write
$$f(r,theta,z)=r^2sin^2theta+2(rcostheta+1)+z.$$ Putting this all together and choosing the order of integration carefully, we finally have our triple integral:
$$int_0^2int_-pi^piint_0^8costhetaf(r,theta,z)dzdtheta dr.$$

share|cite|improve this answer

answered Apr 2 at 16:25

Sriram GopalakrishnanSriram Gopalakrishnan

2245

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the walkthrough. I was overthinking things.
  $endgroup$
  – Uchuuko
  Apr 3 at 1:05
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3171987%2fi-still-cant-find-this-average-value%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Cylindrical coordinates are useful here. Recall how we change to cylindrical coordinates:
$$xmapsto rcostheta$$
$$ymapsto rsintheta$$
$$zmapsto z$$
So, we have $x^2+y^2=r^2(cos^2theta+sin^2theta)=r^2$. Using the equation for the cylinder given, we know that $x^2+y^2=4=r^2$, so $r=2$ (the negative square root corresponds to the portion of the cylinder below the $xy$-plane, which we don't care about). The bounds on our integral for $z$ are then $0$ and $4rcostheta=8costheta$. To find our "lower" integral bound for $r$, we need to solve $4rcostheta=0$ for $r$. We find that $r=0$ is the lower bound for the $r$ portion of our triple integral. Likewise, to find the bounds on $theta$, we need to solve $4rcostheta=0$ for $theta$. That is, we need $costheta=0$. This happens when $theta=pmpi$. So the bounds for the $theta$ portion of our integral are $-pi$ and $pi$. Lastly, we need to change the function we're integrating to cylindrical coordinates. Using the above substitutions, we can write
$$f(r,theta,z)=r^2sin^2theta+2(rcostheta+1)+z.$$ Putting this all together and choosing the order of integration carefully, we finally have our triple integral:
$$int_0^2int_-pi^piint_0^8costhetaf(r,theta,z)dzdtheta dr.$$

share|cite|improve this answer

answered Apr 2 at 16:25

Sriram GopalakrishnanSriram Gopalakrishnan

2245

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the walkthrough. I was overthinking things.
  $endgroup$
  – Uchuuko
  Apr 3 at 1:05
  

  
  
  
  

add a comment | 

2

$begingroup$

Cylindrical coordinates are useful here. Recall how we change to cylindrical coordinates:
$$xmapsto rcostheta$$
$$ymapsto rsintheta$$
$$zmapsto z$$
So, we have $x^2+y^2=r^2(cos^2theta+sin^2theta)=r^2$. Using the equation for the cylinder given, we know that $x^2+y^2=4=r^2$, so $r=2$ (the negative square root corresponds to the portion of the cylinder below the $xy$-plane, which we don't care about). The bounds on our integral for $z$ are then $0$ and $4rcostheta=8costheta$. To find our "lower" integral bound for $r$, we need to solve $4rcostheta=0$ for $r$. We find that $r=0$ is the lower bound for the $r$ portion of our triple integral. Likewise, to find the bounds on $theta$, we need to solve $4rcostheta=0$ for $theta$. That is, we need $costheta=0$. This happens when $theta=pmpi$. So the bounds for the $theta$ portion of our integral are $-pi$ and $pi$. Lastly, we need to change the function we're integrating to cylindrical coordinates. Using the above substitutions, we can write
$$f(r,theta,z)=r^2sin^2theta+2(rcostheta+1)+z.$$ Putting this all together and choosing the order of integration carefully, we finally have our triple integral:
$$int_0^2int_-pi^piint_0^8costhetaf(r,theta,z)dzdtheta dr.$$

share|cite|improve this answer

answered Apr 2 at 16:25

Sriram GopalakrishnanSriram Gopalakrishnan

2245

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the walkthrough. I was overthinking things.
  $endgroup$
  – Uchuuko
  Apr 3 at 1:05
  

  
  
  
  

add a comment | 

$begingroup$

Cylindrical coordinates are useful here. Recall how we change to cylindrical coordinates:
$$xmapsto rcostheta$$
$$ymapsto rsintheta$$
$$zmapsto z$$
So, we have $x^2+y^2=r^2(cos^2theta+sin^2theta)=r^2$. Using the equation for the cylinder given, we know that $x^2+y^2=4=r^2$, so $r=2$ (the negative square root corresponds to the portion of the cylinder below the $xy$-plane, which we don't care about). The bounds on our integral for $z$ are then $0$ and $4rcostheta=8costheta$. To find our "lower" integral bound for $r$, we need to solve $4rcostheta=0$ for $r$. We find that $r=0$ is the lower bound for the $r$ portion of our triple integral. Likewise, to find the bounds on $theta$, we need to solve $4rcostheta=0$ for $theta$. That is, we need $costheta=0$. This happens when $theta=pmpi$. So the bounds for the $theta$ portion of our integral are $-pi$ and $pi$. Lastly, we need to change the function we're integrating to cylindrical coordinates. Using the above substitutions, we can write
$$f(r,theta,z)=r^2sin^2theta+2(rcostheta+1)+z.$$ Putting this all together and choosing the order of integration carefully, we finally have our triple integral:
$$int_0^2int_-pi^piint_0^8costhetaf(r,theta,z)dzdtheta dr.$$

share|cite|improve this answer

answered Apr 2 at 16:25

Sriram GopalakrishnanSriram Gopalakrishnan

2245

$endgroup$

share|cite|improve this answer

answered Apr 2 at 16:25

Sriram GopalakrishnanSriram Gopalakrishnan

2245

answered Apr 2 at 16:25

Sriram GopalakrishnanSriram Gopalakrishnan

2245

answered Apr 2 at 16:25

Sriram GopalakrishnanSriram Gopalakrishnan

2245