Double Integrals: cartesian -> polar and solve

Click For Summary

Homework Help Overview

The discussion revolves around the application of double integrals in cylindrical polar coordinates, specifically focusing on finding the area above a cone and below a sphere. Participants are exploring the setup and limits of integration for the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the limits for the variable r and the integration setup, with some expressing confusion about the correct approach to visualize the region of integration. Questions arise about the minimum and maximum values of r, as well as the reasoning behind the choice of integration variables.

Discussion Status

The discussion is active, with participants providing feedback and clarifications on each other's reasoning. Some have identified potential errors in their approaches, while others are confirming their understanding of the integration limits. There is no explicit consensus, but productive dialogue is ongoing.

Contextual Notes

Participants mention constraints such as the requirement to integrate in cylindrical polar coordinates and the specific geometric conditions imposed by the cone and sphere. There is also a reference to a restriction on the region R, indicating that y must be less than or equal to x.

GreenLantern
Messages
30
Reaction score
0
Double Integrals: cartesian --> polar and solve

here is everything:

HWCALCHELP.jpg


#19: I am stuck...This is to be solved using cylindrical polar coordinates and a double integral. I understand simpler ones such as find the volume of the solid under the cone z= sqrt(x^2 + Y^2) and above the disk (x^2 + y^2 ) <= 4 but i guess what is throwing me on this one is the sphere and that it is asking for area above the cone...

#13: What did i do wrong? My answer is similar to the book but not quite...

Thanks everyone!

-GL
 
Last edited:
Physics news on Phys.org


19:
You have very nearly solved this one. You have found [itex]z \geq r[/itex] and [itex]z \leq \sqrt{1-r^2}[/itex], so all you need to do is find the minimum and maximum values for [itex]r[/itex] and you will have the triple integral
[tex]\int_{\min(r)}^{\max{r}}\int_r^{\sqrt{1-r^2}}\int_0^{2\pi}rd\theta dz dr[/tex]

13:
You forgot to apply the restriction on R that says [itex]y \leq x[/itex]
 


ughhhh okay and how exactly do i find R min/max? i hardly know how i got to where i did to be honest...

im trying to do it by visual inspection which is how my prof. said to do other ones but my mind is turning to mush thinking about it.I'd say that r max would be 1? due to part of the sphere is included and its radius is 1. Is this correct?
and is the minimum r=0 because the point of the cone?
thanks so much for replying to me though!
 


Actually, looking at your work again, you already found [itex]r\leq \sqrt{1/2}[/itex]. Since this is the only restriction on r, the minimum will be 0.
 


okay. and then why are we not integrating anything other than r?
 


WOWWWW
okay so totally disregard me as an asshat because earlier i was totally thinking about doing the integral of [tex](r)[ \sqrt{1-r^{2}}-r][/tex]

with R between 0 and [tex]\sqrt{1/2}[/tex]
and ø between 0 and 2π

as it would make sense to take the volume of the sphere over the cone and subtract the volume of the area under the cone...WOW i suck :(

sorry for wasting space :) and thanks for the crucial help!

-Ben
 


Do you need help with #13 still?
 


no sir i figured it out.

Thanks!
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 24 ·
Replies
24
Views
4K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K