- #1
robertjford80
- 388
- 0
Homework Statement
Do you see how y gets converted to csc? I don't get that. I would y would be converted to sin in polar coordinates.
Converting cartesian to polar coordinates in multiple integrals
- Thread starter robertjford80
- Start date
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Homework Help Overview
The discussion revolves around converting Cartesian coordinates to polar coordinates in the context of multiple integrals. Participants are exploring the transformation of variables and the implications for integration limits.
Discussion Character
- Conceptual clarification, Assumption checking
Approaches and Questions Raised
- The original poster questions the conversion of y to csc in polar coordinates, expressing confusion about the relationship between y and sin. Other participants discuss integration limits and the geometric interpretation of the region of integration.
Discussion Status
Some participants have indicated understanding of the topic, while others are still seeking clarification on specific points. There is an ongoing exploration of the concepts involved in the conversion process.
Contextual Notes
Participants reference a specific integral and the corresponding region of integration, noting the relationship between r, y, and the angle θ. There is mention of a deleted post that may have contained additional context or clarification.
Do you see how y gets converted to csc? I don't get that. I would y would be converted to sin in polar coordinates.
- #2
Karamata
- 59
- 0
<deleted>
- #3
robertjford80
- 388
- 0
thanks, I got it.
- #4
robertjford80
- 388
- 0
I need to see what Karmata wrote again, if anyone knows I would appreciate it.
- #5
robertjford80
- 388
- 0
still needing
- #6
Karamata
- 59
- 0
Hi robertjford80!
I deleted my post because there was error in him (oh, bad English)
But, look at picture.
They said \int_0^6 \int_0^y x \mbox{d}x\mbox{d}y, that is yellow region (x from 0 (parallel y-axes) to x=y, y from 0 to 6). r is moving from r=0 to y=6, so, y= 6= r \sin \theta \Rightarrow r = \dfrac{6}{\sin \theta} = 6 \csc \theta
I deleted my post because there was error in him (oh, bad English)
But, look at picture.
They said \int_0^6 \int_0^y x \mbox{d}x\mbox{d}y, that is yellow region (x from 0 (parallel y-axes) to x=y, y from 0 to 6). r is moving from r=0 to y=6, so, y= 6= r \sin \theta \Rightarrow r = \dfrac{6}{\sin \theta} = 6 \csc \theta
Attachments
- #7
robertjford80
- 388
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ok, thanks, I got it. this so far has been one of the most difficult concepts in calculus to understand but I'm slowly getting it.
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