• Support PF! Buy your school textbooks, materials and every day products Here!

Double Integral problem! What am I suppose to do? Related to polar coordinates.

  • Thread starter DavidAp
  • Start date
  • #1
44
0
The problem and my work is shown in the image below. However, I feel like I did something horrible wrong but I'm not sure where!

24xjrl5.jpg


I'm sorry if my handwriting is illegible. If you're having difficulties please leave a comment and I will not hesitate to type it out as a response. Any guidance is greatly appreciated.

Thank you for taking the time to review my question.
 

Answers and Replies

  • #2
33
0
You've got [itex]\theta[/itex] ranging from [itex]0[/itex] to [itex]cos(\theta)[/itex], that doesn't make any sense. I'm fairly sure you should have [itex]\theta[/itex] going from [itex]0[/itex] to [itex]\frac{\pi}{4}[/itex]. Which you'll be able to see if you draw a picture of [itex]R[/itex].
 
  • #3
44
0
You've got [itex]\theta[/itex] ranging from [itex]0[/itex] to [itex]cos(\theta)[/itex], that doesn't make any sense. I'm fairly sure you should have [itex]\theta[/itex] going from [itex]0[/itex] to [itex]\frac{\pi}{4}[/itex]. Which you'll be able to see if you draw a picture of [itex]R[/itex].
I don't know how to draw 0 <= y <= x. I think that might be part of my problem...

I did something wrong though when using pi/2! The answer is 3/64 pi^2 but I keep getting 3/16 pi^2! Why is my denominator 4x less than the answer? Here's my work.

∫(1 -> 2) ∫(0 -> pi/2) θr drdθ
1/2 ∫(1 -> 2) θ(r^2)(0 -> pi/2) dθ
1/2 ∫(1 -> 2) θ(pi^2)/4 dθ
(pi^2)/8 ∫(1 -> 2) θ dθ
(pi^2)/16 (θ^2)(1 -> 2)
(pi^2)/16 (2^2 - 1^1)
(pi^2)/16 (4 - 1)
(pi^2)/16 (3)
3/16 pi^2

Any insight on what I did wrong?
 
  • #4
33
0
Ah I think I didn't edit my post in time then, the upper limit should have been [itex]\frac{\pi}{4}[/itex], sorry. Your method is fine.

And just draw the lines [itex]y=0[/itex] and [itex]y=x[/itex] and you should be able to see where [itex]y[/itex] satisfies [itex]0{\leq}y{\leq}x[/itex]. Or think about it this way, where does [itex](x,y)[/itex] satisfy [itex]0{\leq}y[/itex] and where does it satisfy [itex]y{\leq}x[/itex]. Then you just find the intersection of those two areas.
 
  • #5
44
0
Ah I think I didn't edit my post in time then, the upper limit should have been [itex]\frac{\pi}{4}[/itex], sorry. Your method is fine.

And just draw the lines [itex]y=0[/itex] and [itex]y=x[/itex] and you should be able to see where [itex]y[/itex] satisfies [itex]0{\leq}y{\leq}x[/itex]. Or think about it this way, where does [itex](x,y)[/itex] satisfy [itex]0{\leq}y[/itex] and where does it satisfy [itex]y{\leq}x[/itex]. Then you just find the intersection of those two areas.
Thank you so much, not just on this problem but on future problem to come! I never thought of visualizing it that way!
 

Related Threads on Double Integral problem! What am I suppose to do? Related to polar coordinates.

Replies
1
Views
1K
Replies
7
Views
9K
Replies
18
Views
6K
  • Last Post
Replies
1
Views
915
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
7
Views
992
  • Last Post
Replies
1
Views
877
  • Last Post
Replies
2
Views
872
Top